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2.37k
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3.5k
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PASSED | df19c70396e6d1827f31e3d6e62eff75 | train_110.jsonl | 1646408100 | You are given a tree of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. A tree is a connected undirected graph without cycles. For each $$$i=1,2, \ldots, n$$$, let $$$w_i$$$ be the weight of the $$$i$$$-th vertex. A vertex is called good if its weight is equal to the sum of the weights of all its neighbors.Initially, the weights of all nodes are unassigned. Assign positive integer weights to each vertex of the tree, such that the number of good vertices in the tree is maximized. If there are multiple ways to do it, you have to find one that minimizes the sum of weights of all vertices in the tree. | 256 megabytes | import com.sun.jmx.remote.internal.ArrayQueue;
import java.util.*;
import java.io.*;
public class Solution {
static List<Integer>[] graph;
static int[] parent;
static Node[][] dp;
static boolean[] isGood;
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(System.in);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuffer out = new StringBuffer();
int T = 1;
OUTER:
while (T-->0) {
int n = in.nextInt();
graph = new ArrayList[n];
for(int i=0; i<n; i++) {
graph[i] = new ArrayList<>();
}
for(int i=0; i<n-1; i++) {
int u = in.nextInt()-1, v = in.nextInt()-1;
graph[u].add(v);
graph[v].add(u);
}
if (n==2) {
out.append(2).append(" ").append(2).append("\n");
out.append(1).append(" ").append(1).append("\n");
break;
}
parent = new int[n];
Arrays.fill(parent, -1);
parent[0] = 0;
dfs(0);
// print(Arrays.toString(parent));
dp = new Node[n][2];
Node r = Node.max(solve(0, 0), solve(0, 1));
isGood = new boolean[n];
build(r, 0);
out.append(r.x).append(" ").append(r.y).append("\n");
for(int i=0; i<n; i++) {
if(isGood[i]) {
out.append(graph[i].size());
} else {
out.append(1);
}
out.append(" ");
}
out.append("\n");
}
System.out.print(out);
}
private static void build(Node node, int i) {
if(node.compareTo(dp[i][0])==0) {
isGood[i] = false;
for(int child: graph[i]) if(child!=parent[i]) {
build(Node.max(solve(child, 0), solve(child, 1)), child);
}
} else {
isGood[i] = true;
for(int child: graph[i]) if(child!=parent[i]) {
build(solve(child, 0), child);
}
}
}
private static void dfs(int i) {
for(int child: graph[i]) if(parent[child]==-1){
parent[child] = i;
dfs(child);
}
}
private static Node solve(int x, int y) {
if(dp[x][y]!=null) {
return dp[x][y];
}
dp[x][y] = new Node(y, y==0?1:graph[x].size());
for(int child: graph[x]) if(child!=parent[x]) {
Node max = solve(child, 0);
if(y==0) {
max = Node.max(max, solve(child, 1));
}
dp[x][y].x += max.x;
dp[x][y].y += max.y;
}
return dp[x][y];
}
private static long gcd(long a, long b) {
if (a==0)
return b;
return gcd(b%a, a);
}
private static int toInt(String s) {
return Integer.parseInt(s);
}
private static long toLong(String s) {
return Long.parseLong(s);
}
private static void print(String s) {
System.out.print(s);
}
private static void println(String s) {
System.out.println(s);
}
}
class Node implements Comparable<Node>{
public int x, y;
Node(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Node obj) {
if (x>obj.x || (x==obj.x && y<obj.y)) {
return +1;
}
if (x<obj.x || (x==obj.x && y>obj.y)) {
return -1;
}
return 0;
}
public static Node max(Node o1, Node o2) {
if(o1.compareTo(o2)>0) {
return o1;
}
return o2;
}
} | Java | ["4\n1 2\n2 3\n2 4", "3\n1 2\n1 3", "2\n1 2", "9\n3 4\n7 6\n2 1\n8 3\n5 6\n1 8\n8 6\n9 6"] | 2 seconds | ["3 4\n1 1 1 1", "2 3\n1 1 1", "2 2\n1 1", "6 11\n1 1 1 1 1 1 1 3 1"] | NoteThis is the tree for the first test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$1$$$, $$$3$$$ and $$$4$$$. It impossible to assign weights so that all vertices are good vertices. The minimum sum of weights in this case is $$$1+1+1+1=4$$$, and it is impossible to have a lower sum because the weights have to be positive integers.This is the tree for the second test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$2$$$ and $$$3$$$. It can be proven that this is an optimal assignment. | Java 8 | standard input | [
"constructive algorithms",
"dfs and similar",
"dp",
"implementation",
"trees"
] | dc3848faf577c5a49273020a14b343e1 | The first line contains one integer $$$n$$$ ($$$2\le n\le 2\cdot 10^5$$$) — the number of vertices in the tree. Then, $$$n−1$$$ lines follow. Each of them contains two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le n$$$) denoting an edge between vertices $$$u$$$ and $$$v$$$. It is guaranteed that the edges form a tree. | 2,000 | In the first line print two integers — the maximum number of good vertices and the minimum possible sum of weights for that maximum. In the second line print $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$1\le w_i\le 10^9$$$) — the corresponding weight assigned to each vertex. It can be proven that there exists an optimal solution satisfying these constraints. If there are multiple optimal solutions, you may print any. | standard output | |
PASSED | a7bbc6ef5ec1da0d904ee8c500a16630 | train_110.jsonl | 1646408100 | You are given a tree of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. A tree is a connected undirected graph without cycles. For each $$$i=1,2, \ldots, n$$$, let $$$w_i$$$ be the weight of the $$$i$$$-th vertex. A vertex is called good if its weight is equal to the sum of the weights of all its neighbors.Initially, the weights of all nodes are unassigned. Assign positive integer weights to each vertex of the tree, such that the number of good vertices in the tree is maximized. If there are multiple ways to do it, you have to find one that minimizes the sum of weights of all vertices in the tree. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
//import java.text.DecimalFormat;
import java.util.*;
public class Codeforces {
static long mod = 1000000007 ;
public static void main(String[] args) throws Exception {
PrintWriter out=new PrintWriter(System.out);
FastScanner fs=new FastScanner();
int t=1;
outer:while(t-->0) {
int n=fs.nextInt();
Node nodes[]=new Node[n];
for(int i=0;i<n;i++) nodes[i]=new Node();
for(int i=0;i<n-1;i++) {
int a=fs.nextInt()-1, b=fs.nextInt()-1;
nodes[a].adj.add(nodes[b]);
nodes[b].adj.add(nodes[a]);
}
Ans ans= nodes[0].goAllowed(null);
if(n==2) {
ans.maxNodes=2;
}
out.println(ans.maxNodes+" "+ans.minCost);
nodes[0].buildAllowed(null);
for(Node nn: nodes) {
if(nn.taken)
out.print(nn.adj.size()+" " );
else
out.print("1 ");
}
out.println();
}
out.close();
}
static class Ans{
int maxNodes, minCost;
}
static Ans better(Ans a,Ans b) {
if(a==null) return b;
if(b==null) return a;
if(a.maxNodes!=b.maxNodes) return a.maxNodes > b.maxNodes ? a : b;
return a.minCost > b.minCost ? b : a;
}
static class Node{
Ans ifAllowed, ifNotAllowed;
List<Node> adj=new ArrayList<>();
boolean taken=false;
public void buildAllowed(Node root) {
Ans ifTake=new Ans();
ifTake.maxNodes+=1;
ifTake.minCost+=adj.size();
for (Node nn:adj) if (nn!=root) {
Ans kidAns=nn.goNotAllowed(this);
ifTake.maxNodes+=kidAns.maxNodes;
ifTake.minCost+=kidAns.minCost;
}
Ans ifDont=goNotAllowed(root);
if(better(ifTake,ifDont)==ifTake) {
taken=true;
for(Node nn: adj) if(nn!=root) {
nn.buildNotAllowed(this);
}
}
else {
for(Node nn: adj) if(nn!=root) {
nn.buildAllowed(this);
}
}
}
public void buildNotAllowed(Node root) {
for(Node nn: adj) if(nn!=root) {
nn.buildAllowed(this);
}
}
public Ans goAllowed(Node root) {
if (ifAllowed!=null) return ifAllowed;
Ans ifTake=new Ans();
ifTake.maxNodes+=1;
ifTake.minCost+=adj.size();
for (Node nn:adj) if (nn!=root) {
Ans kidAns=nn.goNotAllowed(this);
ifTake.maxNodes+=kidAns.maxNodes;
ifTake.minCost+=kidAns.minCost;
}
Ans ifDont=goNotAllowed(root);
return ifAllowed=better(ifTake, ifDont);
}
public Ans goNotAllowed(Node root) {
if (ifNotAllowed!=null) return ifNotAllowed;
ifNotAllowed=new Ans();
ifNotAllowed.minCost++;
for (Node nn:adj) if (nn!=root) {
Ans kidAns=nn.goAllowed(this);
ifNotAllowed.maxNodes+=kidAns.maxNodes;
ifNotAllowed.minCost+=kidAns.minCost;
}
return ifNotAllowed;
}
}
static long pow(long a,long b) {
if(b<0) return 1;
long res=1;
while(b!=0) {
if((b&1)!=0) {
res*=a;
res%=mod;
}
a*=a;
a%=mod;
b=b>>1;
}
return res;
}
static long gcd(long a,long b) {
if(b==0) return a;
return gcd(b,a%b);
}
static long nck(int n,int k) {
if(k>n) return 0;
long res=1;
res*=fact(n);
res%=mod;
res*=modInv(fact(k));
res%=mod;
res*=modInv(fact(n-k));
res%=mod;
return res;
}
static long fact(long n) {
// return fact[(int)n];
long res=1;
for(int i=2;i<=n;i++) {
res*=i;
res%=mod;
}
return res;
}
static long modInv(long n) {
return pow(n,mod-2);
}
static void sort(int[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
int temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
static void sort(long[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
long temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
// Use this to input code since it is faster than a Scanner
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long[] readArrayL(int n) {
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=nextLong();
return a;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["4\n1 2\n2 3\n2 4", "3\n1 2\n1 3", "2\n1 2", "9\n3 4\n7 6\n2 1\n8 3\n5 6\n1 8\n8 6\n9 6"] | 2 seconds | ["3 4\n1 1 1 1", "2 3\n1 1 1", "2 2\n1 1", "6 11\n1 1 1 1 1 1 1 3 1"] | NoteThis is the tree for the first test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$1$$$, $$$3$$$ and $$$4$$$. It impossible to assign weights so that all vertices are good vertices. The minimum sum of weights in this case is $$$1+1+1+1=4$$$, and it is impossible to have a lower sum because the weights have to be positive integers.This is the tree for the second test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$2$$$ and $$$3$$$. It can be proven that this is an optimal assignment. | Java 8 | standard input | [
"constructive algorithms",
"dfs and similar",
"dp",
"implementation",
"trees"
] | dc3848faf577c5a49273020a14b343e1 | The first line contains one integer $$$n$$$ ($$$2\le n\le 2\cdot 10^5$$$) — the number of vertices in the tree. Then, $$$n−1$$$ lines follow. Each of them contains two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le n$$$) denoting an edge between vertices $$$u$$$ and $$$v$$$. It is guaranteed that the edges form a tree. | 2,000 | In the first line print two integers — the maximum number of good vertices and the minimum possible sum of weights for that maximum. In the second line print $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$1\le w_i\le 10^9$$$) — the corresponding weight assigned to each vertex. It can be proven that there exists an optimal solution satisfying these constraints. If there are multiple optimal solutions, you may print any. | standard output | |
PASSED | 27344d89944d1e75dabdf94e9fb95687 | train_110.jsonl | 1646408100 | You are given a tree of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. A tree is a connected undirected graph without cycles. For each $$$i=1,2, \ldots, n$$$, let $$$w_i$$$ be the weight of the $$$i$$$-th vertex. A vertex is called good if its weight is equal to the sum of the weights of all its neighbors.Initially, the weights of all nodes are unassigned. Assign positive integer weights to each vertex of the tree, such that the number of good vertices in the tree is maximized. If there are multiple ways to do it, you have to find one that minimizes the sum of weights of all vertices in the tree. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author real
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
DWeightTheTree solver = new DWeightTheTree();
solver.solve(1, in, out);
out.close();
}
static class DWeightTheTree {
ArrayList<Integer>[] tree;
pair[][] dp;
int[] ar;
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
tree = new ArrayList[n + 1];
for (int i = 0; i < n + 1; i++) {
tree[i] = new ArrayList<>();
}
for (int i = 0; i < n - 1; i++) {
int u = in.nextInt();
int v = in.nextInt();
tree[u].add(v);
tree[v].add(u);
}
dp = new pair[n + 1][2];
ar = new int[n + 1];
// handel n==2 separately.
if (n == 2) {
out.println(2 + " " + 2);
out.println(1 + " " + 1);
return;
}
pair makeGood = dfs(1, 0, 1);
pair notGood = dfs(1, 0, 0);
if (makeGood.cnt > notGood.cnt) {
assign(1, 0, 1);
out.println(makeGood.cnt + " " + makeGood.sum);
} else {
if (makeGood.cnt < notGood.cnt) {
assign(1, 0, 0);
out.println(notGood.cnt + " " + notGood.sum);
} else {
if (makeGood.sum > notGood.sum) {
assign(1, 0, 0);
out.println(notGood.cnt + " " + notGood.sum);
} else {
assign(1, 0, 1);
out.println(makeGood.cnt + " " + makeGood.sum);
}
}
}
for (int i = 1; i <= n; i++) {
out.print(ar[i] + " ");
}
out.println();
}
void assign(int node, int parent, int isGood) {
if (isGood == 1) {
ar[node] = tree[node].size();
for (int c : tree[node]) {
if (c != parent) {
assign(c, node, 0);
}
}
} else {
ar[node] = 1;
for (int c : tree[node]) {
if (c != parent) {
pair makeGood = dfs(c, node, 1);
pair notGood = dfs(c, node, 0);
if (makeGood.cnt > notGood.cnt) {
assign(c, node, 1);
} else {
if (makeGood.cnt < notGood.cnt) {
assign(c, node, 0);
} else {
if (makeGood.sum < notGood.sum) {
assign(c, node, 1);
} else {
assign(c, node, 0);
}
}
}
}
}
}
}
pair dfs(int node, int parent, int isGood) {
if (dp[node][isGood] != null)
return dp[node][isGood];
if (isGood == 1) {
int ct = 1;
int sum = tree[node].size();
for (int c : tree[node]) {
if (c != parent) {
pair ans = dfs(c, node, 0);
ct += ans.cnt;
sum += ans.sum;
}
}
dp[node][isGood] = new pair(ct, sum);
return dp[node][isGood];
}
int ct = 0;
int sum = 1;
for (int c : tree[node]) {
if (c != parent) {
pair makeGood = dfs(c, node, 1);
pair notGood = dfs(c, node, 0);
if (makeGood.cnt > notGood.cnt) {
ct += makeGood.cnt;
sum += makeGood.sum;
} else {
if (makeGood.cnt < notGood.cnt) {
ct += notGood.cnt;
sum += notGood.sum;
} else {
ct += makeGood.cnt;
sum += Math.min(makeGood.sum, notGood.sum);
}
}
}
}
dp[node][isGood] = new pair(ct, sum);
return dp[node][isGood];
}
class pair {
int cnt;
long sum;
pair(int c, long s) {
this.cnt = c;
this.sum = s;
}
}
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar;
private int snumChars;
public InputReader(InputStream st) {
this.stream = st;
}
public int read() {
//*-*------clare-----anjlika---
//remeber while comparing 2 non primitive data type not to use ==
//remember Arrays.sort for primitive data has worst time case complexity of 0(n^2) bcoz it uses quick sort
//again silly mistakes ,yr kb tk krta rhega ye mistakes
//try to write simple codes ,break it into simple things
//knowledge>rating
/*
public class Main
implements Runnable{
public static void main(String[] args) {
new Thread(null,new Main(),"Main",1<<26).start();
}
public void run() {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
libraries.InputReader in = new libraries.InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();//chenge the name of task
solver.solve(1, in, out);
out.close();
}
*/
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
}
| Java | ["4\n1 2\n2 3\n2 4", "3\n1 2\n1 3", "2\n1 2", "9\n3 4\n7 6\n2 1\n8 3\n5 6\n1 8\n8 6\n9 6"] | 2 seconds | ["3 4\n1 1 1 1", "2 3\n1 1 1", "2 2\n1 1", "6 11\n1 1 1 1 1 1 1 3 1"] | NoteThis is the tree for the first test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$1$$$, $$$3$$$ and $$$4$$$. It impossible to assign weights so that all vertices are good vertices. The minimum sum of weights in this case is $$$1+1+1+1=4$$$, and it is impossible to have a lower sum because the weights have to be positive integers.This is the tree for the second test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$2$$$ and $$$3$$$. It can be proven that this is an optimal assignment. | Java 8 | standard input | [
"constructive algorithms",
"dfs and similar",
"dp",
"implementation",
"trees"
] | dc3848faf577c5a49273020a14b343e1 | The first line contains one integer $$$n$$$ ($$$2\le n\le 2\cdot 10^5$$$) — the number of vertices in the tree. Then, $$$n−1$$$ lines follow. Each of them contains two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le n$$$) denoting an edge between vertices $$$u$$$ and $$$v$$$. It is guaranteed that the edges form a tree. | 2,000 | In the first line print two integers — the maximum number of good vertices and the minimum possible sum of weights for that maximum. In the second line print $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$1\le w_i\le 10^9$$$) — the corresponding weight assigned to each vertex. It can be proven that there exists an optimal solution satisfying these constraints. If there are multiple optimal solutions, you may print any. | standard output | |
PASSED | 2b9a4ea8c659efce8893f912226a47b9 | train_110.jsonl | 1646408100 | You are given a tree of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. A tree is a connected undirected graph without cycles. For each $$$i=1,2, \ldots, n$$$, let $$$w_i$$$ be the weight of the $$$i$$$-th vertex. A vertex is called good if its weight is equal to the sum of the weights of all its neighbors.Initially, the weights of all nodes are unassigned. Assign positive integer weights to each vertex of the tree, such that the number of good vertices in the tree is maximized. If there are multiple ways to do it, you have to find one that minimizes the sum of weights of all vertices in the tree. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Random;
import java.util.StringTokenizer;
/*
*/
public class D {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
//TODO: hardcode n==2 case
PrintWriter out=new PrintWriter(System.out);
int T=1;
for (int tt=0; tt<T; tt++) {
int n=fs.nextInt();
Node[] nodes=new Node[n];
for (int i=0; i<n; i++) nodes[i]=new Node(i);
for (int i=1; i<n; i++) {
int a=fs.nextInt()-1, b=fs.nextInt()-1;
nodes[a].adj.add(nodes[b]);
nodes[b].adj.add(nodes[a]);
}
Ans ans=nodes[0].goAllowed(null);
if (n==2) ans.maxNodes=2;
out.println(ans.maxNodes+" "+ans.minCost);
//TODO: rebuild
nodes[0].buildbackAllowed(null);
for (Node nn:nodes) {
if (nn.taken)
out.print(nn.adj.size()+" ");
else
out.print(1+" ");
}
out.println();
}
out.close();
}
static class Ans {
int maxNodes;
int minCost;
public String toString() {
return maxNodes+" $"+minCost;
}
}
static Ans better(Ans a, Ans b) {
if (a==null) return b;
if (b==null) return a;
if (a.maxNodes!=b.maxNodes) return a.maxNodes<b.maxNodes?b:a;
return b.minCost<=a.minCost?b:a;
}
static class Node {
Ans ifAllowed, ifNotAllowed;
ArrayList<Node> adj=new ArrayList<>();
int id;
boolean taken=false;
public Node(int id) {
this.id=id;
}
public void buildbackAllowed(Node root) {
Ans ifTake=new Ans();
ifTake.maxNodes+=1;
ifTake.minCost+=adj.size();
for (Node nn:adj) if (nn!=root) {
Ans kidAns=nn.goNotAllowed(this);
ifTake.maxNodes+=kidAns.maxNodes;
ifTake.minCost+=kidAns.minCost;
}
Ans ifDont=goNotAllowed(root);
if (better(ifTake, ifDont)==ifTake) {
taken=true;
for (Node nn:adj) if (nn!=root) {
nn.buildbackNotAllowed(this);
}
}
else {
for (Node nn:adj) if (nn!=root) {
nn.buildbackAllowed(this);
}
}
}
public void buildbackNotAllowed(Node root) {
for (Node nn:adj) if (nn!=root) {
nn.buildbackAllowed(this);
}
}
public Ans goAllowed(Node root) {
if (ifAllowed!=null) return ifAllowed;
Ans ifTake=new Ans();
ifTake.maxNodes+=1;
ifTake.minCost+=adj.size();
for (Node nn:adj) if (nn!=root) {
Ans kidAns=nn.goNotAllowed(this);
ifTake.maxNodes+=kidAns.maxNodes;
ifTake.minCost+=kidAns.minCost;
}
Ans ifDont=goNotAllowed(root);
return ifAllowed=better(ifTake, ifDont);
}
public Ans goNotAllowed(Node root) {
if (ifNotAllowed!=null) return ifNotAllowed;
ifNotAllowed=new Ans();
ifNotAllowed.minCost++;
for (Node nn:adj) if (nn!=root) {
Ans kidAns=nn.goAllowed(this);
ifNotAllowed.maxNodes+=kidAns.maxNodes;
ifNotAllowed.minCost+=kidAns.minCost;
}
return ifNotAllowed;
}
}
static final Random random=new Random();
static final int mod=1_000_000_007;
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static long add(long a, long b) {
return (a+b)%mod;
}
static long sub(long a, long b) {
return ((a-b)%mod+mod)%mod;
}
static long mul(long a, long b) {
return (a*b)%mod;
}
static long exp(long base, long exp) {
if (exp==0) return 1;
long half=exp(base, exp/2);
if (exp%2==0) return mul(half, half);
return mul(half, mul(half, base));
}
static long[] factorials=new long[2_000_001];
static long[] invFactorials=new long[2_000_001];
static void precompFacts() {
factorials[0]=invFactorials[0]=1;
for (int i=1; i<factorials.length; i++) factorials[i]=mul(factorials[i-1], i);
invFactorials[factorials.length-1]=exp(factorials[factorials.length-1], mod-2);
for (int i=invFactorials.length-2; i>=0; i--)
invFactorials[i]=mul(invFactorials[i+1], i+1);
}
static long nCk(int n, int k) {
return mul(factorials[n], mul(invFactorials[k], invFactorials[n-k]));
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["4\n1 2\n2 3\n2 4", "3\n1 2\n1 3", "2\n1 2", "9\n3 4\n7 6\n2 1\n8 3\n5 6\n1 8\n8 6\n9 6"] | 2 seconds | ["3 4\n1 1 1 1", "2 3\n1 1 1", "2 2\n1 1", "6 11\n1 1 1 1 1 1 1 3 1"] | NoteThis is the tree for the first test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$1$$$, $$$3$$$ and $$$4$$$. It impossible to assign weights so that all vertices are good vertices. The minimum sum of weights in this case is $$$1+1+1+1=4$$$, and it is impossible to have a lower sum because the weights have to be positive integers.This is the tree for the second test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$2$$$ and $$$3$$$. It can be proven that this is an optimal assignment. | Java 8 | standard input | [
"constructive algorithms",
"dfs and similar",
"dp",
"implementation",
"trees"
] | dc3848faf577c5a49273020a14b343e1 | The first line contains one integer $$$n$$$ ($$$2\le n\le 2\cdot 10^5$$$) — the number of vertices in the tree. Then, $$$n−1$$$ lines follow. Each of them contains two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le n$$$) denoting an edge between vertices $$$u$$$ and $$$v$$$. It is guaranteed that the edges form a tree. | 2,000 | In the first line print two integers — the maximum number of good vertices and the minimum possible sum of weights for that maximum. In the second line print $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$1\le w_i\le 10^9$$$) — the corresponding weight assigned to each vertex. It can be proven that there exists an optimal solution satisfying these constraints. If there are multiple optimal solutions, you may print any. | standard output | |
PASSED | 35a25a93d6cba0e72ae611cf53c5ff31 | train_110.jsonl | 1646408100 | You are given a tree of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. A tree is a connected undirected graph without cycles. For each $$$i=1,2, \ldots, n$$$, let $$$w_i$$$ be the weight of the $$$i$$$-th vertex. A vertex is called good if its weight is equal to the sum of the weights of all its neighbors.Initially, the weights of all nodes are unassigned. Assign positive integer weights to each vertex of the tree, such that the number of good vertices in the tree is maximized. If there are multiple ways to do it, you have to find one that minimizes the sum of weights of all vertices in the tree. | 256 megabytes | import java.util.*;
import java.util.function.*;
import java.io.*;
// you can compare with output.txt and expected out
public class Round774D {
MyPrintWriter out;
MyScanner in;
// final static long FIXED_RANDOM;
// static {
// FIXED_RANDOM = System.currentTimeMillis();
// }
final static String IMPOSSIBLE = "IMPOSSIBLE";
final static String POSSIBLE = "POSSIBLE";
final static String YES = "YES";
final static String NO = "NO";
private void initIO(boolean isFileIO) {
if (System.getProperty("ONLINE_JUDGE") == null && isFileIO) {
try{
in = new MyScanner(new FileInputStream("input.txt"));
out = new MyPrintWriter(new FileOutputStream("output.txt"));
}
catch(FileNotFoundException e){
e.printStackTrace();
}
}
else{
in = new MyScanner(System.in);
out = new MyPrintWriter(new BufferedOutputStream(System.out));
}
}
public static void main(String[] args){
// Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
Round774D sol = new Round774D();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = true;
boolean hasMultipleTests = false;
initIO(isFileIO);
int t = hasMultipleTests? in.nextInt() : 1;
for (int i = 1; i <= t; ++i) {
int n = in.nextInt();
int[][] e = in.nextTreeEdges(n, -1);
if(isDebug){
out.printf("Test %d\n", i);
}
Answer ans = solve(n, e);
out.println(ans.numGoodVertices + " " + ans.sumWeights);
out.printlnAns(ans.w);
if(isDebug)
out.flush();
}
in.close();
out.close();
}
private Answer solve(int n, int[][] e) {
if(n == 2) {
return new Answer(2, 2, new int[] {1,1});
}
neighbors = constructNeighborhood(n, e);
// why wasn't choosing leaves of minimum weight and delete its neighbors working well?
nodes = new Node[n][2];
Node node = dfs(0, -1, 0);
w = new int[n];
fillWeight(node);
return new Answer(node.num, node.sum, w);
}
private void fillWeight(Round774D.Node node) {
w[node.index] = node.chosen? neighbors[node.index].length: 1;
for(Node next: node.children) {
fillWeight(next);
}
}
Node[][] nodes;
int[][] neighbors;
int[] w;
private Node dfs(int curr, int parent, int mustNotChosen) {
if(nodes[curr][mustNotChosen] != null)
return nodes[curr][mustNotChosen];
Node node = new Node(curr, 0, 1, false);
for(int next: neighbors[curr]) {
if(next == parent)
continue;
Node ret = dfs(next, curr, 0);
node.num += ret.num;
node.sum += ret.sum;
node.children.add(ret);
}
if(mustNotChosen == 0) {
Node node2 = new Node(curr, 1, neighbors[curr].length, true);
for(int next: neighbors[curr]) {
if(next == parent)
continue;
Node ret = dfs(next, curr, 1);
node2.num += ret.num;
node2.sum += ret.sum;
node2.children.add(ret);
}
if(node2.num > node.num) {
node = node2;
}
else if(node2.num == node.num && node2.sum < node.sum) {
node = node2;
}
}
nodes[curr][mustNotChosen] = node;
return node;
}
class Node{
int num;
long sum;
boolean chosen;
int index;
ArrayList<Node> children;
public Node(int index, int num, long sum, boolean chosen) {
this.index = index;
this.num = num;
this.sum = sum;
this.chosen = chosen;
this.children = new ArrayList<Node>();
}
}
private Answer solve2(int n, int[][] e) {
// assign 1 to leaf
// assign 1 to vertices adjacent to leaves
// assign sum X to vertices
// x = y + (sumX-y) > y
// y = x + (sumY-x) > x
// can't satisfy both
// bi-color the graph
// choose the bigger one
// color all leaves by red
// adjacent to red -> must be black
// adjacent to black -> can be black or red
// black's cost is 1
// color v to red
// -> forces its neighbor to be black
// -> try red or black for each subtree
// this will give you something like 2^(n/x)
// no,
// for each vertex, you will try only 2 choices like dp
// so O(n)
// priority queue
// always choose a min deg vertex to color red
if(n == 2) {
return new Answer(2, 2, new int[] {1,1});
}
boolean[] visited = new boolean[n];
HashSet<Integer>[] neighbors = new HashSet[n];
for(int i=0; i<n; i++)
neighbors[i] = new HashSet<Integer>();
for(int i=0; i<n-1; i++) {
neighbors[e[i][0]].add(e[i][1]);
neighbors[e[i][1]].add(e[i][0]);
}
int[] deg = new int[n];
for(int i=0; i<n; i++)
deg[i] = neighbors[i].size();
int[] w = Arrays.copyOf(deg, n);
// w[i] = deg[i]
// red vertices can't be adjacent
// maximize # red vertices
// then minimize sum w[i]
int num = 0;
long sum = 0;
PriorityQueue<Vertex> pq = new PriorityQueue<>(new Comparator<Vertex>() {
@Override
public int compare(Vertex v, Vertex u) {
int cmp = Integer.compare(v.degree, u.degree);
return cmp!=0? cmp: Integer.compare(w[v.index], w[u.index]);
}
});
for(int i=0; i<n; i++) {
if(deg[i] <= 1)
pq.add(new Vertex(i, deg[i]));
}
// we remove leaves
// and black vertices
// then we again have leaves
//
while(!pq.isEmpty()) {
Vertex vtx = pq.poll();
int v = vtx.index;
if(visited[v])
continue;
visited[v] = true;
ArrayList<Integer> nb = new ArrayList<Integer>(neighbors[v]);
for(int u: nb) {
visited[u] = true;
w[u] = 1;
for(int x: neighbors[u]) {
neighbors[x].remove(u);
//w[x]++;
deg[x]--;
if(deg[x] <= 1)
pq.add(new Vertex(x, deg[x]));
}
neighbors[u] = null;
sum += w[u];
}
num++;
sum += w[v];
}
return new Answer(num, sum, w);
}
class Vertex{
int index, degree;
public Vertex(int index, int degree) {
this.index = index;
this.degree = degree;
}
}
class Answer{
int numGoodVertices;
long sumWeights;
int[] w;
public Answer(int numGoodVertices, long sumWeights, int[] w) {
this.numGoodVertices = numGoodVertices;
this.sumWeights = sumWeights;
this.w = w;
}
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
// 32768?
public MyScanner(InputStream is, int bufferSize) {
br = new BufferedReader(new InputStreamReader(is), bufferSize);
}
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
// br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
}
public void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[][] nextTreeEdges(int n, int offset){
int[][] e = new int[n-1][2];
for(int i=0; i<n-1; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[][] nextMatrix(int n, int m) {
return nextMatrix(n, m, 0);
}
int[][] nextMatrix(int n, int m, int offset) {
int[][] mat = new int[n][m];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[i][j] = nextInt()+offset;
}
}
return mat;
}
int[][] nextPairs(int n){
return nextPairs(n, 0);
}
int[][] nextPairs(int n, int offset) {
int[][] xy = new int[2][n];
for(int i=0; i<n; i++) {
xy[0][i] = nextInt() + offset;
xy[1][i] = nextInt() + offset;
}
return xy;
}
int[][] nextGraphEdges(){
return nextGraphEdges(0);
}
int[][] nextGraphEdges(int offset) {
int m = nextInt();
int[][] e = new int[m][2];
for(int i=0; i<m; i++){
e[i][0] = nextInt()+offset;
e[i][1] = nextInt()+offset;
}
return e;
}
int[] nextIntArray(int len) {
return nextIntArray(len, 0);
}
int[] nextIntArray(int len, int offset){
int[] a = new int[len];
for(int j=0; j<len; j++)
a[j] = nextInt()+offset;
return a;
}
long[] nextLongArray(int len) {
return nextLongArray(len, 0);
}
long[] nextLongArray(int len, int offset){
long[] a = new long[len];
for(int j=0; j<len; j++)
a[j] = nextLong()+offset;
return a;
}
}
public static class MyPrintWriter extends PrintWriter{
public MyPrintWriter(OutputStream os) {
super(os);
}
public void printlnAns(long ans) {
println(ans);
}
public void printlnAns(int ans) {
println(ans);
}
public void printlnAns(boolean ans) {
if(ans)
println(YES);
else
println(NO);
}
public void printAns(long[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(long[] arr){
printAns(arr);
println();
}
public void printAns(int[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(int[] arr){
printAns(arr);
println();
}
public <T> void printAns(ArrayList<T> arr){
if(arr != null && arr.size() > 0){
print(arr.get(0));
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i));
}
}
}
public <T> void printlnAns(ArrayList<T> arr){
printAns(arr);
println();
}
public void printAns(int[] arr, int add){
if(arr != null && arr.length > 0){
print(arr[0]+add);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]+add);
}
}
}
public void printlnAns(int[] arr, int add){
printAns(arr, add);
println();
}
public void printAns(ArrayList<Integer> arr, int add) {
if(arr != null && arr.size() > 0){
print(arr.get(0)+add);
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i)+add);
}
}
}
public void printlnAns(ArrayList<Integer> arr, int add){
printAns(arr, add);
println();
}
public void printlnAnsSplit(long[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public void printlnAnsSplit(int[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public <T> void printlnAnsSplit(ArrayList<T> arr, int split){
if(arr != null && !arr.isEmpty()){
for(int i=0; i<arr.size(); i+=split){
print(arr.get(i));
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr.get(j));
}
println();
}
}
}
}
static private void permutateAndSort(long[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
long temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private void permutateAndSort(int[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
int temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private int[][] constructChildren(int n, int[] parent, int parentRoot){
int[][] childrens = new int[n][];
int[] numChildren = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
numChildren[parent[i]]++;
}
for(int i=0; i<n; i++) {
childrens[i] = new int[numChildren[i]];
}
int[] idx = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
childrens[parent[i]][idx[parent[i]]++] = i;
}
return childrens;
}
static private int[][][] constructDirectedNeighborhood(int n, int[][] e){
int[] inDegree = new int[n];
int[] outDegree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outDegree[u]++;
inDegree[v]++;
}
int[][] inNeighbors = new int[n][];
int[][] outNeighbors = new int[n][];
for(int i=0; i<n; i++) {
inNeighbors[i] = new int[inDegree[i]];
outNeighbors[i] = new int[outDegree[i]];
}
int[] inIdx = new int[n];
int[] outIdx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outNeighbors[u][outIdx[u]++] = v;
inNeighbors[v][inIdx[v]++] = u;
}
return new int[][][] {inNeighbors, outNeighbors};
}
static private int[][] constructNeighborhood(int n, int[][] e) {
int[] degree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
degree[u]++;
degree[v]++;
}
int[][] neighbors = new int[n][];
for(int i=0; i<n; i++)
neighbors[i] = new int[degree[i]];
int[] idx = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
neighbors[u][idx[u]++] = v;
neighbors[v][idx[v]++] = u;
}
return neighbors;
}
static private void drawGraph(int[][] e) {
makeDotUndirected(e);
try {
final Process process = new ProcessBuilder("dot", "-Tpng", "graph.dot")
.redirectOutput(new File("graph.png"))
.start();
} catch (IOException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
}
static private void makeDotUndirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict graph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "--" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
static private void makeDotDirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict digraph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "->" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
}
| Java | ["4\n1 2\n2 3\n2 4", "3\n1 2\n1 3", "2\n1 2", "9\n3 4\n7 6\n2 1\n8 3\n5 6\n1 8\n8 6\n9 6"] | 2 seconds | ["3 4\n1 1 1 1", "2 3\n1 1 1", "2 2\n1 1", "6 11\n1 1 1 1 1 1 1 3 1"] | NoteThis is the tree for the first test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$1$$$, $$$3$$$ and $$$4$$$. It impossible to assign weights so that all vertices are good vertices. The minimum sum of weights in this case is $$$1+1+1+1=4$$$, and it is impossible to have a lower sum because the weights have to be positive integers.This is the tree for the second test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$2$$$ and $$$3$$$. It can be proven that this is an optimal assignment. | Java 17 | standard input | [
"constructive algorithms",
"dfs and similar",
"dp",
"implementation",
"trees"
] | dc3848faf577c5a49273020a14b343e1 | The first line contains one integer $$$n$$$ ($$$2\le n\le 2\cdot 10^5$$$) — the number of vertices in the tree. Then, $$$n−1$$$ lines follow. Each of them contains two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le n$$$) denoting an edge between vertices $$$u$$$ and $$$v$$$. It is guaranteed that the edges form a tree. | 2,000 | In the first line print two integers — the maximum number of good vertices and the minimum possible sum of weights for that maximum. In the second line print $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$1\le w_i\le 10^9$$$) — the corresponding weight assigned to each vertex. It can be proven that there exists an optimal solution satisfying these constraints. If there are multiple optimal solutions, you may print any. | standard output | |
PASSED | 46fc014566e8fdd3971b4cab0391f811 | train_110.jsonl | 1646408100 | You are given a tree of $$$n$$$ vertices numbered from $$$1$$$ to $$$n$$$. A tree is a connected undirected graph without cycles. For each $$$i=1,2, \ldots, n$$$, let $$$w_i$$$ be the weight of the $$$i$$$-th vertex. A vertex is called good if its weight is equal to the sum of the weights of all its neighbors.Initially, the weights of all nodes are unassigned. Assign positive integer weights to each vertex of the tree, such that the number of good vertices in the tree is maximized. If there are multiple ways to do it, you have to find one that minimizes the sum of weights of all vertices in the tree. | 256 megabytes | import java.io.*;
import java.util.*;
public class WeightTheTree {
public static void solve(FastIO io) {
final int N = io.nextInt();
Node[] nodes = new Node[N + 1];
for (int i = 1; i <= N; ++i) {
nodes[i] = new Node(i);
}
for (int i = 1; i < N; ++i) {
final int U = io.nextInt();
final int V = io.nextInt();
nodes[U].next.add(nodes[V]);
nodes[V].next.add(nodes[U]);
}
if (N == 2) {
io.println(2, 2);
io.println(1, 1);
return;
}
nodes[1].root(null);
MemoData[][] memo = new MemoData[2][N + 1];
MemoData happyResult = get(HAPPY, nodes[1], memo);
MemoData unhappyResult = get(UNHAPPY, nodes[1], memo);
MemoData result = betterResult(happyResult, unhappyResult);
boolean[] markHappy = new boolean[N + 1];
if (result == happyResult) {
mark(HAPPY, nodes[1], memo, markHappy);
} else {
mark(UNHAPPY, nodes[1], memo, markHappy);
}
int goodCount = 0;
int weightSum = 0;
int[] nodeValues = new int[N + 1];
for (int i = 1; i <= N; ++i) {
if (markHappy[i]) {
++goodCount;
nodeValues[i] = nodes[i].next.size();
if (nodes[i].parent != null) {
++nodeValues[i];
}
} else {
nodeValues[i] = 1;
}
weightSum += nodeValues[i];
}
io.println(goodCount, weightSum);
io.printlnArray(Arrays.copyOfRange(nodeValues, 1, N + 1));
}
private static final int UNHAPPY = 0;
private static final int HAPPY = 1;
private static MemoData get(int isHappy, Node u, MemoData[][] memo) {
if (memo[isHappy][u.id] == null) {
if (isHappy == HAPPY) {
int maxHappiness = 1;
int minWeight = u.next.size();
if (u.parent != null) {
++minWeight;
}
for (Node v : u.next) {
MemoData result = get(UNHAPPY, v, memo);
maxHappiness += result.maxHappiness;
minWeight += result.minWeight;
}
memo[isHappy][u.id] = new MemoData(maxHappiness, minWeight);
} else {
int maxHappiness = 0;
int minWeight = 1;
for (Node v : u.next) {
MemoData happyResult = get(HAPPY, v, memo);
MemoData unhappyResult = get(UNHAPPY, v, memo);
MemoData result = betterResult(happyResult, unhappyResult);
maxHappiness += result.maxHappiness;
minWeight += result.minWeight;
}
memo[isHappy][u.id] = new MemoData(maxHappiness, minWeight);
}
}
return memo[isHappy][u.id];
}
private static void mark(int isHappy, Node u, MemoData[][] memo, boolean[] outHappy) {
if (isHappy == HAPPY) {
outHappy[u.id] = true;
for (Node v : u.next) {
mark(UNHAPPY, v, memo, outHappy);
}
} else {
for (Node v : u.next) {
MemoData happyResult = get(HAPPY, v, memo);
MemoData unhappyResult = get(UNHAPPY, v, memo);
MemoData result = betterResult(happyResult, unhappyResult);
if (result == happyResult) {
mark(HAPPY, v, memo, outHappy);
} else {
mark(UNHAPPY, v, memo, outHappy);
}
}
}
}
private static MemoData betterResult(MemoData a, MemoData b) {
if (a.maxHappiness > b.maxHappiness) {
return a;
} else if (a.maxHappiness < b.maxHappiness) {
return b;
} else if (a.minWeight < b.minWeight) {
return a;
} else {
return b;
}
}
private static class MemoData {
public int maxHappiness;
public int minWeight;
public MemoData(int maxHappiness, int minWeight) {
this.maxHappiness = maxHappiness;
this.minWeight = minWeight;
}
}
private static class Node {
public int id;
public ArrayList<Node> next = new ArrayList<>();
public Node parent;
public Node(int id) {
this.id = id;
}
public void root(Node p) {
next.remove(p);
parent = p;
for (Node v : next) {
v.root(this);
}
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
// TODO: read this byte-by-byte like the other read functions.
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
public static void main(String[] args) {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
} | Java | ["4\n1 2\n2 3\n2 4", "3\n1 2\n1 3", "2\n1 2", "9\n3 4\n7 6\n2 1\n8 3\n5 6\n1 8\n8 6\n9 6"] | 2 seconds | ["3 4\n1 1 1 1", "2 3\n1 1 1", "2 2\n1 1", "6 11\n1 1 1 1 1 1 1 3 1"] | NoteThis is the tree for the first test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$1$$$, $$$3$$$ and $$$4$$$. It impossible to assign weights so that all vertices are good vertices. The minimum sum of weights in this case is $$$1+1+1+1=4$$$, and it is impossible to have a lower sum because the weights have to be positive integers.This is the tree for the second test case: In this case, if you assign a weight of $$$1$$$ to each vertex, then the good vertices (which are painted black) are $$$2$$$ and $$$3$$$. It can be proven that this is an optimal assignment. | Java 17 | standard input | [
"constructive algorithms",
"dfs and similar",
"dp",
"implementation",
"trees"
] | dc3848faf577c5a49273020a14b343e1 | The first line contains one integer $$$n$$$ ($$$2\le n\le 2\cdot 10^5$$$) — the number of vertices in the tree. Then, $$$n−1$$$ lines follow. Each of them contains two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le n$$$) denoting an edge between vertices $$$u$$$ and $$$v$$$. It is guaranteed that the edges form a tree. | 2,000 | In the first line print two integers — the maximum number of good vertices and the minimum possible sum of weights for that maximum. In the second line print $$$n$$$ integers $$$w_1, w_2, \ldots, w_n$$$ ($$$1\le w_i\le 10^9$$$) — the corresponding weight assigned to each vertex. It can be proven that there exists an optimal solution satisfying these constraints. If there are multiple optimal solutions, you may print any. | standard output | |
PASSED | 85a806f2989dafdb30758dcc3805b40e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.util.function.*;
import java.io.*;
// you can compare with output.txt and expected out
public class RoundDeltixAutumn2021C {
MyPrintWriter out;
MyScanner in;
// final static long FIXED_RANDOM;
// static {
// FIXED_RANDOM = System.currentTimeMillis();
// }
final static String IMPOSSIBLE = "IMPOSSIBLE";
final static String POSSIBLE = "POSSIBLE";
final static String YES = "YES";
final static String NO = "NO";
private void initIO(boolean isFileIO) {
if (System.getProperty("ONLINE_JUDGE") == null && isFileIO) {
try{
in = new MyScanner(new FileInputStream("input.txt"));
out = new MyPrintWriter(new FileOutputStream("output.txt"));
}
catch(FileNotFoundException e){
e.printStackTrace();
}
}
else{
in = new MyScanner(System.in);
out = new MyPrintWriter(new BufferedOutputStream(System.out));
}
}
public static void main(String[] args){
// Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
// the code for the deep recursion (more than 100k or 3~400k or so)
// Thread t = new Thread(null, new RoundEdu130F(), "threadName", 1<<28);
// t.start();
// t.join();
RoundDeltixAutumn2021C sol = new RoundDeltixAutumn2021C();
sol.run();
}
private void run() {
boolean isDebug = false;
boolean isFileIO = true;
boolean hasMultipleTests = true;
initIO(isFileIO);
final int MAX = 1_000_000;
precompute(MAX);
int t = hasMultipleTests? in.nextInt() : 1;
for (int i = 1; i <= t; ++i) {
if(isDebug){
out.printf("Test %d\n", i);
}
getInput();
solve();
printOutput();
}
in.close();
out.close();
}
boolean[] isPrime;
private void precompute(final int MAX) {
isPrime = new boolean[MAX+1];
Arrays.fill(isPrime, true);
isPrime[1] = false;
isPrime[0] = false;
for(int i=2; i<=MAX; i++) {
if(isPrime[i]) {
if(MAX/i >= i) {
for(int j=i*i; j<=MAX; j+=i)
isPrime[j] = false;
}
}
}
}
// use suitable one between matrix(2, n), matrix(n, 2), transposedMatrix(2, n) for graph edges, pairs, ...
int n, e;
int[] a;
void getInput() {
n = in.nextInt();
e = in.nextInt();
a = in.nextIntArray(n);
}
void printOutput() {
out.printlnAns(ans);
}
long ans;
void solve(){
// had understood the problem wrong
// # of pairs (i, k) s.t.
// k >= 1
// a[i]a[i+e]...a[i+ke] is prime
n += e;
a = Arrays.copyOf(a, n);
ans = 0;
for(int r=0; r<e; r++) {
int start = r;
int primePos = -1;
for(int i=r; i<n; i+=e) {
if(isPrime[a[i]]) {
if(primePos == -1)
primePos = i;
else {
ans += (long)(primePos-start+e)/e * (i-primePos)/e - 1;
start = primePos+1;
primePos = i;
}
}
else if(a[i] != 1) {
if(primePos != -1)
ans += (long)(primePos-start+e)/e * (i-primePos)/e - 1;
start = i+e;
primePos = -1;
}
}
}
}
// Optional<T> solve()
// return Optional.empty();
static class Pair implements Comparable<Pair>{
final static long FIXED_RANDOM = System.currentTimeMillis();
int first, second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
@Override
public int hashCode() {
// http://xorshift.di.unimi.it/splitmix64.c
long x = first;
x <<= 32;
x += second;
x += FIXED_RANDOM;
x += 0x9e3779b97f4a7c15l;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9l;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebl;
return (int)(x ^ (x >> 31));
}
@Override
public boolean equals(Object obj) {
// if (this == obj)
// return true;
// if (obj == null)
// return false;
// if (getClass() != obj.getClass())
// return false;
Pair other = (Pair) obj;
return first == other.first && second == other.second;
}
@Override
public String toString() {
return "[" + first + "," + second + "]";
}
@Override
public int compareTo(Pair o) {
int cmp = Integer.compare(first, o.first);
return cmp != 0? cmp: Integer.compare(second, o.second);
}
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
// 32768?
public MyScanner(InputStream is, int bufferSize) {
br = new BufferedReader(new InputStreamReader(is), bufferSize);
}
public MyScanner(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
// br = new BufferedReader(new InputStreamReader(System.in));
// br = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
}
public void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[][] nextMatrix(int n, int m) {
return nextMatrix(n, m, 0);
}
int[][] nextMatrix(int n, int m, int offset) {
int[][] mat = new int[n][m];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[i][j] = nextInt()+offset;
}
}
return mat;
}
int[][] nextTransposedMatrix(int n, int m){
return nextTransposedMatrix(n, m, 0);
}
int[][] nextTransposedMatrix(int n, int m, int offset){
int[][] mat = new int[m][n];
for(int i=0; i<n; i++) {
for(int j=0; j<m; j++) {
mat[j][i] = nextInt()+offset;
}
}
return mat;
}
int[] nextIntArray(int len) {
return nextIntArray(len, 0);
}
int[] nextIntArray(int len, int offset){
int[] a = new int[len];
for(int j=0; j<len; j++)
a[j] = nextInt()+offset;
return a;
}
long[] nextLongArray(int len) {
return nextLongArray(len, 0);
}
long[] nextLongArray(int len, int offset){
long[] a = new long[len];
for(int j=0; j<len; j++)
a[j] = nextLong()+offset;
return a;
}
String[] nextStringArray(int len) {
String[] s = new String[len];
for(int i=0; i<len; i++)
s[i] = next();
return s;
}
}
public static class MyPrintWriter extends PrintWriter{
public MyPrintWriter(OutputStream os) {
super(os);
}
// public <T> void printlnAns(Optional<T> ans) {
// if(ans.isEmpty())
// println(-1);
// else
// printlnAns(ans.get());
// }
public void printlnAns(OptionalInt ans) {
println(ans.orElse(-1));
}
public void printlnAns(long ans) {
println(ans);
}
public void printlnAns(int ans) {
println(ans);
}
public void printlnAns(boolean ans) {
if(ans)
println(YES);
else
println(NO);
}
public void printAns(long[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(long[] arr){
printAns(arr);
println();
}
public void printAns(int[] arr){
if(arr != null && arr.length > 0){
print(arr[0]);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]);
}
}
}
public void printlnAns(int[] arr){
printAns(arr);
println();
}
public <T> void printAns(ArrayList<T> arr){
if(arr != null && arr.size() > 0){
print(arr.get(0));
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i));
}
}
}
public <T> void printlnAns(ArrayList<T> arr){
printAns(arr);
println();
}
public void printAns(int[] arr, int add){
if(arr != null && arr.length > 0){
print(arr[0]+add);
for(int i=1; i<arr.length; i++){
print(" ");
print(arr[i]+add);
}
}
}
public void printlnAns(int[] arr, int add){
printAns(arr, add);
println();
}
public void printAns(ArrayList<Integer> arr, int add) {
if(arr != null && arr.size() > 0){
print(arr.get(0)+add);
for(int i=1; i<arr.size(); i++){
print(" ");
print(arr.get(i)+add);
}
}
}
public void printlnAns(ArrayList<Integer> arr, int add){
printAns(arr, add);
println();
}
public void printlnAnsSplit(long[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public void printlnAnsSplit(int[] arr, int split){
if(arr != null){
for(int i=0; i<arr.length; i+=split){
print(arr[i]);
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr[j]);
}
println();
}
}
}
public <T> void printlnAnsSplit(ArrayList<T> arr, int split){
if(arr != null && !arr.isEmpty()){
for(int i=0; i<arr.size(); i+=split){
print(arr.get(i));
for(int j=i+1; j<i+split; j++){
print(" ");
print(arr.get(j));
}
println();
}
}
}
}
static private void permutateAndSort(long[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
long temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private void permutateAndSort(int[] a) {
int n = a.length;
Random R = new Random(System.currentTimeMillis());
for(int i=0; i<n; i++) {
int t = R.nextInt(n-i);
int temp = a[n-1-i];
a[n-1-i] = a[t];
a[t] = temp;
}
Arrays.sort(a);
}
static private int[][] constructChildren(int n, int[] parent, int parentRoot){
int[][] childrens = new int[n][];
int[] numChildren = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
numChildren[parent[i]]++;
}
for(int i=0; i<n; i++) {
childrens[i] = new int[numChildren[i]];
}
int[] idx = new int[n];
for(int i=0; i<parent.length; i++) {
if(parent[i] != parentRoot)
childrens[parent[i]][idx[parent[i]]++] = i;
}
return childrens;
}
static private int[][][] constructDirectedNeighborhood(int n, int[][] e){
int[] inDegree = new int[n];
int[] outDegree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outDegree[u]++;
inDegree[v]++;
}
int[][] inNeighbors = new int[n][];
int[][] outNeighbors = new int[n][];
for(int i=0; i<n; i++) {
inNeighbors[i] = new int[inDegree[i]];
outNeighbors[i] = new int[outDegree[i]];
}
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
outNeighbors[u][--outDegree[u]] = v;
inNeighbors[v][--inDegree[v]] = u;
}
return new int[][][] {inNeighbors, outNeighbors};
}
static private int[][] constructNeighborhood(int n, int[][] e) {
int[] degree = new int[n];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
degree[u]++;
degree[v]++;
}
int[][] neighbors = new int[n][];
for(int i=0; i<n; i++)
neighbors[i] = new int[degree[i]];
for(int i=0; i<e.length; i++) {
int u = e[i][0];
int v = e[i][1];
neighbors[u][--degree[u]] = v;
neighbors[v][--degree[v]] = u;
}
return neighbors;
}
static private void drawGraph(int[][] e) {
makeDotUndirected(e);
try {
final Process process = new ProcessBuilder("dot", "-Tpng", "graph.dot")
.redirectOutput(new File("graph.png"))
.start();
} catch (IOException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
}
static private void makeDotUndirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict graph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "--" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
static private void makeDotDirected(int[][] e) {
MyPrintWriter out2 = null;
try {
out2 = new MyPrintWriter(new FileOutputStream("graph.dot"));
} catch (FileNotFoundException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
out2.println("strict digraph {");
for(int i=0; i<e.length; i++){
out2.println(e[i][0] + "->" + e[i][1] + ";");
}
out2.println("}");
out2.close();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 17 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 163f36677e7334213a1a19f226f4e7fe | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
// https://codeforces.com/contest/1609/problem/C
public class C {
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int M = 1000010;
boolean[] primes = new boolean[M];
for(int i = 2; i < M; i++)
primes[i] = true;
for(int i = 2; i < M; i++)
{
if(primes[i])
{
for(int j = i + i; j < M; j += i)
primes[j] = false;
}
}
int T = Integer.parseInt(br.readLine());
for(int t = 0; t < T; t++)
{
String[] ne = br.readLine().split(" ");
int n = Integer.parseInt(ne[0]);
int e = Integer.parseInt(ne[1]);
//for(int p = 2; )
int arr[] = new int[n];
int left[] = new int[n];
int right[] = new int[n];
String[] line = br.readLine().split(" ");
for(int i = 0; i < n; i++)
{
arr[i] = Integer.parseInt(line[i]);
}
for(int z = 0; z < e; z++)
{
int i = z;
int cc = 0;
while(i < n)
{
if(arr[i] > 1)
{
left[i] = cc + 1;
cc = 0;
}
else
cc += 1;
i += e;
}
cc = 0;
i -= e;
while(i >= 0)
{
if(arr[i] > 1)
{
right[i] = cc + 1;
cc = 0;
}
else
cc += 1;
i -= e;
}
}
long ans = 0;
for(int i = 0; i < n; i++)
{
if(primes[arr[i]])
// ¿por qué no anda sin este 1L?
ans += left[i] * 1L * right[i] - 1;
}
System.out.println(ans);
}
br.close();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 59c45ac9649cdbe520619b8c5d427a3a | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class C {
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int M = 1000010;
boolean[] primes = new boolean[M];
for(int i = 2; i < M; i++)
primes[i] = true;
for(int i = 2; i < M; i++)
{
if(primes[i])
{
for(int j = i + i; j < M; j += i)
primes[j] = false;
}
}
int T = Integer.parseInt(br.readLine());
for(int t = 0; t < T; t++)
{
String[] ne = br.readLine().split(" ");
int n = Integer.parseInt(ne[0]);
int e = Integer.parseInt(ne[1]);
//for(int p = 2; )
int arr[] = new int[n];
int left[] = new int[n];
int right[] = new int[n];
String[] line = br.readLine().split(" ");
for(int i = 0; i < n; i++)
{
arr[i] = Integer.parseInt(line[i]);
}
for(int z = 0; z < e; z++)
{
int i = z;
int cc = 0;
while(i < n)
{
if(arr[i] > 1)
{
left[i] = cc + 1;
cc = 0;
}
else
cc += 1;
i += e;
}
cc = 0;
i -= e;
while(i >= 0)
{
if(arr[i] > 1)
{
right[i] = cc + 1;
cc = 0;
}
else
cc += 1;
i -= e;
}
}
long ans = 0;
for(int i = 0; i < n; i++)
{
if(primes[arr[i]])
ans += left[i] * 1L * right[i] - 1;
}
System.out.println(ans);
}
br.close();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 04fc66d608e5cb099b3c1232b078b268 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class cses{
static boolean[] primes = new boolean[1000000 + 1];
public static void main(String[] args) {
Arrays.fill(primes, true);
primes[0] = primes[1] = false;
for (int i = 2; i < primes.length; i++) {
if(primes[i] && (long)i * i <= 1000000){
for (int j = i * i; j <= 1000000; j += i) {
primes[j] = false;
}
}
}
/*p.add(2);
primes[2] = true;
for (int i = 3; i < 1000001; i+=2) {
boolean isP = true;
for (int j = 0; j < p.size() && p.get(j) <= (int) Math.sqrt(i); j++) {
if(i % p.get(j) == 0){
isP = false;
break;
}
}
if(isP) {
primes[i] = true;
p.add(i);
}
}*/
int t = io.nextInt();
long[] ans = new long[t];
StringBuilder out = new StringBuilder();
for (int i = 0; i < t; i++) {
int n = io.nextInt();
int e = io.nextInt();
int[] arr = new int[n];
for (int j = 0; j < n; j++) {
arr[j] = io.nextInt();
}
for(int k = 0; k < n; k++) {
if(primes[arr[k]]) {
int l = 0, r = 0;
for (int j = k + e; j < n; j += e) {
if (arr[j] != 1)
break;
r++;
}
for (int j = k - e; j >= 0; j -= e) {
if (arr[j] != 1)
break;
l++;
}
ans[i] += l + r + (long)l*r;
}
}
/*ArrayList<Integer>[] div = new ArrayList[e];
ArrayList<Integer>[] divr = new ArrayList[e];
for (int j = 0; j < e; j++) {
div[j] = new ArrayList<>();
divr[j] = new ArrayList<>();
}
for (int j = 0; j < n; j++) {
int a = io.nextInt();
div[j % e].add(a);
divr[j % e].add(0, a);
}
int[][] arr1 = ways(div);
int[][] arr2 = ways(divr);
for (int j = 0; j < e; j++) {
for (int k = 0; k < arr1[j].length; k++) {
ans[i] += (arr1[j][k] + 1) * (arr2[j][arr2[j].length - k-1] +1) - 1;
}
}*/
}
for (int i = 0; i < t; i++) {
out.append(ans[i]).append('\n');
}
out.deleteCharAt(out.length()-1);
System.out.print(out);
}
public static int[][] ways(ArrayList<Integer>[] arr){
int[][] ret = new int[arr.length][];
for (int i = 0; i < arr.length; i++) {
ArrayList<Integer> cur = arr[i];
ret[i] = new int[cur.size()];
int cnt = 0;
for (int j = 0; j < cur.size(); j++) {
if(cur.get(j) == 1){
cnt++;
}
else{
if(primes[cur.get(j)])
ret[i][j]+=cnt;
cnt = 0;
}
}
}
return ret;
}
static Kattio29 io = new Kattio29();
static class Kattio29 extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio29() {
this(System.in, System.out);
}
public Kattio29(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio29(String problemName) throws IOException {
super(new FileWriter(problemName + ".out"));
r = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) {
}
return null;
}
public int nextInt() { return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 2dbb1207f055d6736d492c157e992808 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.util.ArrayList;
import java.util.Scanner;
import java.io.*;
public class pro3 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int casess = input.nextInt();
ArrayList<Integer> arr = new ArrayList<>();
do {
int num = input.nextInt();
int e = input.nextInt();
arr = new ArrayList<>();
for (int i = 0; i < num; i++) {
arr.add(input.nextInt());
}
if (!arr.contains(1)) {
System.out.println(0);
casess--;
continue;
}
long number = 0;
for (int i = 0; i < num; i++) {
if (isPrime(arr.get(i))) {
long w = 0;
long r = 0;
for (int j = i + e; j < num; j += e) {
if (arr.get(j) == 1)
w++;
else
break;
}
for (int j = i - e; j >= 0; j -= e) {
if (arr.get(j) == 1)
r++;
else
break;
}
number += w + r + (w * r);
}
}
System.out.println(number);
casess--;
} while (casess != 0);
}
public static boolean isPrime(long n) {
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2) {
if (n % i == 0)
return false;
}
return true;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 4addbcfcbec9cd353b517e4e1548a4b9 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class C1 {
static HashSet<Long> primes = new HashSet<Long>();
public static void main (String[] args) throws IOException {
Kattio io = new Kattio();
sieveOfEratosthenes(1000001);
int t = io.nextInt();
//System.out.println(primes.contains(880667));
//System.out.println(primes1.contains(3));
for (int ii=0; ii<t; ii++) {
long ans = 0;
int n = io.nextInt();
long e = io.nextLong();
long[] arr = new long[n];
for (int i=0; i<n; i++) {
arr[i] = io.nextLong();
}
boolean[] visited = new boolean[n];
Arrays.fill(visited, false);
for (int i=0; i<n; i++) {
if (visited[i]) continue; //if already in a cycle
long cur_ones = 0;
ArrayList<Long> ones = new ArrayList<Long>();
for (int j=i; j<n; j+= e) { //start a new cycle
visited[j] = true;
if (!primes.contains(arr[j]) && arr[j] != 1) {
break;
}
if (primes.contains(arr[j])) {
ones.add(cur_ones);
cur_ones = 0;
} else if (arr[j] == 1) {
cur_ones++;
}
}
ones.add(cur_ones);
if (ones.size() == 1) continue;
for (int j=0; j<ones.size()-1; j++) {
ans += (((ones.get(j)+1) * (ones.get(j+1)+1))-1);
}
/*System.out.println("ONES: ");
for (int j : ones) {
System.out.println(j);
}*/
}
System.out.println(ans);
}
}
static void sieveOfEratosthenes(int n) {
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
primes.add((long) i);
}
}
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() { this(System.in, System.out); }
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(new FileWriter(problemName + ".out"));
r = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) { }
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 6f41dd115d4d32ca9dbc3b94eb726a2d | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
public static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
public static int mod = (int)1e9+7;
static boolean[] isprime=new boolean[10000001];
static {
for(int i=2;i<isprime.length;i++) isprime[i]=true;
for(int i=2;i<isprime.length;i++) {
if(isprime[i]) {
for(int k = 2; i*k<isprime.length; k++) {
isprime[i*k] = false;
}
}
}
}
public static void main(String[] args) throws Exception {
String ss = bf.readLine();
if (ss.contains(" ")) ss = ss.substring(0, ss.indexOf(" "));
int t = Integer.parseInt(ss);
while (t-- > 0) {
String s[] = bf.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int e = Integer.parseInt(s[1]);
String s1[] = bf.readLine().split(" ");
int a[] = new int[n];
long ans = 0;
for (int i = 0; i < n; i++) {
a[i] = Integer.parseInt(s1[i]);
}
for (int i = 0; i < n; i++) {
if(isprime[a[i]]){
long cnt1 = 0 ,cnt2 = 0;
for(int j = i-e;j >= 0;j-=e){
if(a[j] == 1) cnt1++;
else break;
}
for(int j = i+e;j < n;j+=e){
if(a[j] == 1) cnt2++;
else break;
}
ans += (cnt1*cnt2+cnt1+cnt2);
}
}
bw.write(ans+"\n");
}
bw.flush();
}
public static long gcd(long a, long b){
return b == 0 ? a : gcd(b,a%b);
}
public static long lcm(long a, long b){
return a*b / gcd(a,b);
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 245d064275f9428c0c1d7d9d22e0e99e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
public static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
public static int mod = (int)1e9+7;
static boolean[] isprime=new boolean[1000001];
static {
for(int i=2;i<isprime.length;i++) isprime[i]=true;
for(int i=2;i<isprime.length;i++) {
if(isprime[i]) {
for(int k = 2; i*k<isprime.length; k++) {
isprime[i*k] = false;
}
}
}
}
public static void main(String[] args) throws Exception {
String ss = bf.readLine();
if (ss.contains(" ")) ss = ss.substring(0, ss.indexOf(" "));
int t = Integer.parseInt(ss);
while (t-- > 0) {
String s[] = bf.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int e = Integer.parseInt(s[1]);
String s1[] = bf.readLine().split(" ");
int a[] = new int[n];
long ans = 0;
for (int i = 0; i < n; i++) {
a[i] = Integer.parseInt(s1[i]);
}
for (int i = 0; i < n; i++) {
if(isprime[a[i]]){
long cnt1 = 0 ,cnt2 = 0;
for(int j = i-e;j >= 0;j-=e){
if(a[j] == 1) cnt1++;
else break;
}
for(int j = i+e;j < n;j+=e){
if(a[j] == 1) cnt2++;
else break;
}
ans += (cnt1*cnt2+cnt1+cnt2);
}
}
bw.write(ans+"\n");
}
bw.flush();
}
public static long gcd(long a, long b){
return b == 0 ? a : gcd(b,a%b);
}
public static long lcm(long a, long b){
return a*b / gcd(a,b);
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | a4807292ff344b0b60bd7c83bdd87d30 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;
public class snackDown3
{
public static class FastReader {
BufferedReader b;
StringTokenizer s;
public FastReader() {
b=new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(s==null ||!s.hasMoreElements()) {
try {
s=new StringTokenizer(b.readLine());
}
catch(IOException e) {
e.printStackTrace();
}
}
return s.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str="";
try {
str=b.readLine();
}
catch(IOException e) {
e.printStackTrace();
}
return str;
}
boolean hasNext() {
if (s != null && s.hasMoreTokens()) {
return true;
}
String tmp;
try {
b.mark(1000);
tmp = b.readLine();
if (tmp == null) {
return false;
}
b.reset();
} catch (IOException e) {
return false;
}
return true;
}
}
public static class pair{
char a;
int b;
public pair(char a,int b) {
this.a=a;
this.b=b;
}
}
public static class pair2{
int a;
int b;
public pair2(int a,int b) {
this.a=a;
this.b=b;
}
public int compareTo(pair2 b) {
return this.b-b.b;
}
}
//public static class trip
public static void main (String[] args) throws Exception
{
// your code goes here
// int a[][][]=new int[3][3][3];
FastReader sc=new FastReader();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
// OutputStream log = new BufferedOutputStream ( System.out );
int t=sc.nextInt();
boolean prime[]=prime(1000001);
while(t--!=0) {
int n=sc.nextInt();
int e=sc.nextInt();
int max=-1;
int a[]=new int[n+1];
for(int i=1;i<=n;i++) {
a[i]=sc.nextInt();
max=Math.max(max, a[i]);
}
long ans=0;
long cnt=0;
boolean pr=false;
long pcnt=0;
for(int i=1;i<=e;i++) {
cnt=0;
pcnt=0;
pr=false;
for(int j=i;j<=n;j+=e) {
if(a[j]==1) {
cnt++;
if(pr) {
ans+=pcnt;
ans+=1;
}
}
else if(prime[a[j]]) {
ans+=cnt;
pcnt=cnt;
cnt=0;
pr=true;
}
else {
cnt=0;
pcnt=0;
pr=false;
}
}
}
log.write(ans+"\n");
log.flush();
}
}
static int bs(LinkedList<Integer> ar,int el) {
int start=0;
int end=ar.size()-1;
while(start<=end) {
int mid=start+(end-start)/2;
if(ar.get(mid)==el)return mid;
else if(ar.get(mid)<el)start=mid+1;
else end=mid-1;
}
return end;
}
static boolean[] prime(int n) {
boolean vis[]=new boolean[n+1];
Arrays.fill(vis, true);
vis[0]=false;
vis[1]=false;
for(int i=2;i*i<=n;i++) {
if(vis[i]) {
for(int j=2*i;j<=n;j+=i) {
vis[j]=false;
}
}
}
return vis;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7465e570c734496f3dfa43f9848f77f6 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.BitSet;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.Set;
import java.util.StringTokenizer;
public class A3y {
static BitSet notPrimes = notPrimes(1_000_000);
void solve() {
int n = cint();
int step = cint();
int[] a = carr_int(n);
int[] d = new int[n]; // длина цепочки только из 1
int[] pos = new int[n]; // позиция первого не 1
int[] ans = new int[n];
long sum = 0;
for (int i = n-1; i >= 0; i--) {
boolean prime = !notPrimes.get(a[i]);
int next = i + step;
if (next >= n) {
if (a[i] == 1) {
d[i] = 1;
} else if (prime) {
pos[i] = i;
}
} else {
if (a[i] == 1) {
d[i] = d[next] + 1;
pos[i] = pos[next];
if (pos[i] > 0) {
ans[i] = ans[pos[i]] + 1;
}
} else if (prime) {
pos[i] = i;
if (d[next] > 0) {
ans[i] = d[next];
}
}
sum += ans[i];
}
}
cout(sum);
}
static BitSet notPrimes(int n) {
BitSet np = new BitSet(n + 1); // not prime
np.set(0);
np.set(1);
for (int i = 2; i <= n; i++) {
if (!np.get(i)) {
for (long j = i * (long)i; j <= n; j += i) {
np.set((int)j);
}
}
}
return np;
}
public static void main(String... args) {
int t = in.nextInt();
for (int test = 0; test < t; test++) {
new A3y().solve();
}
}
static final QuickReader in = new QuickReader(System.in);
static final PrintStream out = System.out;
static int cint() {
return in.nextInt();
}
static long clong() {
return in.nextLong();
}
static String cstr() {
return in.nextLine();
}
static int[] carr_int(int n) {
return in.nextInts(n);
}
static long[] carr_long(int n) {
return in.nextLongs(n);
}
static void cout(int... a) {
StringBuilder buf = new StringBuilder();
for (int i = 0; i < a.length; i++) {
if (i != 0) {
buf.append(' ');
}
buf.append(a[i]);
}
out.println(buf);
}
static void cout(long... a) {
StringBuilder buf = new StringBuilder();
for (int i = 0; i < a.length; i++) {
if (i != 0) {
buf.append(' ');
}
buf.append(a[i]);
}
out.println(buf);
}
static void cout(Object o) {
out.println(o);
}
static class QuickReader {
BufferedReader in;
StringTokenizer token;
public QuickReader(InputStream ins) {
in = new BufferedReader(new InputStreamReader(ins));
token = new StringTokenizer("");
}
public boolean hasNext() {
while (!token.hasMoreTokens()) {
try {
String s = in.readLine();
if (s == null) {
return false;
}
token = new StringTokenizer(s);
} catch (IOException e) {
throw new InputMismatchException();
}
}
return true;
}
public String next() {
hasNext();
return token.nextToken();
}
public String nextLine() {
try {
String s = in.readLine();
token = new StringTokenizer("");
return s;
} catch (IOException e) {
throw new InputMismatchException();
}
}
public int nextInt() {
return Integer.parseInt(next());
}
public int[] nextInts(int n) {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
public long nextLong() {
return Long.parseLong(next());
}
public long[] nextLongs(int n) {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextLong();
}
return res;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 59e2498a40cf0ba1202345243a376f38 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
public class A3x {
void solve() {
int n = cint();
int step = cint();
int[] a = carr_int(n);
int[] left1 = new int[n];
int[] right1 = new int[n];
for (int i = 0; i < n; i++) {
if (a[i] == 1) {
left1[i] = 1;
right1[i] = 1;
}
}
for (int i = step; i < n; i++) {
if (a[i] == 1) {
left1[i] = left1[i - step] + 1;
}
}
for (int i = n-step-1; i >= 0; i--) {
if (a[i] == 1) {
right1[i] = right1[i + step] + 1;
}
}
long ans = 0;
for (int i = 0; i < n; i++) {
if (prime.get(a[i])) {
long k1 = i - step >= 0 ? left1[i - step] : 0;
long k2 = i + step < n ? right1[i + step] : 0;
ans += Math.max(0, (k1 + 1) * (k2 + 1) - 1);
}
}
cout(ans);
}
public static void main(String... args) {
int t = in.nextInt();
for (int test = 0; test < t; test++) {
new A3x().solve();
}
}
static BitSet prime = primeSet(1_000_000);
static BitSet primeSet(int n) {
BitSet p = new BitSet(n + 1);
p.flip(2, n + 1);
for (int i = 2; i <= n; i++) {
if (p.get(i)) {
for (long j = i * (long)i; j <= n; j += i) {
p.clear((int)j);
}
}
}
return p;
}
static final QuickReader in = new QuickReader(System.in);
static final PrintStream out = System.out;
static int cint() {
return in.nextInt();
}
static long clong() {
return in.nextLong();
}
static String cstr() {
return in.nextLine();
}
static int[] carr_int(int n) {
return in.nextInts(n);
}
static long[] carr_long(int n) {
return in.nextLongs(n);
}
static void cout(int... a) {
StringBuilder buf = new StringBuilder();
for (int i = 0; i < a.length; i++) {
if (i != 0) {
buf.append(' ');
}
buf.append(a[i]);
}
out.println(buf);
}
static void cout(long... a) {
StringBuilder buf = new StringBuilder();
for (int i = 0; i < a.length; i++) {
if (i != 0) {
buf.append(' ');
}
buf.append(a[i]);
}
out.println(buf);
}
static void cout(Object o) {
out.println(o);
}
static class QuickReader {
BufferedReader in;
StringTokenizer token;
public QuickReader(InputStream ins) {
in = new BufferedReader(new InputStreamReader(ins));
token = new StringTokenizer("");
}
public boolean hasNext() {
while (!token.hasMoreTokens()) {
try {
String s = in.readLine();
if (s == null) {
return false;
}
token = new StringTokenizer(s);
} catch (IOException e) {
throw new InputMismatchException();
}
}
return true;
}
public String next() {
hasNext();
return token.nextToken();
}
public String nextLine() {
try {
String s = in.readLine();
token = new StringTokenizer("");
return s;
} catch (IOException e) {
throw new InputMismatchException();
}
}
public int nextInt() {
return Integer.parseInt(next());
}
public int[] nextInts(int n) {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
public long nextLong() {
return Long.parseLong(next());
}
public long[] nextLongs(int n) {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextLong();
}
return res;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | f339626808dcdc54a08524a50ef44001 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.BitSet;
import java.util.Random;
public class A3 {
void solve() throws Exception {
String line = in.readLine();
String[] p = line.split("[\\s]");
int n = Integer.parseInt(p[0]);
int step = Integer.parseInt(p[1]);
int[] a = carr_int(n);
int[] d = new int[n]; // длина цепочки только из 1
int[] pos = new int[n]; // позиция первого не 1
long[] ans = new long[n];
long sum = 0;
for (int i = n-1; i >= 0; i--) {
boolean prime = !notPrimes.get(a[i]);
int next = i + step;
if (next >= n) {
if (a[i] == 1) {
d[i] = 1;
} else if (prime) {
pos[i] = i;
}
} else {
if (a[i] == 1) {
d[i] = d[next] + 1;
pos[i] = pos[next];
if (pos[i] > 0) {
ans[i] = ans[pos[i]] + 1;
}
} else if (prime) {
pos[i] = i;
if (d[next] > 0) {
ans[i] = d[next];
}
}
sum += ans[i];
}
}
cout(sum);
}
static BitSet notPrimes(int n) {
BitSet np = new BitSet(n + 1); // not prime
np.set(0);
np.set(1);
for (int i = 2; i <= n; i++) {
if (!np.get(i)) {
for (long j = i * (long)i; j <= n; j += i) {
np.set((int)j);
}
}
}
return np;
}
static int[] randomArray(int n, int maxValue) {
Random rnd = new Random();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = rnd.nextInt(maxValue) + 1;
}
return a;
}
static BitSet notPrimes;
public static void main(String... args) throws Exception {
// long ns = System.currentTimeMillis();
// int[] a = randomArray(200_000, 100_000_000);
// StringBuilder buf = new StringBuilder();
// for (int i = 0; i < a.length; i++) {
// if (i != 0) {
// buf.append(' ');
// }
// buf.append(a[i]);
// }
// int[] b = parse(buf.toString(), 200_000);
// cout(buf.length());
// cout(System.currentTimeMillis() - ns);
notPrimes = notPrimes(1_000_000);
int t = Integer.parseInt(in.readLine());
for (int test = 0; test < t; test++) {
new A3().solve();
}
}
static final BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
static final PrintStream out = System.out;
static int[] carr_int(int n)throws Exception {
String s = in.readLine();
int[] a = new int[n];
for (int i = 0, j = 0; i < s.length() && j < n; i++) {
if (s.charAt(i) == ' ') {
j++;
} else {
a[j] = a[j] * 10 + s.charAt(i) - '0';
}
}
return a;
}
static int[] parse(String s, int n) {
int[] a = new int[n];
for (int i = 0, j = 0; i < s.length() && j < n; i++) {
if (s.charAt(i) == ' ') {
j++;
} else {
a[j] = a[j] * 10 + s.charAt(i) - '0';
}
}
return a;
}
static void cout(Object o) {
out.println(o);
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | ae008c8482c53e331a0b6d67586026ec | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.HashMap;
import java.util.StringTokenizer;
/**
*
* @author eslam
*/
public class ComplexMarketAnalysis {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static HashMap<Integer,Integer> h = new HashMap<>();
public static void main(String[] args) throws IOException {
FastReader input = new FastReader();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
int t = input.nextInt();
for (int i = 2; i < Math.pow(10, 6); i++) {
if(isPrime(i))h.put(i, 1);
}
while (t-- > 0) {
int n = input.nextInt();
int e = input.nextInt();
long ans = 0;
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = input.nextInt();
}
for (int i = 0; i < n; i++) {
if (isPrime(a[i])) {
h.put(a[i], 1);
int index = i + e;
int index2 = i -e;
long b = 0;
while (index < n) {
if (a[index] != 1) {
break;
}
index += e;
b++;
}
while (index2 > -1) {
if (a[index2] != 1) {
break;
}
index2 -= e;
ans+= b+1;
}
ans += b;
}
}
log.write(ans+"\n");
}
log.flush();
}
private static boolean isPrime(int num) {
if(h.get(num)!=null)return true;
if (num == 1) {
return false;
}
if (num == 2) {
return true;
}
if (num % 2 == 0) {
return false;
}
if (num == 3) {
return true;
}
for (int i = 3; i <= Math.sqrt(num) + 1; i += 2) {
if (num % i == 0) {
return false;
}
}
return true;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | e460a9adde1b95cc1e66feac2ab55a96 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.HashMap;
import java.util.StringTokenizer;
/**
*
* @author eslam
*/
public class ComplexMarketAnalysis {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static HashMap<Integer,Integer> h = new HashMap<>();
public static void main(String[] args) throws IOException {
FastReader input = new FastReader();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
int t = input.nextInt();
while (t-- > 0) {
int n = input.nextInt();
int e = input.nextInt();
long ans = 0;
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = input.nextInt();
}
for (int i = 0; i < n; i++) {
if (isPrime(a[i])) {
h.put(a[i], 1);
int index = i + e;
int index2 = i -e;
long b = 0;
while (index < n) {
if (a[index] != 1) {
break;
}
index += e;
b++;
}
while (index2 > -1) {
if (a[index2] != 1) {
break;
}
index2 -= e;
ans+= b+1;
}
ans += b;
}
}
log.write(ans+"\n");
}
log.flush();
}
private static boolean isPrime(int num) {
if(h.get(num)!=null)return true;
if (num == 1) {
return false;
}
if (num == 2) {
return true;
}
if (num % 2 == 0) {
return false;
}
if (num == 3) {
return true;
}
for (int i = 3; i <= Math.sqrt(num) + 1; i += 2) {
if (num % i == 0) {
return false;
}
}
return true;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7d8e0d0caeb0d0e34e483c1a6636a061 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
//Complex Market Analysis
public class ComplexMarketAnalysis {
public static void main(String[] args) {
ComplexMarketAnalysis.solution();
}
public static void solution() {
Scanner scanner = new Scanner(System.in);
int numCases = Integer.parseInt(scanner.nextLine());
for (int i = 0; i < numCases; i++) {
String[] tokens = scanner.nextLine().split(" ");
int e = Integer.parseInt(tokens[1]);
int[] arr = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
int[] pre = new int[arr.length];
long count = 0;
for (int j = 0; j < arr.length; j++) {
pre[j] = arr[j] == 1 ? 1 : 0;
if (j - e >= 0 && arr[j] == 1) {
pre[j] += pre[j - e];
}
}
int[] suf = new int[arr.length];
for (int j = arr.length - 1; j >= 0; j--) {
suf[j] = arr[j] == 1 ? 1 : 0;
if (j + e < arr.length && arr[j] == 1) {
suf[j] += suf[j + e];
}
}
for (int j = 0; j < arr.length; j++) {
if (isPrime(arr[j])) {
long left = j - e >= 0 ? pre[j - e] : 0;
long right = j + e < arr.length ? suf[j + e] : 0;
count = count + left + right + (left * right);
}
}
System.out.println(count);
}
}
private static boolean isPrime(int num) {
int sqrt = (int) Math.sqrt(num);
for (int i = 2; i <= sqrt; i++) {
if (num % i == 0) {
return false;
}
}
return num > 1;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | dfc91a93363a63771ee55d0895000844 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int numCases = Integer.parseInt(scanner.nextLine());
for (int i = 0; i < numCases; i++) {
String[] tokens = scanner.nextLine().split(" ");
int e = Integer.parseInt(tokens[1]);
List<Integer> arr = Arrays.stream(scanner.nextLine().split(" ")).map((Integer::parseInt)).collect(Collectors.toList());
int[] pre = new int[arr.size()];
long count = 0;
for (int j = 0; j < arr.size(); j++) {
pre[j] = arr.get(j) == 1 ? 1 : 0;
if (j - e >= 0 && arr.get(j) == 1) {
pre[j] += pre[j - e];
}
}
int[] suf = new int[arr.size()];
for (int j = arr.size() - 1; j >= 0; j--) {
suf[j] = arr.get(j) == 1 ? 1 : 0;
if (j + e < arr.size() && arr.get(j) == 1) {
suf[j] += suf[j + e];
}
}
for (int j = 0; j < arr.size(); j++) {
if (isPrime(arr.get(j))) {
long left = j - e >= 0 ? pre[j - e] : 0;
long right = j + e < arr.size() ? suf[j + e] : 0;
count = count + left + right + (left * right);
}
}
System.out.println(count);
}
}
private static boolean isPrime(int num) {
int sqrt = (int) Math.sqrt(num);
for (int i = 2; i <= sqrt; i++) {
if (num % i == 0) {
return false;
}
}
return num > 1;
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 15b248eeebfdd0e7f616b8c26bd327e3 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.io.File;
import java.io.FileInputStream;
import java.util.*;
public class Main {
// static final File ip = new File("input.txt");
// static final File op = new File("output.txt");
// static {
// try {
// System.setOut(new PrintStream(op));
// System.setIn(new FileInputStream(ip));
// } catch (Exception e) {
// }
// }
public static void main(String[] args) {
FastReader sc = new FastReader();
int[] primes = getPrimes((int) 1e6);
int test = 1;
test = sc.nextInt();
while (test-- > 0) {
int n = sc.nextInt();
int k = sc.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
int[] pre = new int[n];
int[] suff = new int[n];
// pre[0] = a[0] == 1 ? 1 : 0;
for (int i = 0; i < n; i++) {
if (i >= k && a[i - k] == 1 && a[i] == 1)
pre[i] = pre[i - k];
if (a[i] == 1)
pre[i]++;
}
// suff[n - 1] = a[n - 1] == 1 ? 1 : 0;
for (int i = n - 1; i >= 0; i--) {
if (i + k < n && a[i + k] == 1 && a[i] == 1)
suff[i] = suff[i + k];
if (a[i] == 1)
suff[i]++;
}
long sum = 0;
for (int i = 0; i < n; i++) {
long p = 0, s = 0;
if (primes[a[i]] == 1 && a[i] != 1) {
if (i + k < n) {
s = suff[i + k];
sum += s;
}
if (i >= k) {
p = pre[i - k];
sum += p;
}
sum += p * s;
}
}
System.out.println(sum);
// for(long i : pre) System.out.print(i+" ");
// System.out.println();
// for(long i : suff) System.out.print(i+" ");
}
}
static long power(long x, long y, long p) {
long res = 1;
x = x % p;
if (x == 0)
return 0;
while (y > 0) {
if ((y & 1) != 0)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
public static int countSetBits(long number) {
int count = 0;
while (number > 0) {
++count;
number &= number - 1;
}
return count;
}
static int lower_bound(long target, pair[] a, int pos) {
if (pos >= a.length)
return -1;
int low = pos, high = a.length - 1;
while (low < high) {
int mid = low + (high - low) / 2;
if (a[mid].a < target)
low = mid + 1;
else
high = mid;
}
return a[low].a >= target ? low : -1;
}
private static <T> void swap(int[] a, int i, int j) {
int tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
static class pair {
int a;
int b;
pair(int x, int y) {
this.a = x;
this.b = y;
}
}
static class first implements Comparator<pair> {
public int compare(pair p1, pair p2) {
if (p1.a > p2.a)
return -1;
else if (p1.a < p2.a)
return 1;
return 0;
}
}
static class second implements Comparator<pair> {
public int compare(pair p1, pair p2) {
if (p1.b > p2.b)
return 1;
else if (p1.b < p2.b)
return -1;
return 0;
}
}
private static long getSum(int[] array) {
long sum = 0;
for (int value : array) {
sum += value;
}
return sum;
}
private static boolean isPrime(Long x) {
if (x < 2)
return false;
for (long d = 2; d * d <= x; ++d) {
if (x % d == 0)
return false;
}
return true;
}
static int[] reverse(int a[]) {
int n = a.length;
int i, k, t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
private static boolean isPrimeInt(int x) {
if (x < 2)
return false;
for (int d = 2; d * d <= x; ++d) {
if (x % d == 0)
return false;
}
return true;
}
public static String reverse(String input) {
StringBuilder str = new StringBuilder("");
for (int i = input.length() - 1; i >= 0; i--) {
str.append(input.charAt(i));
}
return str.toString();
}
private static int[] getPrimes(int n) {
boolean[] used = new boolean[n + 1];
used[0] = used[1] = true;
// int size = 0;
for (int i = 2; i <= n; ++i) {
if (!used[i]) {
// ++size;
for (int j = 2 * i; j <= n; j += i) {
used[j] = true;
}
}
}
int[] primes = new int[n + 1];
for (int i = 0; i <= n; ++i) {
if (!used[i]) {
primes[i] = 1;
}
}
return primes;
}
private static long lcm(long a, long b) {
return a / gcd(a, b) * b;
}
private static long gcd(long a, long b) {
return (a == 0 ? b : gcd(b % a, a));
}
static void sortI(int[] arr) {
int n = arr.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
int tmp = arr[i];
int randomPos = i + rnd.nextInt(n - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
Arrays.sort(arr);
}
static void shuffleList(ArrayList<Long> arr) {
int n = arr.size();
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
long tmp = arr.get(i);
int randomPos = i + rnd.nextInt(n - i);
arr.set(i, arr.get(randomPos));
arr.set(randomPos, tmp);
}
}
static void factorize(long n) {
int count = 0;
while (!(n % 2 > 0)) {
n >>= 1;
count++;
}
if (count > 0) {
// System.out.println("2" + " " + count);
}
long i = 0;
for (i = 3; i <= (long) Math.sqrt(n); i += 2) {
count = 0;
while (n % i == 0) {
count++;
n = n / i;
}
if (count > 0) {
// System.out.println(i + " " + count);
}
}
if (n > 2) {
// System.out.println(i + " " + count);
}
}
static void sortL(long[] arr) {
int n = arr.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
long tmp = arr[i];
int randomPos = i + rnd.nextInt(n - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
Arrays.sort(arr);
}
////////////////////////////////// DSU START ///////////////////////////
static class DSU {
int[] parent, rank, total_Elements;
DSU(int n) {
parent = new int[n + 1];
rank = new int[n + 1];
total_Elements = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
rank[i] = 1;
total_Elements[i] = 1;
}
}
int find(int u) {
if (parent[u] == u)
return u;
return parent[u] = find(parent[u]);
}
void unionByRank(int u, int v) {
int pu = find(u);
int pv = find(v);
if (pu != pv) {
if (rank[pu] > rank[pv]) {
parent[pv] = pu;
total_Elements[pu] += total_Elements[pv];
} else if (rank[pu] < rank[pv]) {
parent[pu] = pv;
total_Elements[pv] += total_Elements[pu];
} else {
parent[pu] = pv;
total_Elements[pv] += total_Elements[pu];
rank[pv]++;
}
}
}
boolean unionBySize(int u, int v) {
u = find(u);
v = find(v);
if (u != v) {
parent[u] = v;
total_Elements[v] += total_Elements[u];
total_Elements[u] = 0;
return true;
}
return false;
}
}
////////////////////////////////// DSU END /////////////////////////////
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public boolean hasNext() {
return false;
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 2af029062704a5fa258fb694b62dea1c | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.Scanner;
public class ProblemC {
static boolean[] nPrimes;
public static void main(String[] args) {
nPrimes = new boolean[1000001];
nPrimes[0] = true;
nPrimes[1] = true;
int t1 = 2;
while (t1 < 1000) {
for (int i = 2 * t1; i <= 1000000; i += t1) {
nPrimes[i] = true;
}
t1++;
while (nPrimes[t1]) t1++;
}
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
for (int testNum = 0; testNum < t; testNum++) {
int n = scanner.nextInt();
int e = scanner.nextInt();
int[] a = new int[n];
for (int arrNum = 0; arrNum < n; arrNum++) {
a[arrNum] = scanner.nextInt();
}
long sum = 0;
for (int i = 1; i <= n; i++) {
int pos = i - 1;
if (!nPrimes[a[pos]]) {
long rightSum = 0;
pos += e;
while (pos < n) {
if (a[pos] != 1) break;
rightSum++;
pos += e;
}
sum += rightSum;
pos = i - 1 - e;
rightSum++;
while (pos >= 0) {
if (a[pos] != 1) break;
sum += rightSum;
pos -= e;
}
}
}
System.out.println(sum);
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | df8aa600d668ef2a8ffa01d9fde49811 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class JaiShreeRam{
static Scanner in=new Scanner();
static long mod = 1000000007;
static ArrayList<ArrayList<Integer>> adj;
static int seive[]=new int[1000001];
public static void main(String[] args) throws Exception{
int z=in.readInt();
seive();
while(z-->0) {
solve();
}
}
static void solve() {
int n=in.readInt();
int e=in.readInt();
int a[]=nia(n);
int pre[]=new int[n];
int suff[]=new int[n];
for(int i=0;i<n;i++) {
if(a[i]==1) {
if(i>=e) {
pre[i]=1+pre[i-e];
}
else {
pre[i]=1;
}
}
}
for(int i=n-1;i>=0;i--) {
if(a[i]==1) {
if(i<n-e) {
suff[i]=1+suff[i+e];
}
else {
suff[i]=1;
}
}
}
long count=0;
for(int i=0;i<n;i++) {
if(seive[a[i]]==1) {
long p= i>=e?pre[i-e]:0;
long s= i<n-e?suff[i+e]:0;
count+=p*s+p+s;
}
}
System.out.println(count);
}
static void seive() {
Arrays.fill(seive, 1);
seive[0]=0;
seive[1]=0;
for(int i=2;i*i<1000001;i++) {
if(seive[i]==1) {
for(int j=i*i;j<1000001;j+=i) {
if(seive[j]==1) {
seive[j]=0;
}
}
}
}
}
static int[] nia(int n){
int[] arr= new int[n];
int i=0;
while(i<n){
arr[i++]=in.readInt();
}
return arr;
}
static long[] nla(int n){
long[] arr= new long[n];
int i=0;
while(i<n){
arr[i++]=in.readLong();
}
return arr;
}
static int[] nia1(int n){
int[] arr= new int[n+1];
int i=1;
while(i<=n){
arr[i++]=in.readInt();
}
return arr;
}
static long gcd(long a, long b) {
if (b==0) return a;
return gcd(b, a%b);
}
static class Scanner{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String readString() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
double readDouble() {
return Double.parseDouble(readString());
}
int readInt() {
return Integer.parseInt(readString());
}
long readLong() {
return Long.parseLong(readString());
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 0b371091a18c09a3391faa59edd4b1a6 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;
public class Solution{
public static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static class Pair{
int a,b;
Pair(int x, int y) {
a=x;
b=y;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
static long mod = 998244353;
static StringBuffer sb = new StringBuffer("");
static int ans=0;
public static void main(String[] args) throws Exception
{
//Read input from user
//Scanner scn = new Scanner(System.in);
FastReader scn = new FastReader();
// BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
// int t = scn.nextInt();
//
OutputStream outputStream = System.out;
OutputWriter output = new OutputWriter(outputStream);
int t = scn.nextInt();
while(t>0) {
int n = scn.nextInt();
int e = scn.nextInt();
long[] a= new long[n];
long[] c = new long[n];
for(int i=0;i<n;i++) {
a[i]= scn.nextLong();
if(a[i]==1) {
c[i]= 1+(i-e<0?0:c[i-e]);
}
}
int[] c2= new int[n];
for(int i=n-1;i>=0;i--) {
if(a[i]==1) {
c2[i]= 1+(i+e>=n?0:c2[i+e]);
}
//System.out.println(i+" "+c2[i]);
}
long ans = 0;
for(int i=0;i<n;i++) {
if(a[i]!=1 &&( (i-e>=0 && c[i-e]>0) ||(i+e<n && c2[i+e]>0))) {
//System.out.println("num "+a[i]);
boolean p = true;
for(long j=2;j<=Math.sqrt(a[i]);j++) {
if(a[i]%j==0) {
p= false;
break;
}
}
if(p==true) {
//System.out.println("prime "+a[i]);
long a1=0,a2=0;
if(i-e>=0 && c[i-e]>0) {
a1 = c[i-e];
}
if(i+e<n && c2[i+e]>0) {
a2 = c2[i+e];
}
//System.out.println((i+1)+" "+a[i]+" "+a1+" "+a2);
if(a1==0 || a2==0) {
ans = ans+a1+a2;
}else {
ans = ans+(a1+1)*(a2+1)-1;
}
}
}
}
//output.println(ans);
System.out.println(ans);
t--;
}
// output.close();
}
/* public static int unsortedMex(List<Integer> b, int mex, int si) {
TreeSet<Integer> s = new TreeSet<>();
for(int i=0;i<=mex;i++) {
s.add(i);
}
for(int i= si;i<b.size();i++) {
if(s.contains(b.get(i))) {
s.remove(b.get(i));
}
Integer x = s.ceiling(0);
if(x==null || x>=mex) {
return i;
}
}
return (b.size()-1);
}
public static int mex(List<Integer> a, int si) {
int c=0;
HashSet<Integer> set = new HashSet<>();
for(int i=si;i<a.size();i++) {
set.add(a.get(i));
}
while(set.contains(c)) {
c++;
}
return c;
}
public static int findpow2(long n) {
int c=0;
while(n>0 && n%2==0) {
n = n/2;
c++;
}
return c;
}
public static void countLeaves(int curr, HashMap<Integer, List<Integer>> m1, List<Integer> p, List<List<Integer>> f) {
if(!m1.containsKey(curr)) {
f.add(new ArrayList<>(p));
return;
}
for(int next: m1.get(curr)) {
p.add(next);
countLeaves(next,m1,p,f);
p = new ArrayList<>();
}
}
public static int dfs(int curr, HashMap<Integer, List<Integer>> m1, int[] v) {
if(!m1.containsKey(curr)) {
return v[curr];
}
int val = v[curr];
for(int next: m1.get(curr)) {
val = val+dfs(next,m1,v);
}
// System.out.println("val "+curr+" "+val);
if(val==0) {
ans++;
}
return val;
}
public static long fact(long n) {
if(n<=1) {
return 1L;
}
return ((fact(n-1)%mod)*n)%mod;
}
public static long gcd(long a,long b){
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long lcm( long a,long b)
{
long g = gcd(a,b);
return (a*b)/g;
}
public static void find(long a, List<Long> div) {
for(long i=1;i<=Math.sqrt(a);i++) {
if(a%i==0) {
div.add(i);
if((a/i)!=i) {
div.add(a/i);
}
}
}
}*/
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 1d6f27ae26750e8dd1494fadfc3c6abd | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;
public class Solution{
public static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static class Pair{
int a,b;
Pair(int x, int y) {
a=x;
b=y;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
static long mod = 998244353;
static StringBuffer sb = new StringBuffer("");
static int ans=0;
public static void main(String[] args) throws Exception
{
//Read input from user
//Scanner scn = new Scanner(System.in);
FastReader scn = new FastReader();
// BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
// int t = scn.nextInt();
//
OutputStream outputStream = System.out;
OutputWriter output = new OutputWriter(outputStream);
int t = scn.nextInt();
while(t>0) {
int n = scn.nextInt();
int e = scn.nextInt();
long[] a= new long[n];
long[] c = new long[n];
for(int i=0;i<n;i++) {
a[i]= scn.nextLong();
if(a[i]==1) {
c[i]= 1+(i-e<0?0:c[i-e]);
}
}
int[] c2= new int[n];
for(int i=n-1;i>=0;i--) {
if(a[i]==1) {
c2[i]= 1+(i+e>=n?0:c2[i+e]);
}
//System.out.println(i+" "+c2[i]);
}
long ans = 0;
for(int i=0;i<n;i++) {
if(a[i]!=1 &&( (i-e>=0 && c[i-e]>0) ||(i+e<n && c2[i+e]>0))) {
//System.out.println("num "+a[i]);
boolean p = true;
for(long j=2;j<=Math.sqrt(a[i]);j++) {
if(a[i]%j==0) {
p= false;
break;
}
}
if(p==true) {
//System.out.println("prime "+a[i]);
long a1=0,a2=0;
if(i-e>=0 && c[i-e]>0) {
a1 = c[i-e];
}
if(i+e<n && c2[i+e]>0) {
a2 = c2[i+e];
}
//System.out.println((i+1)+" "+a[i]+" "+a1+" "+a2);
if(a1==0 || a2==0) {
ans = ans+a1+a2;
}else {
ans = ans+(a1+1)*(a2+1)-1;
}
}
}
}
output.println(ans);
t--;
}
output.close();
}
/* public static int unsortedMex(List<Integer> b, int mex, int si) {
TreeSet<Integer> s = new TreeSet<>();
for(int i=0;i<=mex;i++) {
s.add(i);
}
for(int i= si;i<b.size();i++) {
if(s.contains(b.get(i))) {
s.remove(b.get(i));
}
Integer x = s.ceiling(0);
if(x==null || x>=mex) {
return i;
}
}
return (b.size()-1);
}
public static int mex(List<Integer> a, int si) {
int c=0;
HashSet<Integer> set = new HashSet<>();
for(int i=si;i<a.size();i++) {
set.add(a.get(i));
}
while(set.contains(c)) {
c++;
}
return c;
}
public static int findpow2(long n) {
int c=0;
while(n>0 && n%2==0) {
n = n/2;
c++;
}
return c;
}
public static void countLeaves(int curr, HashMap<Integer, List<Integer>> m1, List<Integer> p, List<List<Integer>> f) {
if(!m1.containsKey(curr)) {
f.add(new ArrayList<>(p));
return;
}
for(int next: m1.get(curr)) {
p.add(next);
countLeaves(next,m1,p,f);
p = new ArrayList<>();
}
}
public static int dfs(int curr, HashMap<Integer, List<Integer>> m1, int[] v) {
if(!m1.containsKey(curr)) {
return v[curr];
}
int val = v[curr];
for(int next: m1.get(curr)) {
val = val+dfs(next,m1,v);
}
// System.out.println("val "+curr+" "+val);
if(val==0) {
ans++;
}
return val;
}
public static long fact(long n) {
if(n<=1) {
return 1L;
}
return ((fact(n-1)%mod)*n)%mod;
}
public static long gcd(long a,long b){
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long lcm( long a,long b)
{
long g = gcd(a,b);
return (a*b)/g;
}
public static void find(long a, List<Long> div) {
for(long i=1;i<=Math.sqrt(a);i++) {
if(a%i==0) {
div.add(i);
if((a/i)!=i) {
div.add(a/i);
}
}
}
}*/
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 46c219fc34829d6467e4626d74d227bf | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;
public class Solution{
public static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static class Pair{
int a,b;
Pair(int x, int y) {
a=x;
b=y;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
static long mod = 998244353;
static StringBuffer sb = new StringBuffer("");
static int ans=0;
public static void main(String[] args) throws Exception
{
//Read input from user
//Scanner scn = new Scanner(System.in);
FastReader scn = new FastReader();
// BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
// int t = scn.nextInt();
//
OutputStream outputStream = System.out;
OutputWriter output = new OutputWriter(outputStream);
int t = scn.nextInt();
while(t>0) {
int n = scn.nextInt();
int e = scn.nextInt();
long[] a= new long[n];
long[] c = new long[n];
for(int i=0;i<n;i++) {
a[i]= scn.nextLong();
if(a[i]==1) {
c[i]= 1+(i-e<0?0:c[i-e]);
}
}
int[] c2= new int[n];
for(int i=n-1;i>=0;i--) {
if(a[i]==1) {
c2[i]= 1+(i+e>=n?0:c2[i+e]);
}
//System.out.println(i+" "+c2[i]);
}
long ans = 0;
for(int i=0;i<n;i++) {
if(a[i]!=1 &&( (i-e>=0 && c[i-e]>0) ||(i+e<n && c2[i+e]>0))) {
//System.out.println("num "+a[i]);
boolean p = true;
for(long j=2;j<=Math.sqrt(a[i]);j++) {
if(a[i]%j==0) {
p= false;
break;
}
}
if(p==true) {
//System.out.println("prime "+a[i]);
long a1=0,a2=0;
if(i-e>=0 && c[i-e]>0) {
a1 = c[i-e];
}
if(i+e<n && c2[i+e]>0) {
a2 = c2[i+e];
}
//System.out.println((i+1)+" "+a[i]+" "+a1+" "+a2);
if(a1==0 || a2==0) {
ans = ans+a1+a2;
}else {
ans = ans+(a1+1)*(a2+1)-1;
}
}
}
}
output.println(ans);
t--;
}
output.close();
}
public static int unsortedMex(List<Integer> b, int mex, int si) {
TreeSet<Integer> s = new TreeSet<>();
for(int i=0;i<=mex;i++) {
s.add(i);
}
for(int i= si;i<b.size();i++) {
if(s.contains(b.get(i))) {
s.remove(b.get(i));
}
Integer x = s.ceiling(0);
if(x==null || x>=mex) {
return i;
}
}
return (b.size()-1);
}
public static int mex(List<Integer> a, int si) {
int c=0;
HashSet<Integer> set = new HashSet<>();
for(int i=si;i<a.size();i++) {
set.add(a.get(i));
}
while(set.contains(c)) {
c++;
}
return c;
}
public static int findpow2(long n) {
int c=0;
while(n>0 && n%2==0) {
n = n/2;
c++;
}
return c;
}
public static void countLeaves(int curr, HashMap<Integer, List<Integer>> m1, List<Integer> p, List<List<Integer>> f) {
if(!m1.containsKey(curr)) {
f.add(new ArrayList<>(p));
return;
}
for(int next: m1.get(curr)) {
p.add(next);
countLeaves(next,m1,p,f);
p = new ArrayList<>();
}
}
public static int dfs(int curr, HashMap<Integer, List<Integer>> m1, int[] v) {
if(!m1.containsKey(curr)) {
return v[curr];
}
int val = v[curr];
for(int next: m1.get(curr)) {
val = val+dfs(next,m1,v);
}
// System.out.println("val "+curr+" "+val);
if(val==0) {
ans++;
}
return val;
}
public static long fact(long n) {
if(n<=1) {
return 1L;
}
return ((fact(n-1)%mod)*n)%mod;
}
public static long gcd(long a,long b){
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long lcm( long a,long b)
{
long g = gcd(a,b);
return (a*b)/g;
}
public static void find(long a, List<Long> div) {
for(long i=1;i<=Math.sqrt(a);i++) {
if(a%i==0) {
div.add(i);
if((a/i)!=i) {
div.add(a/i);
}
}
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | af486c83412157cb2489b80dbbcc4cbc | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class c1609C {
private SimpleInputReader in;
private SimpleOutputWriter out;
Set<Integer> primes = sieve();
public c1609C(Reader r, Writer w) {
in = new SimpleInputReader(r);
out = new SimpleOutputWriter(w);
}
public c1609C() {
in = null;
out = null;
}
public void closeIO() throws Exception {
if(in != null) {
in.close();
}
if(out != null){
out.close();
}
}
public void takeInput() throws Exception {
int t = in.nextInt();
for(int i = 0; i < t; i++) {
int n = in.nextInt();
int e = in.nextInt();
int[] arr = new int[n];
for(int j = 0; j < n; j++) {
arr[j] = in.nextInt();
}
long res = solve(arr, e);
out.writeLine(res);
}
}
private long solve(int[] arr, int e) {
int n = arr.length;
long res = 0;
for(int i = 0; i < e; i++) {
int j = i;
int relJ = 0;
Integer lastPrimeFound = null;
int k = j;
int relK = 0;
for(; k < n; k += e, relK += 1) {
if(arr[k] == 1) {
continue;
}
if(primes.contains(arr[k])) {
if(lastPrimeFound == null) {
lastPrimeFound = relK;
}
else {
int c1 = (relK - lastPrimeFound);
int c2 = (lastPrimeFound - relJ);
res += (long) (c1) * (c2) + (c1 - 1);
relJ = lastPrimeFound + 1;
lastPrimeFound = relK;
}
}
else {
if(lastPrimeFound != null) {
int c1 = (relK - lastPrimeFound);
int c2 = (lastPrimeFound - relJ);
res += (long) (c1) * (c2) + (c1 - 1);
}
relJ = relK + 1;
lastPrimeFound = null;
}
}
if(lastPrimeFound != null) {
int c1 = (relK - lastPrimeFound);
int c2 = (lastPrimeFound - relJ);
res += (long) (c1) * (c2) + (c1 - 1);
}
}
return res;
}
private Set<Integer> sieve() {
Set<Integer> res = new TreeSet<>();
int N = (int)1e6 + 1;
boolean[] composite = new boolean[N];
for(int i = 2; i * i < N; i++) {
if(composite[i]) {
continue;
}
for(int j = i * i; j < N; j+=i) {
composite[j] = true;
}
}
for(int i = 2; i < N; i++) {
if(!composite[i]) {
res.add(i);
}
}
return res;
}
public static void main(String[] args) throws Exception {
Reader r = new InputStreamReader(System.in);
Writer w = new PrintWriter(System.out);
c1609C sol = new c1609C(r, w);
sol.takeInput();
sol.closeIO();
}
public static class SimpleInputReader implements AutoCloseable{
private BufferedReader reader;
private StringTokenizer tokenizer;
public SimpleInputReader(Reader reader) {
this.reader = new BufferedReader(reader);
}
public String readLine() throws IOException {
return reader.readLine();
}
public String nextString() throws IOException {
while(tokenizer == null || !tokenizer.hasMoreTokens()) {
String nextLine = readLine();
if(nextLine == null) {
throw new IOException("no more lines left!");
}
tokenizer = new StringTokenizer(nextLine);
}
return tokenizer.nextToken();
}
public Integer nextInt() throws IOException {
return Integer.parseInt(nextString());
}
public Long nextLong() throws IOException {
return Long.parseLong(nextString());
}
public float nextFloat() throws IOException {
return Float.parseFloat(nextString());
}
public double nextDouble() throws IOException {
return Double.parseDouble(nextString());
}
@Override
public void close() throws Exception {
reader.close();
}
}
public static class SimpleOutputWriter implements AutoCloseable{
private final BufferedWriter bufferedWriter;
public SimpleOutputWriter(Writer writer) {
bufferedWriter = new BufferedWriter(writer);
}
public void writeLine(Object s, Object... extras) throws IOException {
writeTokens(s, (Object[]) extras);
bufferedWriter.newLine();
}
public void newLine() throws IOException {
bufferedWriter.newLine();
}
public void flush() throws IOException {
bufferedWriter.flush();
}
public void writeTokens(Object o, Object... extras) throws IOException {
bufferedWriter.write(o.toString());
for (Object extra : extras) {
bufferedWriter.write(extra.toString());
}
}
@Override
public void close() throws Exception {
bufferedWriter.close();
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | e210700004c89acfacdb0728a159616c | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.util.*;
public class cp {
static int mod=(int)1e9+7;
// static Reader sc=new Reader();
static FastReader sc=new FastReader(System.in);
static int[] sp;
static int size=(int)1e6;
static int[] arInt;
static long[] arLong;
public static void main(String[] args) throws IOException {
long tc=sc.nextLong();
// Scanner sc=new Scanner(System.in);
// int tc=1;
primeSet=new HashSet<>();
sieveOfEratosthenes((int)1e6+5);
while(tc-->0)
{
int n=sc.nextInt();
int e=sc.nextInt();
int arr[]=new int[n];
ArrayList<Integer> index=new ArrayList<>();
boolean ones[]=new boolean[n];
for (int i = 0; i < arr.length; i++) {
arr[i]=sc.nextInt();
if(primeSet.contains(arr[i]))
index.add(i);
if(arr[i]==1)
ones[i]=true;
}
int dp1[]=new int[n];
int dp2[]=new int[n];
for(int i=0;i<n;i++)
{
if(i-e>=0 && ones[i])
{
dp1[i]=dp1[i-e]+1;
}
else if (ones[i]) {
dp1[i]=1;
}
}
for(int i=n-1;i>=0;i--)
{
if(i+e<n && ones[i])
{
dp2[i]=dp2[i+e]+1;
}
else if (ones[i]) {
dp2[i]=1;
}
}
long cnt=0;
for(Integer idx:index)
{
long left=idx-e>=0?(long)dp1[idx-e]:0L;
long right=idx+e<n?(long)dp2[idx+e]:0L;
cnt+=left+right+(left*right);
}
out.println(cnt);
}
out.flush();
out.close();
System.gc();
}
/*
...SOLUTION ENDS HERE...........SOLUTION ENDS HERE...
*/
static void arrInt(int n) throws IOException
{
arInt=new int[n];
for (int i = 0; i < arInt.length; i++) {
arInt[i]=sc.nextInt();
}
}
static void arrLong(int n) throws IOException
{
arLong=new long[n];
for (int i = 0; i < arLong.length; i++) {
arLong[i]=sc.nextLong();
}
}
static ArrayList<Integer> add(int id,int c)
{
ArrayList<Integer> newArr=new ArrayList<>();
for(int i=0;i<id;i++)
newArr.add(arInt[i]);
newArr.add(c);
for(int i=id;i<arInt.length;i++)
{
newArr.add(arInt[i]);
}
return newArr;
}
// function to find last index <= y
static int upper(ArrayList<Integer> arr, int n, int y)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr.get(mid) <= y)
l = mid + 1;
else
h = mid - 1;
}
return h;
}
static int lower(ArrayList<Integer> arr, int n, int x)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr.get(mid) >= x)
h = mid - 1;
else
l = mid + 1;
}
return l;
}
static int N = 501;
// Array to store inverse of 1 to N
static long[] factorialNumInverse = new long[N + 1];
// Array to precompute inverse of 1! to N!
static long[] naturalNumInverse = new long[N + 1];
// Array to store factorial of first N numbers
static long[] fact = new long[N + 1];
// Function to precompute inverse of numbers
public static void InverseofNumber(int p)
{
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for(int i = 2; i <= N; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] *
(long)(p - p / i) % p;
}
// Function to precompute inverse of factorials
public static void InverseofFactorial(int p)
{
factorialNumInverse[0] = factorialNumInverse[1] = 1;
// Precompute inverse of natural numbers
for(int i = 2; i <= N; i++)
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p;
}
// Function to calculate factorial of 1 to N
public static void factorial(int p)
{
fact[0] = 1;
// Precompute factorials
for(int i = 1; i <= N; i++)
{
fact[i] = (fact[i - 1] * (long)i) % p;
}
}
// Function to return nCr % p in O(1) time
public static long Binomial(int N, int R, int p)
{
// n C r = n!*inverse(r!)*inverse((n-r)!)
long ans = ((fact[N] * factorialNumInverse[R]) %
p * factorialNumInverse[N - R]) % p;
return ans;
}
static String tr(String s)
{
int now = 0;
while (now + 1 < s.length() && s.charAt(now)== '0')
++now;
return s.substring(now);
}
static ArrayList<Integer> ans;
static void dfs(int node,Graph gg,int cnt,int k,ArrayList<Integer> temp)
{
if(cnt==k)
return;
for(Integer each:gg.list[node])
{
if(each==0)
{
temp.add(each);
ans=new ArrayList<>(temp);
temp.remove(temp.size()-1);
continue;
}
temp.add(each);
dfs(each,gg,cnt+1,k,temp);
temp.remove(temp.size()-1);
}
return;
}
static boolean isPrime(long n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static ArrayList<Integer> commDiv(int a, int b)
{
// find gcd of a, b
int n = gcd(a, b);
// Count divisors of n.
ArrayList<Integer> Div=new ArrayList<>();
for (int i = 1; i <= Math.sqrt(n); i++) {
// if 'i' is factor of n
if (n % i == 0) {
// check if divisors are equal
if (n / i == i)
Div.add(i);
else
{
Div.add(i);
Div.add(n/i);
}
}
}
return Div;
}
static HashSet<Integer> factors(int x)
{
HashSet<Integer> a=new HashSet<Integer>();
for(int i=2;i*i<=x;i++)
{
if(x%i==0)
{
a.add(i);
a.add(x/i);
}
}
return a;
}
static class Node
{
int vertex;
HashSet<Node> adj;
boolean rem;
Node(int ver)
{
vertex=ver;
rem=false;
adj=new HashSet<Node>();
}
@Override
public String toString()
{
return vertex+" ";
}
}
static class Tuple{
int a;
int b;
int c;
public Tuple(int a,int b,int c) {
this.a=a;
this.b=b;
this.c=c;
}
}
//function to find prime factors of n
static HashMap<Long,Long> findFactors(long n2)
{
HashMap<Long,Long> ans=new HashMap<>();
if(n2%2==0)
{
ans.put(2L, 0L);
// cnt++;
while((n2&1)==0)
{
n2=n2>>1;
ans.put(2L, ans.get(2L)+1);
//
}
}
for(long i=3;i*i<=n2;i+=2)
{
if(n2%i==0)
{
ans.put((long)i, 0L);
// cnt++;
while(n2%i==0)
{
n2=n2/i;
ans.put((long)i, ans.get((long)i)+1);
}
}
}
if(n2!=1)
{
ans.put(n2, ans.getOrDefault(n2, (long) 0)+1);
}
return ans;
}
//fenwick tree implementaion
static class fwt
{
int n;
long BITree[];
fwt(int n)
{
this.n=n;
BITree=new long[n+1];
}
fwt(int arr[], int n)
{
this.n=n;
BITree=new long[n+1];
for(int i = 0; i < n; i++)
updateBIT(n, i, arr[i]);
}
long getSum(int index)
{
long sum = 0;
index = index + 1;
while(index>0)
{
sum += BITree[index];
index -= index & (-index);
}
return sum;
}
void updateBIT(int n, int index,int val)
{
index = index + 1;
while(index <= n)
{
BITree[index] += val;
index += index & (-index);
}
}
void print()
{
for(int i=0;i<n;i++)
out.print(getSum(i)+" ");
out.println();
}
}
class sparseTable{
int n;
long[][]dp;
int log2[];
int P;
void buildTable(long[] arr)
{
n=arr.length;
P=(int)Math.floor(Math.log(n)/Math.log(2));
log2=new int[n+1];
log2[0]=log2[1]=0;
for(int i=2;i<=n;i++)
{
log2[i]=log2[i/2]+1;
}
dp=new long[P+1][n];
for(int i=0;i<n;i++)
{
dp[0][i]=arr[i];
}
for(int p=1;p<=P;p++)
{
for(int i=0;i+(1<<p)<=n;i++)
{
long left=dp[p-1][i];
long right=dp[p-1][i+(1<<(p-1))];
dp[p][i]=Math.max(left, right);
}
}
}
long maxQuery(int l,int r)
{
int len=r-l+1;
int p=(int)Math.floor(log2[len]);
long left=dp[p][l];
long right=dp[p][r-(1<<p)+1];
return Math.max(left, right);
}
}
//Function to find number of set bits
static int setBitNumber(long n)
{
if (n == 0)
return 0;
int msb = 0;
n = n / 2;
while (n != 0) {
n = n / 2;
msb++;
}
return msb;
}
static int getFirstSetBitPos(long n)
{
return (int)((Math.log10(n & -n)) / Math.log10(2)) + 1;
}
static ArrayList<Integer> primes;
static HashSet<Integer> primeSet;
static boolean prime[];
static void sieveOfEratosthenes(int n)
{
// Create a boolean array
// "prime[0..n]" and
// initialize all entries
// it as true. A value in
// prime[i] will finally be
// false if i is Not a
// prime, else true.
prime= new boolean[n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
primeSet.add(i);
}
}
static long mod(long a, long b) {
long c = a % b;
return (c < 0) ? c + b : c;
}
static void swap(long arr[],int i,int j)
{
long temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
static boolean util(int a,int b,int c)
{
if(b>a)util(b, a, c);
while(c>=a)
{
c-=a;
if(c%b==0)
return true;
}
return (c%b==0);
}
static void flag(boolean flag)
{
out.println(flag ? "YES" : "NO");
out.flush();
}
static void print(int a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print(long a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print_int(ArrayList<Integer> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void print_long(ArrayList<Long> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void printYesNo(boolean condition)
{
if (condition) {
out.println("YES");
}
else {
out.println("NO");
}
}
static int LowerBound(int a[], int x)
{ // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
static int lowerIndex(int arr[], int n, int x)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr[mid] >= x)
h = mid - 1;
else
l = mid + 1;
}
return l;
}
// function to find last index <= y
static int upperIndex(int arr[], int n, int y)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr[mid] <= y)
l = mid + 1;
else
h = mid - 1;
}
return h;
}
static int upperIndex(long arr[], int n, long y)
{
int l = 0, h = n - 1;
while (l <= h)
{
int mid = (l + h) / 2;
if (arr[mid] <= y)
l = mid + 1;
else
h = mid - 1;
}
return h;
}
static int UpperBound(int a[], int x)
{// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
static int UpperBound(long a[], long x)
{// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
static class DisjointUnionSets
{
int[] rank, parent;
int n;
// Constructor
public DisjointUnionSets(int n)
{
rank = new int[n];
parent = new int[n];
this.n = n;
makeSet();
}
// Creates n sets with single item in each
void makeSet()
{
for (int i = 0; i < n; i++)
parent[i] = i;
}
int find(int x)
{
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
// Unites the set that includes x and the set
// that includes x
void union(int x, int y)
{
int xRoot = find(x), yRoot = find(y);
if (xRoot == yRoot)
return;
if (rank[xRoot] < rank[yRoot])
parent[xRoot] = yRoot;
else if (rank[yRoot] < rank[xRoot])
parent[yRoot] = xRoot;
else // if ranks are the same
{
parent[yRoot] = xRoot;
rank[xRoot] = rank[xRoot] + 1;
}
// if(xRoot!=yRoot)
// parent[y]=x;
}
int connectedComponents()
{
int cnt=0;
for(int i=0;i<n;i++)
{
if(parent[i]==i)
cnt++;
}
return cnt;
}
}
static class Graph
{
int v;
ArrayList<Integer> list[];
Graph(int v)
{
this.v=v;
list=new ArrayList[v+1];
for(int i=1;i<=v;i++)
list[i]=new ArrayList<Integer>();
}
void addEdge(int a, int b)
{
this.list[a].add(b);
}
}
// static class GraphMap{
// Map<String,ArrayList<String>> graph;
// GraphMap() {
// // TODO Auto-generated constructor stub
// graph=new HashMap<String,ArrayList<String>>();
//
// }
// void addEdge(String a,String b)
// {
// if(graph.containsKey(a))
// this.graph.get(a).add(b);
// else {
// this.graph.put(a, new ArrayList<>());
// this.graph.get(a).add(b);
// }
// }
// }
// static void dfsMap(GraphMap g,HashSet<String> vis,String src,int ok)
// {
// vis.add(src);
//
// if(g.graph.get(src)!=null)
// {
// for(String each:g.graph.get(src))
// {
// if(!vis.contains(each))
// {
// dfsMap(g, vis, each, ok+1);
// }
// }
// }
//
// cnt=Math.max(cnt, ok);
// }
static double sum[];
static long cnt;
// static void DFS(Graph g, boolean[] visited, int u)
// {
// visited[u]=true;
//
// for(int i=0;i<g.list[u].size();i++)
// {
// int v=g.list[u].get(i);
//
// if(!visited[v])
// {
// cnt1=cnt1*2;
// DFS(g, visited, v);
//
// }
//
// }
//
//
// }
static class Pair implements Comparable<Pair>
{
int x;
int y;
Pair(int x,int y)
{
this.x=x;
this.y=y;
}
@Override
public int compareTo(Pair o) {
// TODO Auto-generated method stub
return this.x-o.x;
}
}
static long sum_array(int a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static long sum_array(long a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static void sort(int[] a)
{
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void sort(long[] a)
{
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void reverse_array(int a[])
{
int n=a.length;
int i,t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static void reverse_array(long a[])
{
int n=a.length;
int i; long t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
// static long modInverse(long a, long m)
// {
// long g = gcd(a, m);
//
// return power(a, m - 2, m);
//
// }
static long power(long x, long y)
{
long res = 1;
x = x % mod;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % mod;
y = y >> 1; // y = y/2
x = (x * x) % mod;
}
return res;
}
static int power(int x, int y)
{
int res = 1;
x = x % mod;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % mod;
y = y >> 1; // y = y/2
x = (x * x) % mod;
}
return res;
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
//cnt+=a/b;
return gcd(b%a,a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static PrintWriter out=new PrintWriter(System.out);
static int int_max=Integer.MAX_VALUE;
static int int_min=Integer.MIN_VALUE;
static long long_max=Long.MAX_VALUE;
static long long_min=Long.MIN_VALUE;
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 4f150adcb8c8af944e412e6f6e77d8d0 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
int m = 1000_000;
boolean prime[] = new boolean[m + 1];
for (int i = 0; i <= m; i++)
prime[i] = true;
for (int p = 2; p * p <= m; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= m; i += p)
prime[i] = false;
}
}
int inc = 0;
while (t-- > 0) {
inc++;
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int e = Integer.parseInt(st.nextToken());
st = new StringTokenizer(br.readLine());
int arr[] = new int[n];
int onec[] = new int[n];
for (int i = 0; i < n; i++) {
int e2 = Integer.parseInt(st.nextToken());
arr[i] = e2;
if (e2 == 1)
onec[i]++;
if (i >= e) {
if (arr[i - e] == 1) {
onec[i] += onec[i - e];
}
}
}
// if(inc == 119)
// output.write(Arrays.toString(arr) + " " + e + "\n");
// output.write(Arrays.toString(onec) + "\n");
long res = 0;
for (int i = n - 1; i >= n - e; i--) {
int tempc = 0;
for (int j = i; j >= 0; j -= e) {
if (arr[j] == 1) {
tempc++;
continue;
}
else {
if (!prime[arr[j]])
{
tempc = 0;
continue;
}
// output.write("for i = " + i + " j = " + j + " temp = " + tempc + " onec[j] = " + onec[j] + " res before = " + res + " \n");
res += tempc;
res += onec[j];
if (onec[j] > 0 && tempc > 0)
res += ((long)tempc * (long)onec[j]);
tempc = 0;
// output.write("for i = " + i + " j = " + j + " temp = " + tempc + " onec[j] = " + onec[j] + " res before = " + res + " \n");
}
}
}
output.write(res + "\n");
// int k = Integer.parseInt(st.nextToken());
// char arr[] = br.readLine().toCharArray();
// output.write();
// int n = Integer.parseInt(st.nextToken());
// HashMap<Character, Integer> map = new HashMap<Character, Integer>();
// if
// output.write("YES\n");
// else
// output.write("NO\n");
// long a = Long.parseLong(st.nextToken());
// long b = Long.parseLong(st.nextToken());
// if(flag == 1)
// output.write("NO\n");
// else
// output.write("YES\n" + x + " " + y + " " + z + "\n");
// output.write(n+ "\n");
}
output.flush();
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 616a6a95ee22ccfceb85fa8d313e921a | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.lang.*;
import java.util.*;
public class Main {
public static int mod = (int) 1e9 + 7;
public static int imax = Integer.MAX_VALUE;
public static long lmax = Long.MAX_VALUE;
public static PrintWriter out = new PrintWriter(System.out);
// **** -----> Disjoint Set Union(DSU) Start **********
public static int findPar(int node, int[] parent) {
if (parent[node] == node)
return node;
return parent[node] = findPar(parent[node], parent);
}
public static void union(int u, int v, int[] rank, int[] parent) {
u = findPar(u, parent);
v = findPar(v, parent);
if (rank[u] < rank[v])
parent[u] = v;
else if (rank[u] > rank[v])
parent[v] = u;
else {
parent[u] = v;
rank[v]++;
}
}
// **** DSU Ends ***********
public static String toBinary(int decimal) {
StringBuilder sb = new StringBuilder();
while (decimal > 0) {
sb.append(decimal % 2);
decimal = decimal / 2;
}
// System.out.println(sb.reverse().toString());
return sb.reverse().toString();
}
public static boolean isPalindrome(String s) {
int i = 0, j = s.length() - 1;
while (i < j) {
if (s.charAt(i) != s.charAt(j))
return false;
i++;
j--;
}
return true;
}
public static boolean isPalindrome(int[] arr) {
int i = 0, j = arr.length - 1;
while (i < j) {
if (arr[i] != arr[j])
return false;
}
return true;
}
static long pow(long x, long y) {
long res = 1; // Initialize result
x = x % mod; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % mod;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % mod;
}
return res;
}
public static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
public static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
public static void sort(long[] a) {
Random rnd = new Random();
for (int i = 0; i < a.length; i++) {
int pos = i + rnd.nextInt(a.length - i);
long temp = a[i];
a[i] = a[pos];
a[pos] = temp;
}
Arrays.sort(a);
}
static void reverse(int a[]) {
int i, k, t, n = a.length;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
public static void sort(int[] a) {
Random rnd = new Random();
for (int i = 0; i < a.length; i++) {
int pos = i + rnd.nextInt(a.length - i);
int temp = a[i];
a[i] = a[pos];
a[pos] = temp;
}
Arrays.sort(a);
}
public static void revSort(int[] a) {
Random rnd = new Random();
for (int i = 0; i < a.length; i++) {
int pos = i + rnd.nextInt(a.length - i);
int temp = a[i];
a[i] = a[pos];
a[pos] = temp;
}
Arrays.sort(a);
reverse(a);
}
public static long LCM(long a, long b) {
if (a > b) {
long t = a;
a = b;
b = t;
}
a /= gcd(a, b);
return (a * b);
}
static int findMax(int[] a, int left, int right) {
int res = left;
int max = a[left];
for (int i = left + 1; i <= right; i++) {
if (a[i] > max) {
max = a[i];
res = i;
}
}
return res;
}
public static long findClosest(long arr[], long target) {
int n = arr.length;
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
if (target < arr[mid]) {
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1], arr[mid], target);
j = mid;
} else {
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid], arr[mid + 1], target);
i = mid + 1;
}
}
return arr[mid];
}
public static long getClosest(long val1, long val2, long target) {
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
public static int findClosest(int arr[], int target) {
int n = arr.length;
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
if (target < arr[mid]) {
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1], arr[mid], target);
j = mid;
} else {
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid], arr[mid + 1], target);
i = mid + 1;
}
}
return arr[mid];
}
public static int getClosest(int val1, int val2, int target) {
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
public static String reverse(String str) {
String nstr = "";
char ch;
for (int i = 0; i < str.length(); i++) {
ch = str.charAt(i); // extracts each character
nstr = ch + nstr; // adds each character in front of the existing string
}
return nstr;
}
public static boolean isPrime(int n) {
if (n <= 1)
return false;
else if (n == 2)
return true;
else if (n % 2 == 0)
return false;
for (int i = 3; i <= Math.sqrt(n); i += 2) {
if (n % i == 0)
return false;
}
return true;
}
// static boolean isPrime(int n)
// {
// // Corner cases
// if (n <= 1)
// return false;
// if (n <= 3)
// return true;
// // This is checked so that we can skip
// // middle five numbers in below loop
// if (n % 2 == 0 || n % 3 == 0)
// return false;
// for (int i = 5; i * i <= n; i = i + 6)
// if (n % i == 0 || n % (i + 2) == 0)
// return false;
// return true;
// }
static HashMap<Integer, Integer> map;
public static boolean solve(int root, HashMap<Integer, ArrayList<Integer>> tree) {
if(!tree.containsKey(root)) return true;
for(int nbr : tree.get(root)){
if(map.get(nbr) < map.get(root)){
return false;
}
if(!solve(nbr, tree)) return false;
}
return true;
}
/*
* Available Functions :-
*
* toBinary(int), isPalindrome(String), isPalindrome(int[]), pow(long, long),
* gcd(int, int), gcd(long, long), sort(long[]), sort(int[]), reverse(int[]),
* sort(long[]), reverse(int[]), revSort(int[]) LCM(long, long), DSU
* findMax(int[],int,int), findClosest(long[], long),
*
*/
// NOTE :- To use DSU first call makeSet()
public static void main(String[] args) throws java.lang.Exception {
if (System.getProperty("ONLINE_JUDGE") == null) {
PrintStream ps = new PrintStream(new File("output.txt"));
System.setOut(ps);
}
FastScanner sc = new FastScanner("input.txt");
int T = sc.nextInt();
while (T-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
int[] arr = sc.readArray(n);
int[] ansL = new int[n];
int[] ansR = new int[n];
for(int i = n-1; i >= 0; i--){
if(arr[i] != 1) continue;
if(i <= n-e-1 && arr[i+e] == 1) ansR[i] += ansR[i+e]+1;
else ansR[i] = 1;
}
for(int i = 0; i < n; i++){
if(arr[i] != 1) continue;
if(i >= e && arr[i-e] == 1) ansL[i] += ansL[i-e]+1;
else ansL[i] = 1;
}
// for(int i = 0; i < n; i++){
// System.out.print(ansL[i] + " ");
// }
// System.out.println();
// for(int i = 0; i < n; i++){
// System.out.print(ansR[i] + "");
// }
// System.out.println();
long res = 0;
for(int i = 0; i < n; i++){
if(arr[i] != 1 && isPrime(arr[i])){
int L = 0, R = 0;
if(i >= e && arr[i-e] == 1) L = ansL[i-e];
if(i+e < n && arr[i+e] == 1) R = ansR[i+e];
res += (long)(L+1)*(R+1)-1;
}
}
System.out.println(res);
}
out.close();
}
public static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(String s) {
try {
br = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String nextToken() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
if (st == null || !st.hasMoreTokens()) {
try {
return br.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return st.nextToken("\n");
}
public String[] readStringArray(int n) {
String[] a = new String[n];
for (int i = 0; i < n; i++) {
a[i] = next();
}
return a;
}
int nextInt() {
return Integer.parseInt(nextToken());
}
String next() {
return nextToken();
}
long nextLong() {
return Long.parseLong(nextToken());
}
double nextDouble() {
return Double.parseDouble(nextToken());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
int[][] read2dArray(int n, int m) {
int[][] a = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
a[i][j] = nextInt();
}
}
return a;
}
long[][] read2dlongArray(int n, int m) {
long[][] a = new long[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
a[i][j] = nextLong();
}
}
return a;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 961b6d5ce1570009d82d91eff7c249d8 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
public class Sol {
public static void sieveOfEratosthenes(boolean prime[])
{
int n=prime.length;
for (int i = 0; i < n; i++)
prime[i] = true;
for (int p = 2; p * p < n; p++){
if (prime[p] == true)
{
for (int i = p * p; i < n; i += p)
prime[i] = false;
}
}
}
public static void main(String[] args) throws IOException {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
boolean prime[]=new boolean[1000001];
sieveOfEratosthenes(prime);
while(t--> 0) {
int n=sc.nextInt();
int e=sc.nextInt();
int arr[]=new int[n];
ArrayList<Integer> a=new ArrayList<>();
ArrayList<Integer> b=new ArrayList<>();
for(int i=0;i<n;i++) {
arr[i]=sc.nextInt();
if(arr[i]==1) a.add(i);
else if(prime[arr[i]]) b.add(i);
}
if(b.size()==0 || a.size()==0) System.out.println(0);
else {
int dp[]=new int[n];
for(int i=0;i<a.size();i++) {
dp[a.get(i)]+=a.get(i)>=e?dp[a.get(i)-e]+1:1;
}
int dp2[]=new int[n];
for(int i=a.size()-1;i>=0;i--) {
dp2[a.get(i)]+=a.get(i)+e<n?dp2[a.get(i)+e]+1:1;
}
long ans=0;
for(int i=0;i<b.size();i++) {
long h=0l;
if(b.get(i)>=e) {
h=dp[b.get(i)-e];
}
long y=0l;
if(b.get(i)+e<n) {
y=dp2[b.get(i)+e];
}
if(h==0 || y==0) ans=ans+y+h;
else {
ans=ans+(h+1)*(y+1)-1;
}
}
System.out.println(ans);
}
}
}} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 4106afa86a660c9186ad686d7c3e205a | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class Main {
// Graph
// prefix sums
//inputs
public static void main(String args[])throws Exception{
Input sc=new Input();
precalculates p=new precalculates();
StringBuilder sb=new StringBuilder();
long MM=1000005;
boolean prime[]=new boolean[(int)MM];
Arrays.fill(prime,true);
prime[0]=false;
prime[1]=false;
for(long i=2;i<MM;i++){
if(prime[(int)i]){
for(long j=i*i;j<MM;j+=i){
prime[(int)j]=false;
}
}
}
// for(int i=0;i<100;i++){
// if(prime[i])
// System.out.println(i);
// }
int t=sc.readInt();
for(int f=0;f<t;f++){
int d[]=sc.readArray();
int n=d[0],e=d[1];
int a[]=sc.readArray();
int dpl[]=new int[n];
int dpr[]=new int[n];
for(int i=0;i<e-1;i++){
if(a[i]==1){
dpl[i]=1;
}
}
for(int i=e-1;i<n;i++){
if(a[i]==1){
if(i-e>=0 && a[i-e]==1){
dpl[i]=dpl[i-e]+1;
}else
dpl[i]=1;
}
}
// for(int i=0;i<n;i++){
// if(prime[a[i]])
// System.out.println(a[i]);
// }
for(int i=n-e+1;i<n;i++){
if(a[i]==1)
dpr[i]=1;
}
for(int i=n-e;i>=0;i--){
if(a[i]==1){
if(i+e<n && a[i+e]==1){
dpr[i]=dpr[i+e]+1;
}else
dpr[i]=1;
}
}
// for(int i=0;i<n;i++){
// System.out.print(dpl[i]+" ");
// }
// System.out.println();
// for(int i=0;i<n;i++){
// System.out.print(dpr[i]+" ");
// }
// System.out.println();
long ans=0;
for(int i=0;i<n;i++){
if(prime[a[i]]){
long left=0;
long right=0;
if(i-e>=0){
left=dpl[i-e];
}
if(i+e<n){
right=dpr[i+e];
}
if(left==0 && right==0){
}else if(left==0 && right!=0){
ans+=right;
}else if(left!=0 && right==0){
ans+=left;
}else{
long m=left+right;
long val=m;
//System.out.println(left+" "+right);
ans+=(left*(right+1)+right);
}
}
}
sb.append(ans+"\n");
}
System.out.print(sb);
}
}
class Input{
BufferedReader br;
StringTokenizer st;
Input(){
br=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
}
public int[] readArray() throws Exception{
st=new StringTokenizer(br.readLine());
int a[]=new int[st.countTokens()];
for(int i=0;i<a.length;i++){
a[i]=Integer.parseInt(st.nextToken());
}
return a;
}
public long[] readArrayLong() throws Exception{
st=new StringTokenizer(br.readLine());
long a[]=new long[st.countTokens()];
for(int i=0;i<a.length;i++){
a[i]=Long.parseLong(st.nextToken());
}
return a;
}
public int readInt() throws Exception{
st=new StringTokenizer(br.readLine());
return Integer.parseInt(st.nextToken());
}
public long readLong() throws Exception{
st=new StringTokenizer(br.readLine());
return Long.parseLong(st.nextToken());
}
public String readString() throws Exception{
return br.readLine();
}
public int[][] read2dArray(int n,int m)throws Exception{
int a[][]=new int[n][m];
for(int i=0;i<n;i++){
st=new StringTokenizer(br.readLine());
for(int j=0;j<m;j++){
a[i][j]=Integer.parseInt(st.nextToken());
}
}
return a;
}
}
class precalculates{
public int[] prefixSumOneDimentional(int a[]){
int n=a.length;
int dp[]=new int[n];
for(int i=0;i<n;i++){
if(i==0)
dp[i]=a[i];
else
dp[i]=dp[i-1]+a[i];
}
return dp;
}
public int[] postSumOneDimentional(int a[]) {
int n = a.length;
int dp[] = new int[n];
for (int i = n - 1; i >= 0; i--) {
if (i == n - 1)
dp[i] = a[i];
else
dp[i] = dp[i + 1] + a[i];
}
return dp;
}
public int[][] prefixSum2d(int a[][]){
int n=a.length;int m=a[0].length;
int dp[][]=new int[n+1][m+1];
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
dp[i][j]=a[i-1][j-1]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
}
}
return dp;
}
public long pow(long a,long b){
long mod=1000000007;
long ans=0;
if(b<=0)
return 1;
if(b%2==0){
ans=pow(a,b/2)%mod;
return ((ans%mod)*(ans%mod))%mod;
}else{
return ((a%mod)*(ans%mod))%mod;
}
}
}
class GraphInteger{
HashMap<Integer,vertex> vtces;
class vertex{
HashMap<Integer,Integer> children;
public vertex(){
children=new HashMap<>();
}
}
public GraphInteger(){
vtces=new HashMap<>();
}
public void addVertex(int a){
vtces.put(a,new vertex());
}
public void addEdge(int a,int b,int cost){
if(!vtces.containsKey(a)){
vtces.put(a,new vertex());
}
if(!vtces.containsKey(b)){
vtces.put(b,new vertex());
}
vtces.get(a).children.put(b,cost);
// vtces.get(b).children.put(a,cost);
}
public boolean isCyclicDirected(){
boolean isdone[]=new boolean[vtces.size()+1];
boolean check[]=new boolean[vtces.size()+1];
for(int i=1;i<=vtces.size();i++) {
if (!isdone[i] && isCyclicDirected(i,isdone, check)) {
return true;
}
}
return false;
}
private boolean isCyclicDirected(int i,boolean isdone[],boolean check[]){
if(check[i])
return true;
if(isdone[i])
return false;
check[i]=true;
isdone[i]=true;
Set<Integer> set=vtces.get(i).children.keySet();
for(Integer ii:set){
if(isCyclicDirected(ii,isdone,check))
return true;
}
check[i]=false;
return false;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | e7584c50c4e81502fe832e476828fa8b | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader obj = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int len = obj.nextInt();
int N = 1000000;
int[] lp = new int[N + 1];
Vector<Integer> pr = new Vector<>();
for (int i = 2; i <= N; ++i) {
if (lp[i] == 0) {
lp[i] = i;
pr.add(i);
}
for (int j = 0; j < pr.size() && pr.get(j) <= lp[i] && i * pr.get(j) <= N; ++j) {
lp[i * pr.get(j)] = pr.get(j);
}
}
while (len-- != 0) {
int n = obj.nextInt();
int e = obj.nextInt();
int[] num = new int[n];
long c = 0;
for (int i = 0; i < n; i++) num[i] = obj.nextInt();
for (int i = 0; i < n; i++) {
int a = 0, b = 0;
if (lp[num[i]] == num[i]) {
int s = i - e;
int f = i + e;
while (s >= 0) {
if (num[s] == 1) a++;
else break;
s -= e;
}
while (f < n) {
if (num[f] == 1) b++;
else break;
f += e;
}
}
c += (long) a * (b + 1) + b;
}
out.println(c);
}
out.flush();
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7352bf4e37b47f628b00d3e190d5f034 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
public class weird_algrithm {
static BufferedWriter output = new BufferedWriter(
new OutputStreamWriter(System.out));
static BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
static int mod = 1000000007;
static String toReturn = "";
static int steps = Integer.MAX_VALUE;
static int maxlen = 1000005;
/*MATHEMATICS FUNCTIONS START HERE
MATHS
MATHS
MATHS
MATHS*/
static long gcd(long a, long b) {
if(b == 0) return a;
else return gcd(b, a % b);
}
static long powerMod(long x, long y, int mod) {
if(y == 0) return 1;
long temp = powerMod(x, y / 2, mod);
temp = ((temp % mod) * (temp % mod)) % mod;
if(y % 2 == 0) return temp;
else return ((x % mod) * (temp % mod)) % mod;
}
static long modInverse(long n, int p) {
return powerMod(n, p - 2, p);
}
static long nCr(int n, int r, int mod, long [] fact, long [] ifact) {
return ((fact[n] % mod) * ((ifact[r] * ifact[n - r]) % mod)) % mod;
}
static boolean isPrime(long a) {
if(a == 1) return false;
else if(a == 2 || a == 3 || a== 5) return true;
else if(a % 2 == 0 || a % 3 == 0) return false;
for(int i = 5; i * i <= a; i = i + 6) {
if(a % i == 0 || a % (i + 2) == 0) return false;
}
return true;
}
static int [] seive(int a) {
int [] toReturn = new int [a + 1];
for(int i = 0; i < a; i++) toReturn[i] = 1;
toReturn[0] = 0;
toReturn[1] = 0;
toReturn[2] = 1;
for(int i = 2; i * i <= a; i++) {
if(toReturn[i] == 0) continue;
for(int j = 2 * i; j <= a; j += i) toReturn[j] = 0;
}
return toReturn;
}
static long [] fact(int a) {
long [] arr = new long[a + 1];
arr[0] = 1;
for(int i = 1; i < a + 1; i++) {
arr[i] = (arr[i - 1] * i) % mod;
}
return arr;
}
/*MATHS
MATHS
MATHS
MATHS
MATHEMATICS FUNCTIONS END HERE */
/*SWAP FUNCTION START HERE
SWAP
SWAP
SWAP
SWAP
*/
static void swap(int i, int j, long[] arr) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, int[] arr) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, String [] arr) {
String temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int i, int j, char [] arr) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
/*SWAP
SWAP
SWAP
SWAP
SWAP FUNCTION END HERE*/
/*BINARY SEARCH METHODS START HERE
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
*/
static boolean BinaryCheck(long test, long [] arr, long health) {
for(int i = 0; i <= arr.length - 1; i++) {
if(i == arr.length - 1) health -= test;
else if(arr[i + 1] - arr[i] > test) {
health = health - test;
}else {
health = health - (arr[i + 1] - arr[i]);
}
if(health <= 0) return true;
}
return false;
}
static long binarySearchModified(long start1, long n, ArrayList<Long> arr, int a, long r) {
long start = start1, end = n, ans = -1;
while(start < end) {
long mid = (start + end) / 2;
//System.out.println(mid);
if(arr.get((int)mid) + arr.get(a) <= r && mid != start1) {
ans = mid;
start = mid + 1;
}else{
end = mid;
}
}
//System.out.println();
return ans;
}
static int binarySearch(int start, int end, long [] arr, long val) {
while(start < end) {
int mid = (int)Math.ceil((start + end) / 2.0);
//System.out.println(mid);
if(arr[mid] > val) end = mid - 1;
else start = mid;
}
return start;
}
/*BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
* BINARY SEARCH
BINARY SEARCH METHODS END HERE*/
/*RECURSIVE FUNCTION START HERE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
*/
static int recurse(int x, int y, int n, int steps1, Integer [][] dp) {
if(x > n || y > n) return 0;
if(dp[x][y] != null) {
return dp[x][y];
}
else if(x == n || y == n) {
return steps1;
}
return dp[x][y] = Math.max(recurse(x + y, y, n, steps1 + 1, dp), recurse(x, x + y, n, steps1 + 1, dp));
}
/*RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
* RECURSIVE
RECURSIVE FUNCTION END HERE*/
/*GRAPH FUNCTIONS START HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
* */
static class edge{
int from, to;
long weight;
public edge(int x, int y, long weight2) {
this.from = x;
this.to = y;
this.weight = weight2;
}
}
static class sort implements Comparator<pair>{
@Override
public int compare(pair o1, pair o2) {
// TODO Auto-generated method stub
return (int)o1.a - (int)o2.a;
}
}
static void addEdge(ArrayList<ArrayList<edge>> graph, int from, int to, long weight) {
edge temp = new edge(from, to, weight);
edge temp1 = new edge(to, from, weight);
graph.get(from).add(temp);
//graph.get(to).add(temp1);
}
static int ans = 0;
static int dfs(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, int [] childs) {
//System.out.println(graph.get(vertex).size());
if(visited[vertex]) return 0;
visited[vertex] = true;
int ans = 0;
for(int i = 0; i < graph.get(vertex).size(); i++) {
int temp = graph.get(vertex).get(i);
if(!visited[temp]) {
ans += dfs(graph, temp, visited, childs) + 1;
}
}
childs[vertex] = ans;
return ans;
}
static void topoSort(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, ArrayList<Integer> toReturn) {
if(visited[vertex]) return;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(!visited[graph.get(vertex).get(i)]) topoSort(graph, graph.get(vertex).get(i), visited, toReturn);
}
toReturn.add(vertex);
}
static boolean isCyclicDirected(ArrayList<ArrayList<Integer>> graph, int vertex, boolean [] visited, boolean [] reStack) {
if(reStack[vertex]) return true;
if(visited[vertex]) return false;
reStack[vertex] = true;
visited[vertex] = true;
for(int i = 0; i < graph.get(vertex).size(); i++) {
if(isCyclicDirected(graph, graph.get(vertex).get(i), visited, reStack)) return true;
}
reStack[vertex] = false;
return false;
}
static int e = 0;
static long mst(PriorityQueue<edge> pq, int nodes) {
long weight = 0;
while(!pq.isEmpty()) {
edge temp = pq.poll();
int x = parent(parent, temp.to);
int y = parent(parent, temp.from);
if(x != y) {
//System.out.println(temp.weight);
union(x, y, rank, parent);
weight += temp.weight;
e++;
}
}
return weight;
}
static void floyd(long [][] dist) { // to find min distance between two nodes
for(int k = 0; k < dist.length; k++) {
for(int i = 0; i < dist.length; i++) {
for(int j = 0; j < dist.length; j++) {
if(dist[i][j] > dist[i][k] + dist[k][j]) {
dist[i][j] = dist[i][k] + dist[k][j];
}
}
}
}
}
static void dijkstra(ArrayList<ArrayList<edge>> graph, long [] dist, int src) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MAX_VALUE / 2;
dist[src] = 0;
boolean visited[] = new boolean[dist.length];
PriorityQueue<pair> pq = new PriorityQueue<>(new sort());
pq.add(new pair(src, 0));
while(!pq.isEmpty()) {
pair temp = pq.poll();
int index = (int)temp.a;
for(int i = 0; i < graph.get(index).size(); i++) {
if(dist[graph.get(index).get(i).to] > dist[index] + graph.get(index).get(i).weight) {
dist[graph.get(index).get(i).to] = dist[index] + graph.get(index).get(i).weight;
pq.add(new pair(graph.get(index).get(i).to, graph.get(index).get(i).weight));
}
}
}
}
static int parent1 = -1;
static boolean ford(ArrayList<ArrayList<edge>> graph1, ArrayList<edge> graph, long [] dist, int src, int [] parent) {
for(int i = 0; i < dist.length; i++) dist[i] = Long.MIN_VALUE / 2;
dist[src] = 0;
boolean hasNeg = false;
for(int i = 0; i < dist.length - 1; i++) {
for(int j = 0; j < graph.size(); j++) {
int from = graph.get(j).from;
int to = graph.get(j).to;
long weight = graph.get(j).weight;
if(dist[to] < dist[from] + weight) {
dist[to] = dist[from] + weight;
parent[to] = from;
}
}
}
for(int i = 0; i < graph.size(); i++) {
int from = graph.get(i).from;
int to = graph.get(i).to;
long weight = graph.get(i).weight;
if(dist[to] < dist[from] + weight) {
parent1 = from;
hasNeg = true;
/*
* dfs(graph1, parent1, new boolean[dist.length], dist.length - 1);
* //System.out.println(ans); dfs(graph1, 0, new boolean[dist.length], parent1);
*/
//System.out.println(ans);
if(ans == 2) break;
else ans = 0;
}
}
return hasNeg;
}
/*GRAPH FUNCTIONS END HERE
* GRAPH
* GRAPH
* GRAPH
* GRAPH
*/
/*disjoint Set START HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
static int [] rank;
static int [] parent;
static int parent(int [] parent, int x) {
if(parent[x] == x) return x;
else return parent[x] = parent(parent, parent[x]);
}
static boolean union(int x, int y, int [] rank, int [] parent) {
if(parent(parent, x) == parent(parent, y)) {
return true;
}
if(rank[x] > rank[y]) {
swap(x, y, rank);
}
rank[x] += rank[y];
parent[y] = x;
return false;
}
/*disjoint Set END HERE
* disjoint Set
* disjoint Set
* disjoint Set
* disjoint Set
*/
/*INPUT START HERE
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT
*/
static int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(sc.readLine());
}
static long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(sc.readLine());
}
static long [] inputLongArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
long [] toReturn = new long[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Long.parseLong(s[i]);
}
return toReturn;
}
static int max = 0;
static int [] inputIntArr() throws NumberFormatException, IOException{
String [] s = sc.readLine().split(" ");
//System.out.println(s.length);
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
toReturn[i] = Integer.parseInt(s[i]);
}
return toReturn;
}
/*INPUT
* INPUT
* INPUT
* INPUT
* INPUT
* INPUT END HERE
*/
static long [] preCompute(int level) {
long [] toReturn = new long[level];
toReturn[0] = 1;
toReturn[1] = 16;
for(int i = 2; i < level; i++) {
toReturn[i] = ((toReturn[i - 1] % mod) * (toReturn[i - 1] % mod)) % mod;
}
return toReturn;
}
static class pair{
long a;
long b;
long d;
public pair(long in, long y) {
this.a = in;
this.b = y;
this.d = 0;
}
}
static int [] nextGreaterBack(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = 0; i < s.length; i++) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int [] nextGreaterFront(char [] s) {
Stack<Integer> stack = new Stack<>();
int [] toReturn = new int[s.length];
for(int i = s.length - 1; i >= 0; i--) {
if(!stack.isEmpty() && s[stack.peek()] >= s[i]) {
stack.pop();
}
if(stack.isEmpty()) {
stack.push(i);
toReturn[i] = -1;
}else {
toReturn[i] = stack.peek();
stack.push(i);
}
}
return toReturn;
}
static int lps(String s) {
int max = 0;
int [] lps = new int[s.length()];
lps[0] = 0;
int j = 0;
for(int i = 1; i < lps.length; i++) {
j = lps[i - 1];
while(j > 0 && s.charAt(i) != s.charAt(j)) j = lps[j - 1];
if(s.charAt(i) == s.charAt(j)) {
lps[i] = j + 1;
max = Math.max(max, lps[i]);
}
}
return max;
}
static int [][] vectors = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
static String dir = "DRUL";
static boolean check(int i, int j, boolean [][] visited) {
if(i >= visited.length || j >= visited[0].length) return false;
if(i < 0 || j < 0) return false;
return true;
}
static void selectionSort(long arr[], long [] arr1, ArrayList<ArrayList<Integer>> ans)
{
int n = arr.length;
for (int i = 0; i < n-1; i++)
{
int min_idx = i;
for (int j = i+1; j < n; j++)
if (arr[j] < arr[min_idx])
min_idx = j;
else if(arr[j] == arr[min_idx]) {
if(arr1[j] < arr1[min_idx]) min_idx = j;
}
if(i == min_idx) {
continue;
}
ArrayList<Integer> p = new ArrayList<Integer>();
p.add(min_idx + 1);
p.add(i + 1);
ans.add(new ArrayList<Integer>(p));
swap(i, min_idx, arr);
swap(i, min_idx, arr1);
}
}
static int saved = Integer.MAX_VALUE;
static String ans1 = "";
public static boolean isValid(int x, int y, String [] mat) {
if(x >= mat.length || x < 0) return false;
if(y >= mat[0].length() || y < 0) return false;
return true;
}
static boolean recurse1(int i, int j, int [][] arr, int sum, Boolean [][][] dp) {
if(i == arr.length - 1 && j == arr[0].length - 1) {
if(sum + arr[i][j] == 0) return true;
return false;
}
if(i == arr.length) return false;
else if(j == arr[0].length) return false;
if(sum < 0 && dp[i][j][arr.length + arr[0].length + 1 + sum] != null) return dp[i][j][arr.length + arr[0].length + 1 + sum];
if(sum > 0 && dp[i][j][sum] != null) return dp[i][j][sum];
if(sum < 0) dp[i][j][arr.length + arr[0].length + 1 + sum] = recurse1(i + 1, j, arr, sum + arr[i][j], dp) || recurse1(i, j + 1, arr, sum + arr[i][j], dp);
return dp[i][j][sum] = recurse1(i + 1, j, arr, sum + arr[i][j], dp) || recurse1(i, j + 1, arr, sum + arr[i][j], dp);
}
public static void recurse3(ArrayList<Character> arr, int index, String s, int max, ArrayList<String> toReturn) {
if(s.length() == max) {
toReturn.add(s);
return;
}
if(index == arr.size()) return;
recurse3(arr, index + 1, s + arr.get(index), max, toReturn);
recurse3(arr, index + 1, s, max, toReturn);
}
/*
if(arr[i] > q) return Math.max(f(i + 1, q - 1) + 1, f(i + 1, q);
else return f(i + 1, q) + 1
*/
public static long recurse(long [] arr, int i, int n, Long [][] dp) {
if(n == 0) return 0;
if(i >= arr.length - 1) return Integer.MAX_VALUE / 2;
if(dp[i][n] != null) return dp[i][n];
if(arr[i - 1] < arr[i] && arr[i + 1] < arr[i]) {
return dp[i][n] = Math.min(recurse(arr, i + 2, n - 1, dp), recurse(arr, i + 1, n, dp));
}
long cost = Math.max(arr[i - 1], arr[i + 1]) - arr[i] + 1;
long taken = recurse(arr, i + 2, n - 1, dp);
long notTaken = recurse(arr, i + 1, n, dp);
return dp[i][n] = Math.min(taken + cost, notTaken);
}
static void solve() throws IOException {
int [] n = inputIntArr();
int [] arr = inputIntArr();
int k = n[1];
//int [] prime = seive(1000001);
long count = 0;
for(int i = 0; i < arr.length; i++) {
if(arr[i] != 1 && isPrime(arr[i]) ) {
int t = i;
long left = 0;
while(t + k < arr.length && arr[t + k] == 1) {
count++;
left++;
t += k;
}
t = i;
long right = 0;
while(t - k >= 0 && arr[t - k] == 1) {
count++;
right++;
t -= k;
}
count += left * right;
}
}
output.write(count + "\n");
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = Integer.parseInt(sc.readLine()); for(int i = 0; i < t; i++)
solve();
output.flush();
}
}
class TreeNode {
int val;
TreeNode next;
public TreeNode(int x, TreeNode y) {
this.val = x;
this.next = y;
}
}
/*
1
10
6 10 7 9 11 99 45 20 88 31
*/
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 21fd8a5d5bf630c206e52d9f386faf79 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
public class Main{
static boolean [] prime;
public static void sieveOfEratosthenes(int n)
{
prime = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
sieveOfEratosthenes((int) 1e6);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int e = sc.nextInt();
int [] arr = new int[n+1];
for (int i = 1; i <= n; i++) {
arr[i] = sc.nextInt();
}
long ans = 0;
for (int i = 1; i < n+1; i++) {
long l = 0;
long r = 0;
if(prime[arr[i]] && arr[i]!=1){
for (int j = i+e; j <= n ; j+=e) {
if(arr[j]==1)r++;
else break;
}
for (int j = i-e; j >=1 ; j-=e) {
if(arr[j]==1)l++;
else break;
}
ans += (r*l)+(r+l);
}
}
System.out.println(ans);
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 5018be9ea923f3adcc557895b3ea1453 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class C {
private static final int MAX = (int)1e6;
private boolean primes[] = new boolean[MAX + 1];
void go() {
// add code
int n = Reader.nextInt();
int e = Reader.nextInt();
int[] arr = new int[n];
long ans = 0;
int max = 0;
for(int i = 0; i < n; i++) {
arr[i] = Reader.nextInt();
}
for(int i = 0; i < n; i++) {
boolean pri = primes[arr[i]];
if(pri) {
long cntL = 0;
long cntR = 0;
// to right
for(int j = i + e; j < n; j += e) {
if(arr[j] == 1) {
cntR++;
} else {
break;
}
}
// to left
for(int j = i - e; j >= 0; j -= e) {
if(arr[j] == 1) {
cntL += cntR + 1;
} else {
break;
}
}
ans += cntL + cntR;
}
}
Writer.println(ans);
}
void solve() {
sieve();
for(int T = Reader.nextInt(); T > 0; T--) go();
}
void run() throws Exception {
Reader.init(System.in);
Writer.init(System.out);
solve();
Writer.close();
}
private void sieve() {
Arrays.fill(primes, true);
primes[0] = false;
primes[1] = false;
for(int i = 2; i * i <= MAX; i++) {
if(primes[i]) {
for(int j = i * i; j <= MAX; j += i) {
primes[j] = false;
}
}
}
}
public static void main(String[] args) throws Exception {
new C().run();
}
public static class Reader {
public static StringTokenizer st;
public static BufferedReader br;
public static void init(InputStream in) {
br = new BufferedReader(new InputStreamReader(in));
st = new StringTokenizer("");
}
public static String next() {
while(!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new InputMismatchException();
}
}
return st.nextToken();
}
public static int nextInt() {
return Integer.parseInt(next());
}
public static long nextLong() {
return Long.parseLong(next());
}
public static double nextDouble() {
return Double.parseDouble(next());
}
}
public static class Writer {
public static PrintWriter pw;
public static void init(OutputStream os) {
pw = new PrintWriter(new BufferedOutputStream(os));
}
public static void print(String s) {
pw.print(s);
}
public static void print(int x) {
pw.print(x);
}
public static void print(long x) {
pw.print(x);
}
public static void println(String s) {
pw.println(s);
}
public static void println(int x) {
pw.println(x);
}
public static void println(long x) {
pw.println(x);
}
public static void close() {
pw.close();
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 94d3f384fd487f876f5002ac1b7d560d | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
CComplexMarketAnalysis solver = new CComplexMarketAnalysis();
solver.solve(1, in, out);
out.close();
}
static class CComplexMarketAnalysis {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int[] s = new int[1000001];
s[1] = 1;
for (int i = 2; i <= 1000000; i++) {
if (s[i] == 1) continue;
for (int j = 2 * i; j <= 1000000; j += i) s[j] = 1;
}
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int e = in.nextInt();
int[] arr = in.nextIntArray(n);
int[] primepos = new int[n];
for (int i = n - 1; i >= 0; i--) {
if (s[arr[i]] == 0) primepos[i] = i;
else if (arr[i] == 1) {
if (i + e < n) {
primepos[i] = primepos[i + e];
} else {
primepos[i] = n;
}
} else {
primepos[i] = i;
}
}
long ans = 0;
// out.println(primepos);
for (int i = 0; i < n; i++) {
if (s[arr[i]] == 0) {
if (i + e >= n) continue;
int pos = (i + e < n ? primepos[i + e] : n);
if (pos == n)
pos -= 1;
else pos -= e;
int count = (pos - i) / e;
ans += count;
} else if (arr[i] == 1) {
if (primepos[i] == n) continue;
int j = primepos[i];
if (s[arr[j]] == 1) continue;
int pos = (j + e < n ? primepos[j + e] : n);
// if(pos==n)
// pos-=1;
int count = (pos - j + e - 1) / e;
// if(count==0) count++;
ans += count;
// out.println(t,count);
}
}
out.println(ans);
}
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void println(long i) {
writer.println(i);
}
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 72e451733ee90db5958a8c1a3266ef68 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class c {
public static void main(String[] args) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int numcases = Integer.parseInt(in.readLine());
StringBuilder b = new StringBuilder();
for (int casenum = 0; casenum < numcases; casenum++) {
StringTokenizer tokenizer = new StringTokenizer(in.readLine());
int size = Integer.parseInt(tokenizer.nextToken());
int e = Integer.parseInt(tokenizer.nextToken());
int[][] arr = new int[e][size / e + 1];
boolean[][] prime = new boolean[e][size / e + 1];
boolean[][] one = new boolean[e][size / e + 1];
int[][] left1 = new int[e][size / e + 1];
int[][] right1 = new int[e][size / e + 1];
tokenizer = new StringTokenizer(in.readLine());
for (int i = 0; i < size; i++) {
arr[i % e][i / e] = Integer.parseInt(tokenizer.nextToken());
prime[i % e][i / e] = isPrime(arr[i % e][i / e]);
one[i % e][i / e] = arr[i % e][i / e] == 1;
}
for (int i = 0; i < e; i++) {
for (int j = 1; j < arr[i].length; j++) {
left1[i][j] = one[i][j - 1] ? left1[i][j - 1] + 1 : 0;
}
}
for (int i = 0; i < arr.length; i++) {
for (int j = arr[i].length - 2; j >= 0; j--) {
right1[i][j] = one[i][j + 1] ? right1[i][j + 1] + 1 : 0;
}
}
long answer = 0;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
if (prime[i][j] && 1 + right1[i][j] + left1[i][j] >= 2) {
answer += (long) (right1[i][j] + 1) * (left1[i][j] + 1) - 1;
}
}
}
b.append(answer + "\n");
}
System.out.print(b);
in.close();
out.close();
}
public static boolean isPrime(int k) {
if (k < 2)
return false;
if (k <= 3)
return true;
if (k % 2 == 0)
return false;
for (int i = 3; i * i <= k; i += 2) {
if (k % i == 0)
return false;
}
return true;
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 3a4a11c263347395e5c33250d832a36f | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.Math.ceil;
import static java.util.Arrays.sort;
public class C {
static boolean[] isPrime = new boolean[1000055];
public static void main(String[] args) throws IOException {
int t = ri();
setTrue(1000055);
prime(1000055);
while (t-- > 0) {
int n = rni();
int e = ni();
int a[] = ria(n);
long ans = 0;
for (int i = 0 ; i < n ; i++){
if (isPrime[a[i]] && a[i] != 1){
int l = 0 , r = 0;
for(int j = i-e; j>=0 ; j-=e){
if (a[j] == 1){
l++;
}else{
break;
}
}
for (int j = i+e;j<n;j+=e){
if (a[j] == 1){
r++;
}else{
break;
}
}
ans+= (long) r+l+((long) r *l);
}
}
prln(ans);
}
close();
}
static boolean check(int[] a, int low, int high, int val) {
while (low <= high) {
if (a[low] != a[high] && a[low] != val && a[high] != val) {
return false;
}
if (a[low] == val) {
low++;
}
if (a[high] == val) {
high--;
}
if (a[high] == a[low]) {
high--;
low++;
}
}
return true;
}
static BufferedReader __i = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter __o = new PrintWriter(new OutputStreamWriter(System.out));
static StringTokenizer input;
static Random __r = new Random();
// references
// IBIG = 1e9 + 7
// IMAX ~= 2e9
// LMAX ~= 9e18
// constants
static final int IBIG = 1000000007;
static final int IMAX = 2147483647;
static final long LMAX = 9223372036854775807L;
// math util
static int minof(int a, int b, int c) {
return min(a, min(b, c));
}
static int minof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
int min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static long minof(long a, long b, long c) {
return min(a, min(b, c));
}
static long minof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return min(x[0], x[1]);
if (x.length == 3) return min(x[0], min(x[1], x[2]));
long min = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i];
return min;
}
static int maxof(int a, int b, int c) {
return max(a, max(b, c));
}
static int maxof(int... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
int max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static long maxof(long a, long b, long c) {
return max(a, max(b, c));
}
static long maxof(long... x) {
if (x.length == 1) return x[0];
if (x.length == 2) return max(x[0], x[1]);
if (x.length == 3) return max(x[0], max(x[1], x[2]));
long max = x[0];
for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i];
return max;
}
static int powi(int a, int b) {
if (a == 0) return 0;
int ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static long powl(long a, int b) {
if (a == 0) return 0;
long ans = 1;
while (b > 0) {
if ((b & 1) > 0) ans *= a;
a *= a;
b >>= 1;
}
return ans;
}
static int fli(double d) {
return (int) d;
}
static int cei(double d) {
return (int) ceil(d);
}
static long fll(double d) {
return (long) d;
}
static long cel(double d) {
return (long) ceil(d);
}
static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
static long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
static int[] exgcd(int a, int b) {
if (b == 0) return new int[]{1, 0};
int[] y = exgcd(b, a % b);
return new int[]{y[1], y[0] - y[1] * (a / b)};
}
static long[] exgcd(long a, long b) {
if (b == 0) return new long[]{1, 0};
long[] y = exgcd(b, a % b);
return new long[]{y[1], y[0] - y[1] * (a / b)};
}
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static long mix(long x) {
x += 0x9e3779b97f4a7c15L;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L;
x = (x ^ (x >> 27)) * 0x94d049bb133111ebL;
return x ^ (x >> 31);
}
static void setTrue(int n) {
for (int i = 0; i < n; i++) {
isPrime[i] = true;
}
}
static void prime(int n) {
for (int i = 2; i * i < n; i++) {
if (isPrime[i]) {
for (int j = i * i; j < n; j += i) isPrime[j] = false;
}
}
}
// array util
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static void resetBoolean(boolean[] vis, int n) {
for (int i = 0; i < n; i++) {
vis[i] = false;
}
}
static void setMinusOne(int[][] matrix) {
int row = matrix.length;
int col = matrix[0].length;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
matrix[i][j] = -1;
}
}
}
// input
static void r() throws IOException {
input = new StringTokenizer(rline());
}
static int ri() throws IOException {
return Integer.parseInt(rline().split(" ")[0]);
}
static long rl() throws IOException {
return Long.parseLong(rline());
}
static double rd() throws IOException {
return Double.parseDouble(rline());
}
static int[] ria(int n) throws IOException {
int[] a = new int[n];
r();
for (int i = 0; i < n; ++i) a[i] = ni();
return a;
}
static void ria(int[] a) throws IOException {
int n = a.length;
r();
for (int i = 0; i < n; ++i) a[i] = ni();
}
static int[] riam1(int n) throws IOException {
int[] a = new int[n];
r();
for (int i = 0; i < n; ++i) a[i] = ni() - 1;
return a;
}
static void riam1(int[] a) throws IOException {
int n = a.length;
r();
for (int i = 0; i < n; ++i) a[i] = ni() - 1;
}
static long[] rla(int n) throws IOException {
long[] a = new long[n];
r();
for (int i = 0; i < n; ++i) a[i] = nl();
return a;
}
static void rla(long[] a) throws IOException {
int n = a.length;
r();
for (int i = 0; i < n; ++i) a[i] = nl();
}
static double[] rda(int n) throws IOException {
double[] a = new double[n];
r();
for (int i = 0; i < n; ++i) a[i] = nd();
return a;
}
static void rda(double[] a) throws IOException {
int n = a.length;
r();
for (int i = 0; i < n; ++i) a[i] = nd();
}
static char[] rcha() throws IOException {
return rline().toCharArray();
}
static void rcha(char[] a) throws IOException {
int n = a.length, i = 0;
for (char c : rline().toCharArray()) a[i++] = c;
}
static String rline() throws IOException {
return __i.readLine();
}
static String n() {
return input.nextToken();
}
static int rni() throws IOException {
r();
return ni();
}
static int ni() {
return Integer.parseInt(n());
}
static long rnl() throws IOException {
r();
return nl();
}
static long nl() {
return Long.parseLong(n());
}
static double rnd() throws IOException {
r();
return nd();
}
static double nd() {
return Double.parseDouble(n());
}
// output
static void pr(int i) {
__o.print(i);
}
static void prln(int i) {
__o.println(i);
}
static void pr(long l) {
__o.print(l);
}
static void prln(long l) {
__o.println(l);
}
static void pr(double d) {
__o.print(d);
}
static void prln(double d) {
__o.println(d);
}
static void pr(char c) {
__o.print(c);
}
static void prln(char c) {
__o.println(c);
}
static void pr(char[] s) {
__o.print(new String(s));
}
static void prln(char[] s) {
__o.println(new String(s));
}
static void pr(String s) {
__o.print(s);
}
static void prln(String s) {
__o.println(s);
}
static void pr(Object o) {
__o.print(o);
}
static void prln(Object o) {
__o.println(o);
}
static void prln() {
__o.println();
}
static void pryes() {
prln("yes");
}
static void pry() {
prln("Yes");
}
static void prY() {
prln("YES");
}
static void prno() {
prln("no");
}
static void prn() {
prln("No");
}
static void prN() {
prln("NO");
}
static boolean pryesno(boolean b) {
prln(b ? "yes" : "no");
return b;
}
;
static boolean pryn(boolean b) {
prln(b ? "Yes" : "No");
return b;
}
static boolean prYN(boolean b) {
prln(b ? "YES" : "NO");
return b;
}
static void prln(int... a) {
for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i) ;
if (a.length > 0) prln(a[a.length - 1]);
else prln();
}
static void prln(long... a) {
for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i) ;
if (a.length > 0) prln(a[a.length - 1]);
else prln();
}
static void prln(double... a) {
for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i) ;
if (a.length > 0) prln(a[a.length - 1]);
else prln();
}
static <T> void prln(Collection<T> c) {
int n = c.size() - 1;
Iterator<T> iter = c.iterator();
for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i) ;
if (n >= 0) prln(iter.next());
else prln();
}
static void h() {
prln("hlfd");
flush();
}
static void flush() {
__o.flush();
}
static void close() {
__o.close();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 65a62a625cc62deefb5d0c53cca36aeb | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
public class C {
// -- static variables --- //
static FastReader sc = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static int mod = (int) 1000000007;
public static void main(String[] args) throws Exception {
sieve();
int t = sc.nextInt();
while (t-- > 0)
C.go();
// out.println();
out.flush();
}
// >>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<< //
static class pair {
int x, y;
pair(int x, int y) {
this.x = x;
this.y = y;
}
}
static void go() throws Exception {
int n=sc.nextInt();
int e=sc.nextInt();
int a[]=sc.intArray(n);
ArrayList<Integer> aa=new ArrayList<>();
for(int i=0;i<n;i++) {
if(prime[a[i]]==1) {
aa.add(i);
}
}
int left[]=new int[n];
for(int i=0;i<n;i++) {
if(a[i]==1) {
if(i-e>=0 && left[i-e]>0) {
left[i]=left[i-e]+1;
}else {
left[i]=1;
}
}
}
int right[]=new int[n];
for(int i=n-1;i>=0;i--) {
if(a[i]==1) {
if(i+e<n && right[i+e]>0) {
right[i]=right[i+e]+1;
}else {
right[i]=1;
}
}
}
long ans=0;
for(int i :aa) {
long val1=0;
long val2=0;
if(i-e>=0)
val1=left[i-e];
if(i+e<n)
val2=right[i+e];
// out.println(i+" "+val1+" "+val2);
ans=ans+val1*(val2+1)+val2;
}
out.println(ans);
}
static int left(int a[],int ind,int e) {
int c=0;
for(int i=ind-e;i>=0;i-=e) {
if(a[i]==1) {
c++;
}else {
break;
}
}
return c;
}
static int right(int a[],int ind,int e) {
int c=0;
for(int i=ind+e;i<a.length;i+=e) {
if(a[i]==1) {
c++;
}else {
break;
}
}
return c;
}
// static int ceil(ArrayList<Integer> x,int key) {
// int ans=0;
// for(int i:x) {
//
// }
// return ans;
// }
// static int floor(ArrayList<Integer> x,int key) {
// int l=0,r=x.size()-1;
// int ans=0;
// while(l<=r) {
// int mid=(l+r)/2;
// if(x.get(mid)>key) {
// ans=mid+1;
// r=mid-1;
// }else {
// l=mid+1;
// }
// }
// return ans;
// }
static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
// >>>>>>>>>>> Code Ends <<<<<<<<< //
// --For Rounding--//
static double round(double value, int places) {
if (places < 0)
throw new IllegalArgumentException();
BigDecimal bd = new BigDecimal(Double.toString(value));
bd = bd.setScale(places, RoundingMode.HALF_UP);
return bd.doubleValue();
}
// ----Greatest Common Divisor-----//
static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
// --- permutations and Combinations ---//
static long fact[];
static long invfact[];
static long ncr(int n, int k) {
if (k < 0 || k > n) {
return 0;
}
long x = fact[n];
long y = fact[k];
long yy = fact[n - k];
long ans = (x / y);
ans = (ans / yy);
return ans;
}
// ---sieve---//
static int prime[] = new int[1000006];
static void sieve() {
Arrays.fill(prime, 1);
prime[0] = 0;
prime[1] = 0;
for (int i = 2; i * i <= 1000005; i++) {
if (prime[i] == 1)
for (int j = i * i; j <= 1000005; j += i) {
prime[j] = 0;
}
}
}
// ---- Manual sort ------//
static void sort(long[] a) {
ArrayList<Long> aa = new ArrayList<>();
for (long i : a) {
aa.add(i);
}
Collections.sort(aa);
for (int i = 0; i < a.length; i++)
a[i] = aa.get(i);
}
static void sort(int[] a) {
ArrayList<Integer> aa = new ArrayList<>();
for (int i : a) {
aa.add(i);
}
Collections.sort(aa);
for (int i = 0; i < a.length; i++)
a[i] = aa.get(i);
}
// --- Fast exponentiation ---//
static long pow(long x, long y) {
long res = 1l;
while (y != 0) {
if (y % 2 == 1) {
res = (x * res);
}
y /= 2;
x = (x * x);
}
return res;
}
// >>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<< //
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] intArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
return a;
}
long[] longArray(int n) {
long a[] = new long[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextLong();
return a;
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | f5df7bfc44b370c74c195ef79740fc8d | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
public class C {
// -- static variables --- //
static FastReader sc = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static int mod = (int) 1000000007;
public static void main(String[] args) throws Exception {
sieve();
int t = sc.nextInt();
while (t-- > 0)
C.go();
// out.println();
out.flush();
}
// >>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<< //
static class pair {
int x, y;
pair(int x, int y) {
this.x = x;
this.y = y;
}
}
static void go() throws Exception {
int n=sc.nextInt();
int e=sc.nextInt();
int a[]=sc.intArray(n);
ArrayList<Integer> aa=new ArrayList<>();
for(int i=0;i<n;i++) {
if(prime[a[i]]==1) {
aa.add(i);
}
}
long ans=0;
for(int i :aa) {
long val1=left(a,i,e);
long val2=right(a,i,e);
// out.println(i+" "+val1+" "+val2);
ans=ans+val1*(val2+1)+val2;
}
out.println(ans);
}
static int left(int a[],int ind,int e) {
int c=0;
for(int i=ind-e;i>=0;i-=e) {
if(a[i]==1) {
c++;
}else {
break;
}
}
return c;
}
static int right(int a[],int ind,int e) {
int c=0;
for(int i=ind+e;i<a.length;i+=e) {
if(a[i]==1) {
c++;
}else {
break;
}
}
return c;
}
// static int ceil(ArrayList<Integer> x,int key) {
// int ans=0;
// for(int i:x) {
//
// }
// return ans;
// }
// static int floor(ArrayList<Integer> x,int key) {
// int l=0,r=x.size()-1;
// int ans=0;
// while(l<=r) {
// int mid=(l+r)/2;
// if(x.get(mid)>key) {
// ans=mid+1;
// r=mid-1;
// }else {
// l=mid+1;
// }
// }
// return ans;
// }
static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
// >>>>>>>>>>> Code Ends <<<<<<<<< //
// --For Rounding--//
static double round(double value, int places) {
if (places < 0)
throw new IllegalArgumentException();
BigDecimal bd = new BigDecimal(Double.toString(value));
bd = bd.setScale(places, RoundingMode.HALF_UP);
return bd.doubleValue();
}
// ----Greatest Common Divisor-----//
static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
// --- permutations and Combinations ---//
static long fact[];
static long invfact[];
static long ncr(int n, int k) {
if (k < 0 || k > n) {
return 0;
}
long x = fact[n];
long y = fact[k];
long yy = fact[n - k];
long ans = (x / y);
ans = (ans / yy);
return ans;
}
// ---sieve---//
static int prime[] = new int[1000006];
static void sieve() {
Arrays.fill(prime, 1);
prime[0] = 0;
prime[1] = 0;
for (int i = 2; i * i <= 1000005; i++) {
if (prime[i] == 1)
for (int j = i * i; j <= 1000005; j += i) {
prime[j] = 0;
}
}
}
// ---- Manual sort ------//
static void sort(long[] a) {
ArrayList<Long> aa = new ArrayList<>();
for (long i : a) {
aa.add(i);
}
Collections.sort(aa);
for (int i = 0; i < a.length; i++)
a[i] = aa.get(i);
}
static void sort(int[] a) {
ArrayList<Integer> aa = new ArrayList<>();
for (int i : a) {
aa.add(i);
}
Collections.sort(aa);
for (int i = 0; i < a.length; i++)
a[i] = aa.get(i);
}
// --- Fast exponentiation ---//
static long pow(long x, long y) {
long res = 1l;
while (y != 0) {
if (y % 2 == 1) {
res = (x * res);
}
y /= 2;
x = (x * x);
}
return res;
}
// >>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<< //
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] intArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
return a;
}
long[] longArray(int n) {
long a[] = new long[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextLong();
return a;
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 8d603485a73757aa0a160b6fe8d92d1f | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long mod = 1000000007;
static long max ;
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main(String[] args) throws IOException {
FastReader sc = new FastReader();
int t = sc.nextInt();
boolean prime[] = sieveOfEratosthenes((int)1e6);
while( t-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
int arr[] = new int[n];
for( int i = 0 ;i < n ;i++) {
arr[i] = sc.nextInt();
}
long ans =0;
boolean check[] = new boolean[n];
for( int a =0 ; a< n; a++) {
if( check[a]) {
continue;
}
int i = a;
if( arr[i] == 1) {
long count = 1;
check[i] = true;
while( i+e < n && arr[i+e] == 1) {
i+=e;
check[i] = true;
count++;
}
if( i+e < n && prime[arr[i+e]]) {
// out.println(arr[i+e]);
ans+=count;
long fst = count;
count = 0;
i+=e;
while( i+e < n && arr[i+e] == 1) {
i+=e;
count++;
}
ans+=count*fst;
// out.println(ans);
}
}
else if(prime[arr[i]] ) {
// out.println(arr[i] + " start");
long count = 0;
while( i+e < n && arr[i+e] == 1) {
count++;
i+=e;
}
ans+=count;
// out.println(ans);
}
}
out.println(ans);
}
out.flush();
}
/*
* FIRST AND VERY IMP -> READ AND UNDERSTAND THE QUESTION VERY CAREFULLY.
* WARNING -> DONT CODE BULLSHIT , ALWAYS CHECK THE LOGIC ON MULTIPLE TESTCASES AND EDGECASES BEFORE.
* SECOND -> TRY TO FIND RELEVENT PATTERN SMARTLY.
* WARNING -> IF YOU THINK YOUR SOLUION IS JUST DUMB DONT SUBMIT IT BEFORE RECHECKING ON YOUR END.
*/
public static boolean ifpowof2(long n ) {
return ((n&(n-1)) == 0);
}
static boolean isprime(long x ) {
if( x== 2) {
return true;
}
if( x%2 == 0) {
return false;
}
for( long i = 3 ;i*i <= x ;i+=2) {
if( x%i == 0) {
return false;
}
}
return true;
}
static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int)n + 1];
for (int i = 0; i <= n; i++) {
prime[i] = true;
}
for (long p = 2; p * p <= n; p++) {
if (prime[(int)p] == true) {
for (long i = p * p; i <= n; i += p)
prime[(int)i] = false;
}
}
return prime;
}
public static int[] nextLargerElement(int[] arr, int n) {
Stack<Integer> stack = new Stack<>();
int rtrn[] = new int[n];
rtrn[n-1] = -1;
stack.push( n-1);
for( int i = n-2 ;i >= 0 ; i--){
int temp = arr[i];
int lol = -1;
while( !stack.isEmpty() && arr[stack.peek()] <= temp){
if(arr[stack.peek()] == temp ) {
lol = stack.peek();
}
stack.pop();
}
if( stack.isEmpty()){
if( lol != -1) {
rtrn[i] = lol;
}
else {
rtrn[i] = -1;
}
}
else{
rtrn[i] = stack.peek();
}
stack.push( i);
}
return rtrn;
}
static void mysort(int[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
int loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static void mySort(long[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
long loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b)
{
return (a / gcd(a, b)) * b;
}
static long rightmostsetbit(long n) {
return n&-n;
}
static long leftmostsetbit(long n)
{
long k = (long)(Math.log(n) / Math.log(2));
return 1 << k;
}
static HashMap<Long,Long> primefactor( long n){
HashMap<Long ,Long> hm = new HashMap<>();
long temp = 0;
while( n%2 == 0) {
temp++;
n/=2;
}
if( temp!= 0) {
hm.put( 2L, temp);
}
long c = (long)Math.sqrt(n);
for( long i = 3 ; i <= c ; i+=2) {
temp = 0;
while( n% i == 0) {
temp++;
n/=i;
}
if( temp!= 0) {
hm.put( i, temp);
}
}
if( n!= 1) {
hm.put( n , 1L);
}
return hm;
}
static ArrayList<Integer> allfactors(int abs) {
HashMap<Integer,Integer> hm = new HashMap<>();
ArrayList<Integer> rtrn = new ArrayList<>();
for( int i = 2 ;i*i <= abs; i++) {
if( abs% i == 0) {
hm.put( i , 0);
hm.put(abs/i, 0);
}
}
for( int x : hm.keySet()) {
rtrn.add(x);
}
if( abs != 0) {
rtrn.add(abs);
}
return rtrn;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 5f9a1f726adba92751ab2eeab7559229 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long mod = 1000000007;
static long max ;
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main(String[] args) throws IOException {
FastReader sc = new FastReader();
int t = sc.nextInt();
while( t-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
int arr[] = new int[n];
for( int i = 0 ;i < n ;i++) {
arr[i] = sc.nextInt();
}
long ans =0;
boolean check[] = new boolean[n];
for( int a =0 ; a< n; a++) {
if( check[a]) {
continue;
}
int i = a;
if( arr[i] == 1) {
long count = 1;
check[i] = true;
while( i+e < n && arr[i+e] == 1) {
i+=e;
check[i] = true;
count++;
}
if( i+e < n && isprime(arr[i+e])) {
// out.println(arr[i+e]);
ans+=count;
long fst = count;
count = 0;
i+=e;
while( i+e < n && arr[i+e] == 1) {
i+=e;
count++;
}
ans+=count*fst;
// out.println(ans);
}
}
else if(isprime(arr[i]) ) {
// out.println(arr[i] + " start");
long count = 0;
while( i+e < n && arr[i+e] == 1) {
count++;
i+=e;
}
ans+=count;
// out.println(ans);
}
}
out.println(ans);
}
out.flush();
}
/*
* FIRST AND VERY IMP -> READ AND UNDERSTAND THE QUESTION VERY CAREFULLY.
* WARNING -> DONT CODE BULLSHIT , ALWAYS CHECK THE LOGIC ON MULTIPLE TESTCASES AND EDGECASES BEFORE.
* SECOND -> TRY TO FIND RELEVENT PATTERN SMARTLY.
* WARNING -> IF YOU THINK YOUR SOLUION IS JUST DUMB DONT SUBMIT IT BEFORE RECHECKING ON YOUR END.
*/
public static boolean ifpowof2(long n ) {
return ((n&(n-1)) == 0);
}
static boolean isprime(long x ) {
if( x== 2) {
return true;
}
if( x%2 == 0) {
return false;
}
for( long i = 3 ;i*i <= x ;i+=2) {
if( x%i == 0) {
return false;
}
}
return true;
}
public static int[] nextLargerElement(int[] arr, int n) {
Stack<Integer> stack = new Stack<>();
int rtrn[] = new int[n];
rtrn[n-1] = -1;
stack.push( n-1);
for( int i = n-2 ;i >= 0 ; i--){
int temp = arr[i];
int lol = -1;
while( !stack.isEmpty() && arr[stack.peek()] <= temp){
if(arr[stack.peek()] == temp ) {
lol = stack.peek();
}
stack.pop();
}
if( stack.isEmpty()){
if( lol != -1) {
rtrn[i] = lol;
}
else {
rtrn[i] = -1;
}
}
else{
rtrn[i] = stack.peek();
}
stack.push( i);
}
return rtrn;
}
static void mysort(int[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
int loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static void mySort(long[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
long loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b)
{
return (a / gcd(a, b)) * b;
}
static long rightmostsetbit(long n) {
return n&-n;
}
static long leftmostsetbit(long n)
{
long k = (long)(Math.log(n) / Math.log(2));
return 1 << k;
}
static HashMap<Long,Long> primefactor( long n){
HashMap<Long ,Long> hm = new HashMap<>();
long temp = 0;
while( n%2 == 0) {
temp++;
n/=2;
}
if( temp!= 0) {
hm.put( 2L, temp);
}
long c = (long)Math.sqrt(n);
for( long i = 3 ; i <= c ; i+=2) {
temp = 0;
while( n% i == 0) {
temp++;
n/=i;
}
if( temp!= 0) {
hm.put( i, temp);
}
}
if( n!= 1) {
hm.put( n , 1L);
}
return hm;
}
static ArrayList<Integer> allfactors(int abs) {
HashMap<Integer,Integer> hm = new HashMap<>();
ArrayList<Integer> rtrn = new ArrayList<>();
for( int i = 2 ;i*i <= abs; i++) {
if( abs% i == 0) {
hm.put( i , 0);
hm.put(abs/i, 0);
}
}
for( int x : hm.keySet()) {
rtrn.add(x);
}
if( abs != 0) {
rtrn.add(abs);
}
return rtrn;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | d8c05c0489b519f65ba34ebdf598343d | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
static boolean prime[];
public static void main(String[] args) {
sieveOfEratosthenes(1000009);
// System.out.println(prime[1]=false);
prime[1]=false;
FastScanner fs=new FastScanner();
/****** CODE STARTS HERE *****/
//------------------------------------------------------------------------------------------------------------
int t = fs.nextInt();
while(t-->0) {
int n=fs.nextInt(), e=fs.nextInt();
int[] a = new int[n+1];
for(int i=1; i<=n; i++)a[i]=fs.nextInt();
long ans = 0;
for(int i=1; i<=n; i++) {
if(!prime[a[i]])continue;
long cnt1 = 0, cnt2 = 0;
int j = i+e;
while(j<=n && a[j]==1) {
cnt1++;
j+=e;
}
j=i-e;
while(j>=0 && a[j]==1) {
cnt2++;
j-=e;
}
ans += cnt1*cnt2 + cnt1 + cnt2;
}
System.out.println(ans);
}
}
static void sieveOfEratosthenes(int n)
{
prime = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
//****** CODE ENDS HERE *****
//----------------------------------------------------------------------------------------------------------------
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//----------- FastScanner class for faster input---------------------------
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 41b6b73c2e0cac8763425c31e951a829 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.StringTokenizer;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.lang.reflect.Array;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Collections;
import java.util.ArrayList;
public class p3
{
BufferedReader br;
StringTokenizer st;
BufferedWriter bw;
public static void main(String[] args)throws Exception
{
new p3().run();
}
void run()throws IOException
{
br = new BufferedReader(new InputStreamReader(System.in));
bw=new BufferedWriter(new OutputStreamWriter(System.out));
solve();
}
void solve() throws IOException
{
int prime[]=primeNO(1000000000000L);
short t=ns();
while(t-->0)
{
int n=ni();int e=ni();
int a[]=nai(n);
if(e==n)
System.out.println(0);
else
{
long ans=0;
for(int i=-1;++i<e;)
{
for(int j=i;j<n;j+=e)
{
if(a[j]==1)
{
int c=0;
for(;j<n && a[j]==1;j+=e)c++;
if(j<n)
{
if(Arrays.binarySearch(prime, a[j])>=0)
{
ans+=c;
int set=j;
int d=0;
for(j+=e;j<n && a[j]==1;j+=e)d++;
ans+=(long)c*d;
j=set-e;
// System.out.println("v"+ans);
}
}
}
else if(Arrays.binarySearch(prime, a[j])>=0)
{
int c=0;
int set=j;
j+=e;
for(;j<n && a[j]==1;j+=e)c++;
ans+=c;
if(j<n && Arrays.binarySearch(prime, a[j])>=0)j=set;
// j-=e;
// System.out.println("v"+ans);
}
}
}
System.out.println(ans);
}
}
}
public int[] primeNO(long limit)
{
int size=(int)Math.sqrt(limit)+1;
int size1=size/2;
boolean composite[]=new boolean[size1];
int zz=(int)Math.sqrt(size-1);
for(int i=3;i<=zz;)
{
for(int j=(i*i)/2;j<size1;j+=i)
composite[j]=true;
for(i+=2;composite[i/2];i+=2);
}
int prime[]=new int[78498];
prime[0]=2;
int p=1;
for(int i=1;i<size1;i++)
{
if(!composite[i])
prime[p++]=2*i+1;
}
////////////////////////////////////////////////////////////////////////////////////////////////////////
// System.out.println(p);
////////////////////////////////////////////////////////////////////////////////////////////////////////
prime=Arrays.copyOf(prime, p);
return prime;
}
public int gcd(int a, int b)
{
if(a==0)
return b;
else
return gcd(b%a,a);
}
int[] nai(int n) { int a[]=new int[n]; for(int i=-1;++i<n;)a[i]=ni(); return a;}
long[] nal(int n) { long a[]=new long[n]; for(int i=-1;++i<n;)a[i]=nl(); return a;}
char[] nac() {char c[]=nextLine().toCharArray(); return c;}
char [][] nmc(int n) {char c[][]=new char[n][]; for(int i=-1;++i<n;)c[i]=nac(); return c;}
int[][] nmi(int r, int c) {int a[][]=new int[r][c]; for(int i=-1;++i<r;)a[i]=nai(c); return a;}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
byte nb() { return Byte.parseByte(next()); }
short ns() { return Short.parseShort(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int countDigits(int l) {
if (l >= 1000000000) return 10;
if (l >= 100000000) return 9;
if (l >= 10000000) return 8;
if (l >= 1000000) return 7;
if (l >= 100000) return 6;
if (l >= 10000) return 5;
if (l >= 1000) return 4;
if (l >= 100) return 3;
if (l >= 10) return 2;
return 1;
}
int countDigits(long l) {
if (l >= 1000000000000000000L) return 19;
if (l >= 100000000000000000L) return 18;
if (l >= 10000000000000000L) return 17;
if (l >= 1000000000000000L) return 16;
if (l >= 100000000000000L) return 15;
if (l >= 10000000000000L) return 14;
if (l >= 1000000000000L) return 13;
if (l >= 100000000000L) return 12;
if (l >= 10000000000L) return 11;
if (l >= 1000000000L) return 10;
if (l >= 100000000L) return 9;
if (l >= 10000000L) return 8;
if (l >= 1000000L) return 7;
if (l >= 100000L) return 6;
if (l >= 10000L) return 5;
if (l >= 1000L) return 4;
if (l >= 100L) return 3;
if (l >= 10L) return 2;
return 1;
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 78636fe8bc0fe6aa68dc4ba76237629e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
import java.lang.*;
public class Main {
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
public static int cnt;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static boolean isPrime(int n){
if(n<4){
return true;
}
if(n%2==0){
return false;
}
int m = (int)Math.sqrt(n);
for(int i=3;i<=m;i+=2){
if(n%i==0){
return false;
}
}
return true;
}
public static void main(String args[]) {
int mod = 1000000007;
MyScanner sc = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int e = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
long ans = 0;
for(int i=0;i<n;i++){
if(a[i]>1 && isPrime(a[i])){
long cnt1 = 0,cnt2 = 0;
int j = i-e;
while(j>=0 && a[j]==1){
j -= e;
++cnt1;
}
j = i + e;
while(j<n && a[j]==1){
j += e;
++cnt2;
}
ans += cnt1 + cnt2 + cnt1 * cnt2;
}
}
out.println(ans);
}
out.close();
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 9643c6f64d64d7c0c79562f324639615 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution {
public static void main(String[] args) throws IOException {
int[] primes = new int[1000001];
for (int i = 2; i < primes.length; i++) {
for (int j = i; j < primes.length; j += i) {
if (primes[j] == 0)
primes[j] = i;
}
}
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int k = sc.nextInt();
int[] arr = sc.nextIntArray(n);
ArrayList<Integer> ones = new ArrayList<>();
boolean[] chained = new boolean[n];
long ans = 0;
for (int i = 0; i < arr.length; i++) {
ones.clear();
int countOnes = 0;
for (int j = i; j < arr.length; j += k) {
if ((arr[j] == 1 || arr[j] == primes[arr[j]]) && !chained[j]) {
chained[j] = true;
if (arr[j] == 1)
countOnes++;
else {
ones.add(countOnes);
countOnes = 0;
}
} else {
break;
}
}
if (ones.isEmpty())
continue;
ones.add(countOnes);
for (int j = 0; j < ones.size(); j++) {
ans += ones.get(j);
if (j > 0 && j < ones.size() - 1) {
ans += ones.get(j);
}
if (j < ones.size() - 1) {
ans += 1l * ones.get(j) * ones.get(j + 1);
}
}
}
pw.println(ans);
}
pw.close();
}
static boolean check(int[] arr, long health, long damage) {
long sum = damage;
for (int i = 0; i < arr.length - 1; i++) {
sum += Math.min(damage, arr[i + 1] - arr[i]);
}
return sum >= health;
}
static double f(double num, double deno) {
if (num == 1 && deno == 3)
return 1 / 6.0;
return ((num / deno) * (num - 1) / (deno - 1) + f(num - 1, deno - 1));
}
private static int mergeAndCount(int[] arr, int l,
int m, int r) {
// Left subarray
int[] left = Arrays.copyOfRange(arr, l, m + 1);
// Right subarray
int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
int i = 0, j = 0, k = l, swaps = 0;
while (i < left.length && j < right.length) {
if (left[i] <= right[j])
arr[k++] = left[i++];
else {
arr[k++] = right[j++];
swaps += (m + 1) - (l + i);
}
}
while (i < left.length)
arr[k++] = left[i++];
while (j < right.length)
arr[k++] = right[j++];
return swaps;
}
// Merge sort function
private static int mergeSortAndCount(int[] arr, int l,
int r) {
// Keeps track of the inversion count at a
// particular node of the recursion tree
int count = 0;
if (l < r) {
int m = (l + r) / 2;
// Total inversion count = left subarray count
// + right subarray count + merge count
// Left subarray count
count += mergeSortAndCount(arr, l, m);
// Right subarray count
count += mergeSortAndCount(arr, m + 1, r);
// Merge count
count += mergeAndCount(arr, l, m, r);
}
return count;
}
static HashMap Hash(int[] arr) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int i : arr) {
map.put(i, map.getOrDefault(i, 0) + 1);
}
return map;
}
static HashMap Hash(char[] arr) {
HashMap<Character, Integer> map = new HashMap<>();
for (char i : arr) {
map.put(i, map.getOrDefault(i, 0) + 1);
}
return map;
}
static boolean isPrime(int n) {
if (n <= 1)
return false;
for (int i = 2; i <= Math.sqrt(n); i++)
if (n % i == 0)
return false;
return true;
}
public static long combination(long n, long r) {
return factorial(n) / (factorial(n - r) * factorial(r));
}
static long factorial(Long n) {
if (n == 0)
return 1;
return n * factorial(n - 1);
}
static boolean isPalindrome(char[] str, int i, int j) {
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str[i] != str[j])
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
public static double log2(long l) {
double result = (Math.log(l) / Math.log(2));
return result;
}
public static double log4(int N) {
double result = (Math.log(N) / Math.log(4));
return result;
}
public static int setBit(int mask, int idx) {
return mask | (1 << idx);
}
public static int clearBit(int mask, int idx) {
return mask ^ (1 << idx);
}
public static boolean checkBit(int mask, int idx) {
return (mask & (1 << idx)) != 0;
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
long x;
long y;
public pair(long x, long y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return Long.compare(this.y, other.y);
}
return Long.compare(this.x, other.x);
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
public static String toString(int[] arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
return sb.toString().trim();
}
public static String toString(long[] arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
return sb.toString().trim();
}
public static String toString(ArrayList arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.size(); i++) {
sb.append(arr.get(i) + " ");
}
return sb.toString().trim();
}
public static String toString(int[][] arr) {
StringBuilder sb = new StringBuilder();
for (int[] i : arr) {
sb.append(toString(i) + "\n");
}
return sb.toString();
}
public static String toString(boolean[] arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
return sb.toString().trim();
}
public static String toString(Integer[] arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
return sb.toString().trim();
}
public static String toString(String[] arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
return sb.toString().trim();
}
public static String toString(char[] arr) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
return sb.toString().trim();
}
static long mod = 1000000007;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 8dd0e6070e8e7ff2d517d2ec0e6a291e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.math.BigInteger;
import java.util.*;
import java.io.*;
public class CodeForces {
boolean isPrime(int x) {
if (x < 2) {
return false;
}
int xq = (int)Math.sqrt((double)x);
for (int i = 2; i <= xq; i++) {
if (x % i == 0) {
return false;
}
}
return true;
}
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public void solve() throws Exception {
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int e = Integer.parseInt(st.nextToken());
int[] a = new int[n];
st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
a[i] = Integer.parseInt(st.nextToken());
}
long ans = 0;
for (int i = 0; i < n; i++) {
if (isPrime(a[i])) {
long l = 0;
long r = 0;
int pos = i;
while (pos - e >= 0) {
pos -= e;
if (a[pos] == 1) {
l++;
} else {
break;
}
}
pos = i;
while (pos + e < n) {
pos += e;
if (a[pos] == 1) {
r++;
} else {
break;
}
}
ans += l + r + l * r;
}
}
System.out.println(ans);
}
public void run() throws Exception {
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
solve();
}
// solve();
}
public static void main(String[] args) throws Exception {
new CodeForces().run();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | cc12f47dacaaf9a9535080207806bd48 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main{
private static final boolean console = true;
private static final String name = "closing";
private static PrintWriter out;
private static FastIO sc;
private static final int mod = 1_000_000_007;
public static void main(String[] args) throws Exception {
if(console) sc = new FastIO(System.in);
else sc = new FastIO(name+".in");
if(console) out = new PrintWriter(new BufferedOutputStream(System.out));
else out = new PrintWriter(name+".out");
boolean[] primes = new boolean[1000001];
Arrays.fill(primes, true);
primes[1] = false;
for(int i = 2; i*i<=primes.length; i++)
for(int j = i<<1; j<primes.length; j+=i) primes[j] = false;
int t = sc.nextInt();
for(int i = 0; i<t; i++) {
int[] nums = new int[sc.nextInt()];
int e = sc.nextInt();
for(int j = 0; j<nums.length; j++) nums[j] = sc.nextInt();
out.println(solve(nums, e, primes));
}
out.close();
}
private static long solve(int[] nums, int e, boolean[] primes) {
long[] left = new long[nums.length], right = new long[nums.length];
for(int i = 0; i<nums.length; i++) if(nums[i] == 1) left[i] = right[i] = 1;
for(int i = e; i<nums.length; i++) if(nums[i] == 1) left[i]+=left[i-e];
for(int i = nums.length-e-1; i>=0; i--) if(nums[i] == 1) right[i]+=right[i+e];
long ans = 0;
for(int i = 0; i<nums.length; i++) if(primes[nums[i]]) ans+=((i < e ? 0:left[i-e])+1)*((i > nums.length-1-e ? 0:right[i+e])+1)-1;
return ans;
}
private static class FastIO {
InputStream dis;
byte[] buffer = new byte[1 << 17];
int pointer = 0, end = 0;
public FastIO(String fileName) throws Exception {
dis = new FileInputStream(fileName);
}
public FastIO(InputStream is) throws Exception {
dis = is;
}
int nextInt() throws Exception {
int ret = 0;
byte b;
do {
b = nextByte();
} while (b <= ' ');
boolean negative = false;
if (b == '-') {
negative = true;
b = nextByte();
}
while (b >= '0' && b <= '9') {
ret = 10 * ret + b - '0';
b = nextByte();
}
return (negative) ? -ret : ret;
}
long nextLong() throws Exception {
long ret = 0;
byte b;
do {
b = nextByte();
} while (b <= ' ');
boolean negative = false;
if (b == '-') {
negative = true;
b = nextByte();
}
while (b >= '0' && b <= '9') {
ret = 10 * ret + b - '0';
b = nextByte();
}
return (negative) ? -ret : ret;
}
byte nextByte() throws Exception {
while(pointer >= end) {
end = dis.read(buffer, 0, buffer.length);
pointer = 0;
}
return buffer[pointer++];
}
double nextDouble() throws Exception {
return Double.parseDouble(next());
}
String next() throws Exception {
StringBuffer ret = new StringBuffer();
byte b;
do {
b = nextByte();
} while (b <= ' ');
while (b > ' ') {
ret.appendCodePoint(b);
b = nextByte();
}
return ret.toString();
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 35a225188f2d4dd3f0ec7c0ebc2b3352 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
static Set<Integer> sieve;
public static void main(String[] args) {
MyScanner s = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int testcases=s.nextInt();
sieve();
while(testcases-->0) {
int n=s.nextInt();
int e=s.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++)arr[i]=s.nextInt();
if(n==e) {
out.println(0);
continue;
}
long ans=0;
long[] dp=new long[n];
for(int i=n-1;i>=0;i--) {
if(arr[i]!=1)dp[i]=0;
else {
if(i+e<n)dp[i]=dp[i+e] + 1;
else dp[i]=1;
}
}
// for(int i=0;i<n;i++)out.print(dp[i] + " ");
// out.println();
for(int i=n-1;i>=0;i--) {
if(dp[i]==0) {
if(!sieve.contains(arr[i]))continue;
long r;
if(i+e<n)r=dp[i+e];
else r=0;
long l=0;
for(int j=i-e;j>=0;j-=e) {
if(dp[j]==0)break;
l=dp[j];
}
ans+=((l+1)*(r+1) - 1);
}
}
out.println(ans);
}
out.close();
}
private static void sieve() {
sieve=new HashSet<>();
boolean[] arr=new boolean[1000009];
int n=arr.length;
for(int i=2;i<n;i++) {
if(arr[i])continue;
for(int j=i+i;j<n;j+=i) {
arr[j]=true;
}
}
for(int i=2;i<n;i++) {
if(!arr[i])sieve.add(i);
}
// System.out.println(sieve.size());
}
static long gcd(long a, long b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a - b, b);
return gcd(a, b - a);
}
public static PrintWriter out;
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | cd8cf2bc9b21b8b134f947f779d60006 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
// Ctrl + Alt + F for formatting code
public class setup {
static int mod = (int)1e9 + 7;
static long gcd(long a, long b, long n) {
if (a == b) {
return (power(a, n, mod) + power(b, n, mod)) % mod;
}
long num = a - b, max = 1;
for (int i = 1; i * i <= num; i++) {
if (num % i == 0) {
if ((power(a, n, i) + power(b, n, i)) % i == 0) {
max = Math.max(max, i);
}
if ((power(a, n, num / i) + power(b, n, num / i)) % (num / i) == 0) {
max = Math.max(max, num / i);
break;
}
}
}
return (max % mod);
}
static long power(long a, long n, long d) {
a = a % d;
long res = 1;
while (n > 0) {
if (n % 2 == 1) {
res = ((res % d) * (a % d)) % d;
n--;
} else {
a = ((a % d) * (a % d)) % d;
n /= 2;
}
}
return res % d;
}
public static int findParent(int u, int[] parent) {
if (u == parent[u])return u;
int x = findParent(parent[u], parent);
parent[u] = x;
return x;
}
public static boolean equi(String s1, String s2) {
int[] freq = new int[26];
for (int i = 0; i < s1.length(); i++) {
freq[s1.charAt(i) - 'a']++;
}
for (int i = 0; i < s2.length(); i++) {
freq[s2.charAt(i) - 'a']++;
if (freq[s2.charAt(i) - 'a'] >= 2)return true;
}
return false;
}
static class pair {
long count;
int p;
pair(int count, int p) {
this.count = count;
this.p = p;
}
}
public static int cnt = 0;
public static int[] c;
public static void main(String[] args) throws IOException {
try {
System.setIn(new FileInputStream("input1.txt"));
System.setOut(new PrintStream(new FileOutputStream("output1.txt")));
} catch (Exception e) {
System.err.println(e);
}
FastScanner fs = new FastScanner();
PrintWriter output = new PrintWriter(System.out);
int t = fs.nextInt();
//precalculate primes (2*(int)1e5+1)
int maxN = (int)1e6;
boolean[] notPrime = new boolean[maxN + 1];
notPrime[0] = true;
notPrime[1] = true;
for (int i = 2; i * i <= maxN; i++) {
if (!notPrime[i]) {
for (int j = i * i; j <= maxN; j += i) {
notPrime[j] = true;
}
}
}
while (t-- > 0) {
int[] ne = fs.intArray();
int n = ne[0], e = ne[1];
int[] arr = fs.intArray();
boolean[] visited = new boolean[n + 1];
List<List<Integer>> groups = new ArrayList<>();
for (int i = 0; i < n; i++) {
if (!visited[i]) {
List<Integer> g=new ArrayList<>();
int ones = 0;
for (int j = i; j < n; j += e) {
visited[j]=true;
if (arr[j] != 1 && notPrime[arr[j]])break;
if (arr[j] == 1)ones++;
else if (!notPrime[arr[j]]) {
g.add(ones);
ones = 0;
g.add(-1);
}
}
if(ones>0)g.add(ones);
groups.add(g);
}
}
// System.out.println(groups);
long ans=0;
for(List<Integer> g:groups){
for(int i=0;i<g.size();i++){
if(g.get(i)==-1){
if(i==g.size()-1){
ans+=g.get(i-1);
}else{
ans+=((long)(g.get(i-1)+1))*((long)(g.get(i+1)+1))-1;
}
}
}
}
output.println(ans);
}
// output.println(max);
output.flush();
}
public static void dfs(List<List<Integer>> graph, int src, int par, int[] color, int pc) {
if (color[src - 1] != pc) {
pc = color[src - 1];
cnt++;
}
for (int child : graph.get(src)) {
if (child != par) {
dfs(graph, child, src, color, pc);
}
}
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer("");
}
boolean hasNext() {
return st.hasMoreTokens();
}
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {}
return st.nextToken();
}
String[] nextArray() throws IOException {
return br.readLine().split(" ");
}
int[] intArray() throws IOException {
return Arrays.stream(br.readLine().split("\\s")).mapToInt(Integer::parseInt).toArray();
}
long[] longArray() throws IOException {
return Arrays.stream(br.readLine().split("\\s")).mapToLong(Long::parseLong).toArray();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | e155dab8c1543b39c8eb9bd4b2ebb605 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.util.*;
import java.io.*;
import java.lang.*;
public class Main{
public static void main(String args[]){
InputReader in=new InputReader(System.in);
TASK solver = new TASK();
int t=1;
t = in.nextInt();
for(int i=1;i<=t;i++)
{
solver.solve(in,i);
}
}
static class TASK {
static int mod = 1000000007;
static boolean isPrime[] = new boolean[1000000 + 10];
static {
for (int i = 0; i <= 1000000+9; i++) {
isPrime[i] = true;
}
isPrime[0] = false;
isPrime[1] = false;
for (int i = 2; i * i <= 1000000+9; i++) {
if (isPrime[i] == true) {
for (int j = i * i; j <= 1000000+9; j += i) {
isPrime[j] = false;
}
}
}
}
static void solve(InputReader in, int testNumber) {
int n = in.nextInt();
int e = in.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++)
{
a[i]=in.nextInt();
}
// for(int i=0;i<n;i++)
// {
// System.out.print(isPrime[a[i]]+" ");
// }
// System.out.println();
long dp1[] = new long[n];
long dp2[] = new long[n];
for(int j=0;j<e;j++)
{
for(int i=n-j-1;i>=0;i-=e)
{
if(a[i]==1)
{
dp2[i]=1;
}
if(i+e<n && a[i]==1)
{
dp2[i]+=dp2[i+e];
}
}
int next=n;
for(int i=n-j-1;i>=0;i-=e)
{
if(!isPrime[a[i]]) {
dp1[i] = 0;
}
if(a[i]!=1)
{
next=i;
}
if(i+e<n && isPrime[a[i]])
{
dp1[i]=dp2[i+e];
}
if(next<n && a[i]==1)
{
if(isPrime[a[next]])
{
dp1[i]=dp1[next]+1;
}
}
}
// for(int i=0;i<n;i++)
// {
// System.out.print(dp1[i]+" ");
// }
// System.out.println();
}
long ans=0;
for(int i=0;i<n;i++)
{
ans+=dp1[i];
}
System.out.println(ans);
}
}
static class pair{
int x;
int y;
pair(int x,int y)
{
this.x=x;
this.y=y;
}
}
static class Maths {
static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static long factorial(int n) {
long fact = 1;
for (int i = 1; i <= n; i++) {
fact *= i;
}
return fact;
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
BufferedReader br = new BufferedReader(new
InputStreamReader(System.in));
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c
== -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | db753235f9e3933aaad359a8657e04f4 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.*;
public class Codeforces {
static int mod= 1000000007;
static boolean isp[]=new boolean[1000005];
public static void main(String[] args) throws Exception {
PrintWriter out=new PrintWriter(System.out);
FastScanner fs=new FastScanner();
int t=fs.nextInt();
Arrays.fill(isp, true);
isp[0]=false;
isp[1]=false;
for(int i=2;i<isp.length;i++) {
if(isp[i]) {
for(int j=2*i;j<isp.length;j+=i) {
isp[j]=false;
}
}
}
outer:while(t-->0) {
long ans=0;
int n=fs.nextInt(),e=fs.nextInt();
int arr[]=fs.readArray(n);
long left[]=new long[n+1];
long right[]=new long[n+1];
for(int i=1;i<=n;i++) {
if(arr[i-1]==1) {
left[i]++;
if(i-e-1>=0&&arr[i-e-1]==1) left[i]+=left[i-e];
}
}
for(int i=n-1;i>=0;i--) {
if(arr[i]==1) {
right[i]++;
if(i+e<n&&arr[i+e]==1) right[i]+=right[i+e];
}
}
for(int i=0;i<n;i++) {
long l=((i+1-e>=0) ? left[i+1-e] : 0 ) + 1 ;
long r=((i+e<=n) ? right[i+e] : 0) + 1;
if(isp[arr[i]]) {
// System.out.println(arr[i]+" "+l+" "+r);
ans+= (l*r);
ans--;
}
}
out.println(ans);
}
out.close();
}
static long pow(long a,long b) {
if(b<=0) return 1;
long res=1;
while(b!=0) {
if((b&1)!=0) {
res*=a;
res%=mod;
}
a*=a;
a%=mod;
b=b>>1;
}
return res;
}
static int gcd(int a,int b) {
if(b==0) return a;
return gcd(b,a%b);
}
static long nck(int n,int k) {
if(k>n) return 0;
long res=1;
res*=fact(n);
res%=mod;
res*=modInv(fact(k));
res%=mod;
res*=modInv(fact(n-k));
res%=mod;
return res;
}
static long fact(long n) {
// return fact[(int)n];
long res=1;
for(int i=2;i<=n;i++) {
res*=i;
res%=mod;
}
return res;
}
static long modInv(long n) {
return pow(n,mod-2);
}
static void sort(int[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
int temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
// Use this to input code since it is faster than a Scanner
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long[] lreadArray(int n) {
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=nextLong();
return a;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | f088b81e7a4af775bdf41de54f2a73c9 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | // package faltu;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//import java.security.KeyStore.Entry;
import java.util.*;
import java.util.Map.Entry;
public class Main {
static long LowerBound(long[] a2, long x) { // x is the target value or key
int l=-1,r=a2.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a2[m]>=x) r=m;
else l=m;
}
return r;
}
static int UpperBound(int a[], int x) {// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
public static long getClosest(long val1, long val2,long target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
public static long findClosest(long arr[], long target)
{
int n = arr.length;
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n-1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1], target);
i = mid + 1; // update i
}
}
// Only single element left after search
return arr[mid];
}
static void ruffleSort(long[] a) {
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
long oi=r.nextInt(n), temp=a[i];
a[i]=a[(int)oi];
a[(int)oi]=temp;
}
Arrays.sort(a);
}
static void ruffleSort(int[] a) {
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n), temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
Arrays.sort(a);
}
private int ceilIndex(int input[], int T[], int end, int s){
int start = 0;
int middle;
int len = end;
while(start <= end){
middle = (start + end)/2;
if(middle < len && input[T[middle]] < s && s <= input[T[middle+1]]){
return middle+1;
}else if(input[T[middle]] < s){
start = middle+1;
}else{
end = middle-1;
}
}
return -1;
}
public static int findIndex(long arr[], long t)
{
// if array is Null
if (arr == null) {
return -1;
}
// find length of array
int len = arr.length;
int i = 0;
// traverse in the array
while (i < len) {
if (arr[i] == t) {
return i;
}
else {
i = i + 1;
}
}
return -1;
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b)
{
return (a / gcd(a, b)) * b;
}
public static int[] swap(int a[], int left, int right)
{
int temp = a[left];
a[left] = a[right];
a[right] = temp;
return a;
}
public static void swap(int max0,int max1)
{
int temp=max0;
max0=max1;
max1=temp;
}
public static int[] reverse(int a[], int left, int right)
{
// Reverse the sub-array
while (left < right) {
int temp = a[left];
a[left++] = a[right];
a[right--] = temp;
}
return a;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
// *******----segement tree implement---******
// -------------START--------------------------
void buildTree (int[] arr,int[] tree,int start,int end,int treeNode)
{
if(start==end)
{
tree[treeNode]=arr[start];
return;
}
buildTree(arr,tree,start,end,2*treeNode);
buildTree(arr,tree,start,end,2*treeNode+1);
tree[treeNode]=tree[treeNode*2]+tree[2*treeNode+1];
}
void updateTree(int[] arr,int[] tree,int start,int end,int treeNode,int idx,int value)
{
if(start==end)
{
arr[idx]=value;
tree[treeNode]=value;
return;
}
int mid=(start+end)/2;
if(idx>mid)
{
updateTree(arr,tree,mid+1,end,2*treeNode+1,idx,value);
}
else
{
updateTree(arr,tree,start,mid,2*treeNode,idx,value);
}
tree[treeNode]=tree[2*treeNode]+tree[2*treeNode+1];
}
static int[][] dp;
private static long getClosetNumberOfTarget(long[] dataArray, long target) {
// Edge case - Check if array length is zero
if (dataArray.length == 0)
System.exit(1);
// Straight case - If target is smaller or equal to first element in array
if (target <= dataArray[0])
return dataArray[0];
// Straight case - If target is greater or equal to last element in array
if (target >= dataArray[dataArray.length - 1])
return dataArray[dataArray.length - 1];
// Start binary search algorithm
int start = 0;
int end = dataArray.length-1;
int mid = 0;
// Keep dividing array till further division is not possible
while (end - start != 1) {
// Calculate middle index
mid = (start + end) / 2;
// Check if middle element is equal to target
if (target == dataArray[mid])
// If yes return middle element as middle element = target
return dataArray[mid];
// Check if target is smaller to middle element. If yes, take first half of array including mid
if (target < dataArray[mid])
end = mid;
// Check if target is greater to middle element. If yes, take first half of array including mid
if (target > dataArray[mid])
start = mid;
}
// Now you will have only two numbers in array in which target number will fall
return Math.abs(target - dataArray[start]) <= Math.abs(target - dataArray[end]) ? dataArray[start]
: dataArray[end];
}
private static int getClosetNumberOfTarget(int[] dataArray, int target) {
// Edge case - Check if array length is zero
if (dataArray.length == 0)
System.exit(1);
// Straight case - If target is smaller or equal to first element in array
if (target <= dataArray[0])
return dataArray[0];
// Straight case - If target is greater or equal to last element in array
if (target >= dataArray[dataArray.length - 1])
return dataArray[dataArray.length - 1];
// Start binary search algorithm
int start = 0;
int end = dataArray.length - 1;
int mid = 0;
// Keep dividing array till further division is not possible
while (start < end) {
// Calculate middle index
mid = (start + end) / 2;
// Check if middle element is equal to target
if (target == dataArray[mid])
// If yes return middle element as middle element = target
return dataArray[mid];
// Check if target is smaller to middle element. If yes, take first half of
// array including mid
if (target < dataArray[mid]) {
// If target is greater than previous element of middle element, we have reached the deciding range
if (target > dataArray[mid - 1])
return Math.abs(target - dataArray[mid]) >= Math.abs(target - dataArray[mid - 1]) ? dataArray[mid-1]
: dataArray[mid];
end = mid - 1;
}
// Check if target is greater to middle element. If yes, take first half of
// array including mid
if (target > dataArray[mid]) {
// If target is smaller than next element of middle element, we have reached the deciding range
if (target < dataArray[mid + 1])
return Math.abs(target - dataArray[mid]) <= Math.abs(target - dataArray[mid + 1]) ? dataArray[mid]
: dataArray[mid + 1];
start = mid + 1;
}
}
// after exiting form loop, middle element will be closer
return dataArray[mid];
}
// disjoint set implementation --start
static void makeSet(int n)
{
parent=new int[n];
rank=new int[n];
for(int i=0;i<n;i++)
{
parent[i]=i;
rank[i]=0;
}
}
static void union(int u,int v)
{
u=findpar(u);
v=findpar(v);
if(rank[u]<rank[v])parent[u]=v;
else if(rank[v]<rank[u])parent[v]=u;
else
{
// System.out.println(v+" "+parent[v]);
parent[v]=u;
// System.out.println(v+" "+parent[v]);
rank[u]++;
}
}
private static int findpar(int node)
{
if(node==parent[node])return node;
return parent[node]=findpar(parent[node]);
}
static int parent[];
static int rank[];
// *************end
static Long MOD=(long) (1e9+7);
private static int lps(int m ,int n,String s1,String s2,int[][]mat)
{
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(s1.charAt(i-1)==s2.charAt(j-1))mat[i][j]=1+mat[i-1][j-1];
else mat[i][j]=Math.max(mat[i-1][j],mat[i][j-1]);
}
}
return mat[m][n];
}
static long power1(long a,long b) {
if(b == 0){
return 1;
}
long ans = power(a,b/2);
ans *= ans;
if(b % 2!=0){
ans *= a;
}
return ans;
}
public static void main(String[] args)
{
FastReader s = new FastReader();
dp=new int[1001][1001];
sieve();
long tt = s.nextLong();
while (tt-->0)
{
long ans=0;
int n=s.nextInt();
int e=s.nextInt();
long[] a=new long[n];
for(int i=0;i<n;i++)a[i]=s.nextLong();
for(int i=0;i<n;i++)
{
long l=0,r=0;
if(sieve[(int) a[i]]==1&&a[i]!=1)
{
for(int j=i+e;j<n;j+=e)
{
if(a[j]==1)r++;
else break;
}
for(int j=i-e;j>=0;j-=e)
{
if(a[j]==1)l++;
else break;
}
}
ans+=((l+r)+(l*r));
}
System.out.println(ans);
}
}
static String lcs(String X, String Y, int m, int n)
{
int[][] L = new int[m+1][n+1];
// Following steps build L[m+1][n+1] in bottom up fashion. Note
// that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i-1) == Y.charAt(j-1))
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
}
}
// Following code is used to print LCS
int index = L[m][n];
int temp = index;
// Create a character array to store the lcs string
char[] lcs = new char[index+1];
lcs[index] = '\u0000'; // Set the terminating character
// Start from the right-most-bottom-most corner and
// one by one store characters in lcs[]
int i = m;
int j = n;
while (i > 0 && j > 0)
{
// If current character in X[] and Y are same, then
// current character is part of LCS
if (X.charAt(i-1) == Y.charAt(j-1))
{
// Put current character in result
lcs[index-1] = X.charAt(i-1);
// reduce values of i, j and index
i--;
j--;
index--;
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (L[i-1][j] > L[i][j-1])
i--;
else
j--;
}
return String.valueOf(lcs);
// Print the lcs
// System.out.print("LCS of "+X+" and "+Y+" is ");
// for(int k=0;k<=temp;k++)
// System.out.print(lcs[k]);
}
static long lis(long[] aa2, int n)
{
long lis[] = new long[n];
int i, j;
long max = 0;
for (i = 0; i < n; i++)
lis[i] = 1;
for (i = 1; i < n; i++)
for (j = 0; j < i; j++)
if (aa2[i] >= aa2[j] && lis[i] <= lis[j] + 1)
lis[i] = lis[j] + 1;
for (i = 0; i < n; i++)
if (max < lis[i])
max = lis[i];
return max;
}
static boolean isPalindrome(String str)
{
int i = 0, j = str.length() - 1;
while (i < j) {
if (str.charAt(i) != str.charAt(j))
return false;
i++;
j--;
}
return true;
}
static boolean issafe(int i, int j, int r,int c, char ch)
{
if (i < 0 || j < 0 || i >= r || j >= c|| ch!= '1')return false;
else return true;
}
static long power(long a, long b)
{
a %=MOD;
long out = 1;
while (b > 0) {
if((b&1)!=0)out = out * a % MOD;
a = a * a % MOD;
b >>= 1;
a*=a;
}
return out;
}
static long[] sieve;
public static void sieve()
{
int nnn=(int) 1e6+1;
long nn=(int) 1e6;
sieve=new long[(int) nnn];
int[] freq=new int[(int) nnn];
sieve[0]=0;
sieve[1]=1;
for(int i=2;i<=nn;i++)
{
sieve[i]=1;
freq[i]=1;
}
for(int i=2;i*i<=nn;i++)
{
if(sieve[i]==1)
{
for(int j=i*i;j<=nn;j+=i)
{
if(sieve[j]==1)
{
sieve[j]=0;
}
}
}
}
}
// static long power(long x, long y)
// {
// long res = 1; // Initialize result
//
// while (y > 0)
// {
//
// // If y is odd, multiply x with result
// if ((y & 1) != 0)
// res = res * x;
//
// // y must be even now
// y = y >> 1; // y = y/2
// x = x * x; // Change x to x^2
// }
// return res;
// }
}
class pair{
long x,y;
public pair(long x,long y)
{
this.x=x;
this.y=y;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | f7ef8a36d473cf7a57e0e85937fd1889 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class C_Complex_Market_Analysis{
public static void main (String[] args){
final int MAXN = 1000001;
FastReader s = new FastReader();
int t=1;t=s.ni();
boolean[] isPrime = new boolean[MAXN];
Arrays.fill(isPrime, true);
isPrime[0] = isPrime[1] = false;
for (int i = 2; i < MAXN; i++) {
for (int j = i * 2; j < MAXN && isPrime[i]; j += i) {
isPrime[j] = false;
}
}
for(int test=1;test<=t;test++){
int n=s.ni(),e=s.ni();int ans=0;
int a[]=s.readArray(n);
long ret = 0;
for (int i = 0; i < e; i++) {
ArrayList<Integer> pos = new ArrayList<>();
ArrayList<Integer> data = new ArrayList<>();
pos.add(-1);
data.add(-1);
int p = 0;
for (int j = i; j < n; j += e) {
if (a[j] != 1) {
pos.add(p);
data.add(a[j]);
}
p++;
}
pos.add(p);
data.add(p);
for (int j = 1; j < pos.size() - 1; j++) {
long left = (long) pos.get(j) - pos.get(j - 1);
long right = (long) pos.get(j + 1) - pos.get(j);
if (isPrime[data.get(j)])
ret += left * right - 1;
}
}
System.out.println(ret);
}
}
static class FastReader {
BufferedReader br; StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));}
int ni() { return Integer.parseInt(next()); }
int[] readArray(int n){
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] =Integer.parseInt(next());
return a;
}
String next(){
while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());}
catch (IOException e){e.printStackTrace();}}return st.nextToken();}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | fbc425f8482715b4b5b98cb3c7eb6979 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
public class Solutionc {
static boolean isPrime(int num) {
if(num == 1) return false;
if(num == 2) return true;
for(int i=2;i*i<=num;++i) {
if(num%i == 0) return false;
}
return true;
}
public static void main(String arg[]) throws IOException {
CustomInputReader sc = new CustomInputReader();
PrintWriter pw = new PrintWriter(System.out);
int max = (int)1e6 + 5;
boolean isPrime[] = new boolean[max];
Arrays.fill(isPrime, true);
isPrime[1] = false;
for(int i=2;i<max;++i) {
if(isPrime[i]) {
for(int j=i*2;j<max;j+=i) {
isPrime[j] = false;
}
}
}
int tc = sc.nextInt();
while(tc-->0) {
int in[] = sc.nextIntArr(2);
int n = in[0], e = in[1];
long ans = 0;
int a[] = sc.nextIntArr(n);
for(int i=0;i<n;++i) {
if(isPrime[a[i]]) {
long left = 0, right = 0;
for(int j=i-e;j>=0;j-=e) {
if(a[j] == 1) left++;
else break;
}
for(int j=i+e;j<n;j+=e) {
if(a[j] == 1) right++;
else break;
}
ans += (left * (right+1));
ans += right;
}
}
pw.println(ans);
}
//flush output
pw.flush();
pw.close();
}
static class CustomInputReader {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public int nextInt() throws IOException {
return Integer.parseInt(br.readLine());
}
public int[] nextIntArr(int n) throws IOException {
String in[] = br.readLine().split(" ");
int arr[] = new int[n];
for(int i=0;i<n;++i) arr[i]=Integer.parseInt(in[i]);
return arr;
}
public long nextLong() throws IOException {
return Long.parseLong(br.readLine());
}
public long[] nextLongArr(int n) throws IOException {
String in[] = br.readLine().split(" ");
long arr[] = new long[n];
for(int i=0;i<n;++i) arr[i]=Long.parseLong(in[i]);
return arr;
}
public String next() throws IOException {
return br.readLine();
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | c536de6278f81dcdee3949d86a3582bd | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.Scanner;
public class PS1609C {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int cases = Integer.parseInt(scanner.next());
for (int i = 0; i < cases; i++) {
int n = Integer.parseInt(scanner.next());
int e = Integer.parseInt(scanner.next());
int[] arr = new int[n];
for (int j = 0; j < n; j++) {
arr[j] = Integer.parseInt(scanner.next());
}
System.out.println(solve(e, arr));
}
}
private static long solve(int e, int[] arr) {
long[] a = new long[arr.length]; // pairs with current index as i
int[] b = new int[arr.length]; // consecutive 1's AFTER this entry
boolean[] c = new boolean[arr.length]; // is prime or not
long sum = 0;
for (int i = arr.length - 1; i >= 0; i--) {
c[i] = isPrime(arr[i]);
if (i >= arr.length - e) {
a[i] = b[i] = 0;
} else {
b[i] = arr[i + e] == 1 ? b[i + e] + 1 : 0;
if (arr[i] == 1) {
if (arr[i + e] == 1) {
a[i] = a[i + e];
} else {
a[i] = a[i + e] > 0 ? a[i + e] + 1 : (c[i + e] ? 1 : 0);
}
} else if (isPrime(arr[i])) {
if (arr[i + e] == 1) {
a[i] = b[i];
} else {
a[i] = 0;
}
} else {
a[i] = 0;
}
}
sum += a[i];
}
return sum;
}
static byte[] cache = new byte[1000001];
private static boolean isPrime(int n) {
if (n <= 1) return false;
if (cache[n] != 0) {
return cache[n] == 1;
}
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
cache[n] = -1;
return false;
}
}
cache[n] = 1;
return true;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 0cb2f62d86d8e482f97019ecd7af05af | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class C{
public static void main(String[] args){
FastReader sc = new FastReader();
boolean prime[]=sieve_of_Eratosthenes(1000001);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int e=sc.nextInt();
int a[]=sc.fastArray(n);
int p[]=new int [n];
int s[]=new int [n];
for(int i=0;i<n;i++) {
if(a[i]==1) {
p[i]=i-e>=0?p[i-e]+1:1;
}
}
for(int i=n-1;i>=0;i--) {
if(a[i]==1) {
s[i]=i+e<n?s[i+e]+1:1;
}
}
// print(p);
// print(s);
long ans=0;
for(int i=0;i<n;i++) {
if(prime[a[i]]) {
int pr=0;
int sa=0;
if(i-e>=0) {
pr=p[i-e];
ans+=(long)pr;
}
if(i+e<n) {
sa=s[i+e];
ans+=(long)sa;
}
// System.out.println(a[i]+" "+pr+" "+sa);
ans+=((long)pr*(long)sa);
}
}
System.out.println(ans);
}
}
static class pair {
int x;int y;
pair(int x,int y){
this.x=x;
this.y=y;
}
}
static void sort(int a[]) {
ArrayList<Integer> aa= new ArrayList<>();
for(int i:a)aa.add(i);
Collections.sort(aa);
int j=0;
for(int i:aa)a[j++]=i;
}
static ArrayList<Integer> primeFac(int n){
ArrayList<Integer>ans = new ArrayList<Integer>();
int lp[]=new int [n+1];
Arrays.fill(lp, 0); //0-prime
for(int i=2;i<=n;i++) {
if(lp[i]==0) {
for(int j=i;j<=n;j+=i) {
if(lp[j]==0) lp[j]=i;
}
}
}
int fac=n;
while(fac>1) {
ans.add(lp[fac]);
fac=fac/lp[fac];
}
print(ans);
return ans;
}
static ArrayList<Long> prime_in_given_range(long l,long r){
ArrayList<Long> ans= new ArrayList<>();
int n=(int)Math.sqrt(r)+1;
boolean prime[]=sieve_of_Eratosthenes(n);
long res[]=new long [(int)(r-l)+1];
for(int i=0;i<=r-l;i++) {
res[i]=i+l;
}
for(int i=0;i<prime.length;i++) {
if(prime[i]) {
System.out.println(2);
for(int j=Math.max((int)i*i, (int)(l+i-1)/i*i);j<=r;j+=i) {
res[j-(int)l]=0;
}
}
}
for(long i:res) if(i!=0)ans.add(i);
return ans;
}
static boolean [] sieve_of_Eratosthenes(int n) {
boolean prime[]=new boolean [n];
Arrays.fill(prime, true); // 1-prime | 0-not prime
prime[0]=prime[1]=false;
for(int i=2;i<n;i++) {
if(prime[i]==true) {
for(long j=(long)i*(long)i;j<n;j+=i) {
int k=(int)j;
prime[k]=false;
}
}
}
return prime;
}
static long binpow(long a,long b) {
long res=1;
if(b==0)return a;
if(a==0)return 1;
while(b>0) {
if((b&1)==1) {
res*=a;
}
a*=a;
b>>=1;
}
return res;
}
static void print(int a[]) {
System.out.println(a.length);
for(int i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long a[]) {
System.out.println(a.length);
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(ArrayList<Integer> a) {
System.out.println(a.size());
for(long i:a) {
System.out.print(i+" ");
}
System.out.println();
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
int [] fastArray(int n) {
int a[]=new int [n];
for(int i=0;i<n;i++) {
a[i]=nextInt();
}
return a;
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 5dad0f7e6145ef4e8cbf02d638a45532 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class CodeForces{
/*-------------------------------------------EDITING CODE STARTS HERE-------------------------------------------*/
public static void main(String[] args) throws IOException{
openIO();
int testCase = 1;
testCase = sc.nextInt();
preCompute();
for (int i = 1; i <= testCase; i++) solve(i);
closeIO();
}
public static void solve(int tCase)throws IOException {
int n = sc.nextInt();
int e = sc.nextInt();
int[] arr = new int[n];
long sum = 0;
for(int i = 0;i<n;i++)arr[i] = sc.nextInt();
for(int i = 0;i<e;i++){
int oneCount = 0;
List<Integer> list = new ArrayList<>();
for(int j = i ;j<n;j+=e){
if(arr[j] == 1){
oneCount++;
}else {
list.add(oneCount);
list.add(arr[j]);
oneCount = 0;
}
}
list.add(oneCount);
sum += find(list);
}
out.println(sum);
}
static int[] seive = _seive(2000000);
private static long find(List<Integer> list){
long sum = 0;
for(int i=1;i<list.size()-1;i+=2){
if(seive[list.get(i)] !=list.get(i))continue;
int oneBefore = list.get(i-1);
int oneAfter = list.get(i+1);
sum+=(oneBefore + 1L) * (oneAfter+1) - 1;
}
return sum;
}
private static void preCompute(){}
/*-------------------------------------------EDITING CODE ENDS HERE-------------------------------------------*/
static FastestReader sc;
static PrintWriter out;
private static void openIO() throws IOException{
sc = new FastestReader();
out = new PrintWriter(System.out);
}
/*------------------------------------------HELPER FUNCTION STARTS HERE------------------------------------------*/
public static final int mod = (int) 1e9 +7;
private static final int mod2 = 998244353;
public static final int inf_int = (int) 2e9 + 10;
public static final long inf_long = (long) 4e18;
public static int _digitCount(long num,int base){
return (int)(1 + Math.log(num)/Math.log(base));
}
// time : O(r)
public static int _ncr(int n,int r){
if (r > n) return 0;
long[] inv = new long[r + 1];
inv[0] = 1;
if(r+1>=2)
inv[1] = 1;
// Getting the modular inversion
// for all the numbers
// from 2 to r with respect to mod
for (int i = 2; i <= r; i++)
inv[i] = mod - (mod / i) * inv[(mod % i)] % mod;
int ans = 1;
// for 1/(r!) part
for (int i = 2; i <= r; i++)
ans = (int) (((ans % mod) * (inv[i] % mod)) % mod);
// for (n)*(n-1)*(n-2)*...*(n-r+1) part
for (int i = n; i >= (n - r + 1); i--)
ans = (((ans % mod) * (i % mod)) % mod);
return ans;
}
// euclidean algorithm time O(max (loga ,logb))
public static long _gcd(long a, long b) {
if (a == 0)
return b;
return _gcd(b % a, a);
}
public static long _lcm(long a, long b) {
// lcm(a,b) * gcd(a,b) = a * b
return (a / _gcd(a, b)) * b;
}
// binary exponentiation time O(logn)
public static long _power(long x, long n) {
long ans = 1;
while (n > 0) {
if ((n & 1) == 1) {
ans *= x;
ans %= mod;
n--;
} else {
x *= x;
x %= mod;
n >>= 1;
}
}
return ans;
}
//sieve/first divisor time : O(mx * log ( log (mx) ) )
public static int[] _seive(int mx){
int[] firstDivisor = new int[mx+1];
for(int i=0;i<=mx;i++)firstDivisor[i] = i;
for(int i=2;i*i<=mx;i++)
if(firstDivisor[i] == i)
for(int j = i*i;j<=mx;j+=i)
firstDivisor[j] = i;
return firstDivisor;
}
private static boolean _isPrime(long x){
for(long i=2;i*i<=x;i++)
if(x%i==0)return false;
return true;
}
public static void _sort(int[] arr,boolean isAscending){
int n = arr.length;
List<Integer> list = new ArrayList<>();
for(int ele : arr)list.add(ele);
Collections.sort(list);
if(!isAscending)Collections.reverse(list);
for(int i=0;i<n;i++)arr[i] = list.get(i);
}
public static void _sort(long[] arr,boolean isAscending){
int n = arr.length;
List<Long> list = new ArrayList<>();
for(long ele : arr)list.add(ele);
Collections.sort(list);
if(!isAscending)Collections.reverse(list);
for(int i=0;i<n;i++)arr[i] = list.get(i);
}
/*------------------------------------------HELPER FUNCTION ENDS HERE-------------------------------------------*/
/*-------------------------------------------FAST INPUT STARTS HERE---------------------------------------------*/
public static void closeIO() throws IOException{
out.flush();
out.close();
sc.close();
}
private static final class FastestReader {
private static final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public FastestReader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastestReader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private int skip() throws IOException {
int b;
//noinspection StatementWithEmptyBody
while ((b = read()) != -1 && isSpaceChar(b)) {}
return b;
}
public String next() throws IOException {
int b = skip();
final StringBuilder sb = new StringBuilder();
while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = read();
}
return sb.toString();
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) { c = read(); }
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) { return -ret; }
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') { c = read(); }
final boolean neg = c == '-';
if (neg) { c = read(); }
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) { return -ret; }
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') { c = read(); }
final boolean neg = c == '-';
if (neg) { c = read(); }
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) { return -ret; }
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) { buffer[0] = -1; }
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) { fillBuffer(); }
return buffer[bufferPointer++];
}
public void close() throws IOException {
din.close();
}
}
/*---------------------------------------------FAST INPUT ENDS HERE ---------------------------------------------*/
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 4457841d8f1fb81fb3c42169d7e8401b | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
public class c {
// static int mod = 998244353;
static int mod = 1000000007;
public static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String[] args) throws Exception {
Reader scn = new Reader();
PrintWriter pw = new PrintWriter(System.out);
int t = scn.nextInt();
outer : while(t-->0){
int n = scn.nextInt();
int e = scn.nextInt();
int[] arr = new int[n];
for(int i=0; i<n; i++){
arr[i] = scn.nextInt();
}
Set<Integer> hs = new HashSet<>();
for(int i=0; i<n; i++){
boolean res = isPrime(arr[i]);
if(res){
hs.add(i);
}
}
// for(int val : primes.keySet()){
// pw.print(val + " ");
// }
// pw.println();
if(hs.size() == 0){
pw.println(0);
continue outer;
}
HashMap<Integer, Integer> onesback = new HashMap<>();
int i = n-1;
while(i >= 0){
if(arr[i] == 1 && onesback.containsKey(i) == false){
int count = 1, j = i;
while(j >= 0 && arr[j] == 1){
onesback.put(j, count);
count++;
j -= e;
}
i--;
}else{
i--;
}
}
HashMap<Integer, Integer> onesfwd = new HashMap<>();
i = 0;
while(i < n){
if(arr[i] == 1 && onesfwd.containsKey(i) == false){
int count = 1, j = i;
while(j < n && arr[j] == 1){
onesfwd.put(j, count);
count++;
j += e;
}
i++;
}else{
i++;
}
}
long ans = 0L;
for(int val : hs){
int index1 = val - e;
if(onesfwd.containsKey(index1)){
ans += onesfwd.get(index1);
}
int index2 = val + e;
if(onesback.containsKey(index2)){
ans += onesback.get(index2);
}
if(onesfwd.containsKey(index1) && onesback.containsKey(index2)){
ans += ((long)(onesfwd.get(index1) * (long)onesback.get(index2)));
}
}
pw.println(ans);
}
pw.close();
}
private static boolean isPrime(int n){
if(n == 1){
return false;
}
for(int div=2; div*div<=n; div++){
if(n % div == 0){
return false;
}
}
return true;
}
public static class Pair{
int x;
int y;
Pair(int x, int y){
this.x = x;
this.y = y;
}
}
private static void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
private static void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for(int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
// private static void sort(Pair[] arr) {
// List<Pair> list = new ArrayList<>();
// for(int i=0; i<arr.length; i++){
// list.add(arr[i]);
// }
// Collections.sort(list); // collections.sort uses nlogn in backend
// for (int i = 0; i < arr.length; i++){
// arr[i] = list.get(i);
// }
// }
private static void reverseSort(int[] arr){
List<Integer> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list, Collections.reverseOrder()); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
private static void reverseSort(long[] arr){
List<Long> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list, Collections.reverseOrder()); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
private static void reverseSort(ArrayList<Integer> list){
Collections.sort(list, Collections.reverseOrder());
}
private static int lowerBound(int[] arr, int x){
int n = arr.length, si = 0, ei = n - 1;
while(si <= ei){
int mid = si + (ei - si)/2;
if(arr[mid] == x){
if(mid-1 >= 0 && arr[mid-1] == arr[mid]){
ei = mid-1;
}else{
return mid;
}
}else if(arr[mid] > x){
ei = mid - 1;
}else{
si = mid+1;
}
}
return si;
}
private static int upperBound(int[] arr, int x){
int n = arr.length, si = 0, ei = n - 1;
while(si <= ei){
int mid = si + (ei - si)/2;
if(arr[mid] == x){
if(mid+1 < n && arr[mid+1] == arr[mid]){
si = mid+1;
}else{
return mid + 1;
}
}else if(arr[mid] > x){
ei = mid - 1;
}else{
si = mid+1;
}
}
return si;
}
private static int upperBound(ArrayList<Integer> list, int x){
int n = list.size(), si = 0, ei = n - 1;
while(si <= ei){
int mid = si + (ei - si)/2;
if(list.get(mid) == x){
if(mid+1 < n && list.get(mid+1) == list.get(mid)){
si = mid+1;
}else{
return mid + 1;
}
}else if(list.get(mid) > x){
ei = mid - 1;
}else{
si = mid+1;
}
}
return si;
}
// (x^y)%p in O(logy)
private static long power(long x, long y){
long res = 1;
x = x % mod;
while(y > 0){
if ((y & 1) == 1){
res = (res * x) % mod;
}
y = y >> 1;
x = (x * x) % mod;
}
return res;
}
public static boolean nextPermutation(int[] arr) {
if(arr == null || arr.length <= 1){
return false;
}
int last = arr.length-2;
while(last >= 0){
if(arr[last] < arr[last+1]){
break;
}
last--;
}
if (last < 0){
return false;
}
if(last >= 0){
int nextGreater = arr.length-1;
for(int i=arr.length-1; i>last; i--){
if(arr[i] > arr[last]){
nextGreater = i;
break;
}
}
swap(arr, last, nextGreater);
}
reverse(arr, last+1, arr.length-1);
return true;
}
private static void swap(int[] arr, int i, int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
private static void reverse(int[] arr, int i, int j){
while(i < j){
swap(arr, i++, j--);
}
}
private static String reverseStr(String s){
StringBuilder sb = new StringBuilder(s);
return sb.reverse().toString();
}
// TC- O(logmax(a,b))
private static int gcd(int a, int b) {
if(a == 0){
return b;
}
return gcd(b%a, a);
}
private static long gcd(long a, long b) {
if(a == 0L){
return b;
}
return gcd(b%a, a);
}
private static long lcm(long a, long b) {
return a / gcd(a, b) * b;
}
// TC- O(logmax(a,b))
private static int lcm(int a, int b) {
return a / gcd(a, b) * b;
}
private static long inv(long x){
return power(x, mod - 2);
}
private static long summation(long n){
return (n * (n + 1L)) >> 1;
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 8b647fdbd51784a2332e9253f55831cb | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
// static int mod = 998244353 ;
// static int N = 200005;
// static long factorial_num_inv[] = new long[N+1];
// static long natual_num_inv[] = new long[N+1];
// static long fact[] = new long[N+1];
// static void InverseofNumber()
//{
// natual_num_inv[0] = 1;
// natual_num_inv[1] = 1;
// for (int i = 2; i <= N; i++)
// natual_num_inv[i] = natual_num_inv[mod % i] * (mod - mod / i) % mod;
//}
//static void InverseofFactorial()
//{
// factorial_num_inv[0] = factorial_num_inv[1] = 1;
// for (int i = 2; i <= N; i++)
// factorial_num_inv[i] = (natual_num_inv[i] * factorial_num_inv[i - 1]) % mod;
//}
//static long nCrModP(long N, long R)
//{
// long ans = ((fact[(int)N] * factorial_num_inv[(int)R]) % mod * factorial_num_inv[(int)(N - R)]) % mod;
// return ans%mod;
//}
//static boolean prime[];
//static void sieveOfEratosthenes(int n)
//{
// prime = new boolean[n+1];
// for (int i = 0; i <= n; i++)
// prime[i] = true;
// for (int p = 2; p * p <= n; p++)
// {
// // If prime[p] is not changed, then it is a
// // prime
// if (prime[p] == true)
// {
// // Update all multiples of p
// for (int i = p * p; i <= n; i += p)
// prime[i] = false;
// }
// }
//}
static HashSet<Integer> hs = new HashSet<>();
public static void main (String[] args) throws java.lang.Exception
{
// InverseofNumber();
// InverseofFactorial();
// fact[0] = 1;
// for (long i = 1; i <= 2*100000; i++)
// {
// fact[(int)i] = (fact[(int)i - 1] * i) % mod;
// }
FastReader scan = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
int t = scan.nextInt();
sieveOfEratosthenes(1000000);
while(t-->0){
long n = scan.nextLong();
long e = scan.nextLong();
long a[] = new long[(int)n];
for(int i=0;i<(int)n;i++){
a[i] = scan.nextLong();
}
long prefix[] = new long[(int)n];
long suffix[] = new long[(int)n];
for(int i=0;i<(int)n;i++){
if(a[i]==1){
if((i-(int)e)>=0){
prefix[i] = 1+prefix[i-(int)e];
}
else
prefix[i] = 1;
}
}
for(int i=(int)n-1;i>=0;i--){
if(a[i]==1){
if((i+(int)e)<n){
suffix[i] = 1+suffix[i+(int)e];
}
else
suffix[i] = 1;
}
}
long count = 0;
for(int i=0;i<(int)n;i++){
if(hs.contains((int)a[i])){
long count1=0,count2=0;
if((i-(int)e)>=0){
count1 = prefix[i-(int)e];
}
if((i+(int)e)<(int)n){
count2 = suffix[i+(int)e];
}
count = count+count1+count2+count1*count2;
}
}
pw.println(count);
pw.flush();}
}
// static boolean isPrime(long n){
// if(n==1)
// return false;
// for(long i=2;i<=(long)Math.sqrt(n);i++){
// if(n%i==0)
// return false;
// }
// return true;
// }
static void sieveOfEratosthenes(int n)
{
// Create a boolean array
// "prime[0..n]" and
// initialize all entries
// it as true. A value in
// prime[i] will finally be
// false if i is Not a
// prime, else true.
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
hs.add(i);
}
}
//static long bin_exp_mod(long a,long n){
// long res = 1;
// if(a==0)
// return 0;
// while(n!=0){
// if(n%2==1){
// res = ((res)*(a));
// }
// n = n/2;
// a = ((a)*(a));
// }
// return res;
//}
//static long bin_exp_mod(long a,long n){
// long mod = 998244353;
// long res = 1;
// a = a%mod;
// if(a==0)
// return 0;
// while(n!=0){
// if(n%2==1){
// res = ((res%mod)*(a%mod))%mod;
// }
// n = n/2;
// a = ((a%mod)*(a%mod))%mod;
// }
// res = res%mod;
// return res;
//}
// static long gcd(long a,long b){
// if(a==0)
// return b;
// return gcd(b%a,a);
// }
// static long lcm(long a,long b){
// return (a/gcd(a,b))*b;
// }
}
//class Pair{
// Integer x,y;
// Pair(int x,int y){
// this.x = x;
// this.y = y;
// }
// public boolean equals(Object obj) {
// // TODO Auto-generated method stub
// if(obj instanceof Pair)
// {
// Pair temp = (Pair) obj;
// if(this.x.equals(temp.x) && this.y.equals(temp.y))
// return true;
// }
// return false;
// }
// @Override
// public int hashCode() {
// // TODO Auto-generated method stub
// return (this.x.hashCode() + this.y.hashCode());
// }
//}
//class Compar implements Comparator<Pair>{
// public int compare(Pair p1,Pair p2){
// return p1.x-p2.x;
// }
//}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 91179b521ec67f442ca11a85ca2b2884 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.Arrays;
import java.util.Random;
import java.io.FileWriter;
import java.io.PrintWriter;
/*
Solution Created: 18:00:58 28/11/2021
Custom Competitive programming helper.
*/
public class Main {
public static void solve() {
int n = in.nextInt(), e = in.nextInt();
int[] a = in.na(n);
int[] state = new int[n]; //0 = nothing, 1 = have 1, 2 = have prime, 3 = have both
int[] aces = new int[n]; //0 = nothing, 1 = have 1, 2 = have prime, 3 = have both
long[] gain = new long[n];
long ans = 0;
for(int i = 0; i<n; i++) {
if(a[i]==1) {
if(i-e>=0) {
aces[i] = aces[i-e] + 1;
if(state[i-e]==0) { //nothing
state[i] = 1;
gain[i] = 1;
}else if(state[i-e] == 1) { //we have 1
state[i] = 1;
gain[i] = gain[i-e] + 1;
}else if(state[i-e] == 2) { // we have prime
state[i] = 3;
gain[i] = gain[i-e];
ans += gain[i];
}else { //we have both
state[i] = 3;
gain[i] = gain[i-e];
ans += gain[i];
}
}else {
state[i] = 1;
gain[i] = 1;
aces[i] = 1;
}
}else if(Util.isPrime(a[i])) {
if(i-e>=0) {
if(state[i-e]==0) { //nothing
state[i] = 2;
gain[i] = 1;
}else if(state[i-e] == 1) { //we have 1
state[i] = 3;
gain[i] = gain[i-e] + 1;
ans += gain[i] - 1;
}else if(state[i-e] == 2) { // we have prime
state[i] = 2;
gain[i] = 1;
}else { //we have both
if(a[i-e]==1) {
state[i] = 3;
gain[i] = aces[i-e] + 1;
ans+= gain[i] - 1;
}
else {
state[i] = 2;
gain[i] = 1;
}
}
}else {
state[i] = 2;
gain[i] = 1;
}
}
}
out.println(ans);
}
public static void main(String[] args) {
in = new Reader();
out = new Writer();
int t = in.nextInt();
while(t-->0) solve();
out.exit();
}
static Reader in; static Writer out;
static class Reader {
static BufferedReader br;
static StringTokenizer st;
public Reader() {
this.br = new BufferedReader(new InputStreamReader(System.in));
}
public Reader(String f){
try {
this.br = new BufferedReader(new FileReader(f));
} catch (IOException e) {
e.printStackTrace();
}
}
public int[] na(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
public double[] nd(int n) {
double[] a = new double[n];
for (int i = 0; i < n; i++) a[i] = nextDouble();
return a;
}
public long[] nl(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
public char[] nca() {
return next().toCharArray();
}
public String[] ns(int n) {
String[] a = new String[n];
for (int i = 0; i < n; i++) a[i] = next();
return a;
}
public int nextInt() {
ensureNext();
return Integer.parseInt(st.nextToken());
}
public double nextDouble() {
ensureNext();
return Double.parseDouble(st.nextToken());
}
public Long nextLong() {
ensureNext();
return Long.parseLong(st.nextToken());
}
public String next() {
ensureNext();
return st.nextToken();
}
public String nextLine() {
try {
return br.readLine();
} catch (Exception e) {
e.printStackTrace();
return null;
}
}
private void ensureNext() {
if (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
static class Util{
private static Random random = new Random();
static long[] fact;
public static void initFactorial(int n, long mod) {
fact = new long[n+1];
fact[0] = 1;
for (int i = 1; i < n+1; i++) fact[i] = (fact[i - 1] * i) % mod;
}
public static long modInverse(long a, long MOD) {
long[] gcdE = gcdExtended(a, MOD);
if (gcdE[0] != 1) return -1; // Inverted doesn't exist
long x = gcdE[1];
return (x % MOD + MOD) % MOD;
}
public static long[] gcdExtended(long p, long q) {
if (q == 0) return new long[] { p, 1, 0 };
long[] vals = gcdExtended(q, p % q);
long tmp = vals[2];
vals[2] = vals[1] - (p / q) * vals[2];
vals[1] = tmp;
return vals;
}
public static long nCr(int n, int r, long MOD) {
if (r == 0) return 1;
return (fact[n] * modInverse(fact[r], MOD) % MOD * modInverse(fact[n - r], MOD) % MOD) % MOD;
}
public static long nCr(int n, int r) {
return (fact[n]/fact[r])/fact[n-r];
}
public static long nPr(int n, int r, long MOD) {
if (r == 0) return 1;
return (fact[n] * modInverse(fact[n - r], MOD) % MOD) % MOD;
}
public static long nPr(int n, int r) {
return fact[n]/fact[n-r];
}
public static boolean isPrime(int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static boolean[] getSieve(int n) {
boolean[] isPrime = new boolean[n+1];
for (int i = 2; i <= n; i++) isPrime[i] = true;
for (int i = 2; i*i <= n; i++) if (isPrime[i])
for (int j = i; i*j <= n; j++) isPrime[i*j] = false;
return isPrime;
}
static long pow(long x, long pow, long mod){
long res = 1;
x = x % mod;
if (x == 0) return 0;
while (pow > 0){
if ((pow & 1) != 0) res = (res * x) % mod;
pow >>= 1;
x = (x * x) % mod;
}
return res;
}
public static int gcd(int a, int b) {
int tmp = 0;
while(b != 0) {
tmp = b;
b = a%b;
a = tmp;
}
return a;
}
public static long gcd(long a, long b) {
long tmp = 0;
while(b != 0) {
tmp = b;
b = a%b;
a = tmp;
}
return a;
}
public static int random(int min, int max) {
return random.nextInt(max-min+1)+min;
}
public static void dbg(Object... o) {
System.out.println(Arrays.deepToString(o));
}
public static void reverse(int[] s, int l , int r) {
for(int i = l; i<=(l+r)/2; i++) {
int tmp = s[i];
s[i] = s[r+l-i];
s[r+l-i] = tmp;
}
}
public static void reverse(int[] s) {
reverse(s, 0, s.length-1);
}
public static void reverse(long[] s, int l , int r) {
for(int i = l; i<=(l+r)/2; i++) {
long tmp = s[i];
s[i] = s[r+l-i];
s[r+l-i] = tmp;
}
}
public static void reverse(long[] s) {
reverse(s, 0, s.length-1);
}
public static void reverse(float[] s, int l , int r) {
for(int i = l; i<=(l+r)/2; i++) {
float tmp = s[i];
s[i] = s[r+l-i];
s[r+l-i] = tmp;
}
}
public static void reverse(float[] s) {
reverse(s, 0, s.length-1);
}
public static void reverse(double[] s, int l , int r) {
for(int i = l; i<=(l+r)/2; i++) {
double tmp = s[i];
s[i] = s[r+l-i];
s[r+l-i] = tmp;
}
}
public static void reverse(double[] s) {
reverse(s, 0, s.length-1);
}
public static void reverse(char[] s, int l , int r) {
for(int i = l; i<=(l+r)/2; i++) {
char tmp = s[i];
s[i] = s[r+l-i];
s[r+l-i] = tmp;
}
}
public static void reverse(char[] s) {
reverse(s, 0, s.length-1);
}
public static <T> void reverse(T[] s, int l , int r) {
for(int i = l; i<=(l+r)/2; i++) {
T tmp = s[i];
s[i] = s[r+l-i];
s[r+l-i] = tmp;
}
}
public static <T> void reverse(T[] s) {
reverse(s, 0, s.length-1);
}
public static void shuffle(int[] s) {
for (int i = 0; i < s.length; ++i) {
int j = random.nextInt(i + 1);
int t = s[i];
s[i] = s[j];
s[j] = t;
}
}
public static void shuffle(long[] s) {
for (int i = 0; i < s.length; ++i) {
int j = random.nextInt(i + 1);
long t = s[i];
s[i] = s[j];
s[j] = t;
}
}
public static void shuffle(float[] s) {
for (int i = 0; i < s.length; ++i) {
int j = random.nextInt(i + 1);
float t = s[i];
s[i] = s[j];
s[j] = t;
}
}
public static void shuffle(double[] s) {
for (int i = 0; i < s.length; ++i) {
int j = random.nextInt(i + 1);
double t = s[i];
s[i] = s[j];
s[j] = t;
}
}
public static void shuffle(char[] s) {
for (int i = 0; i < s.length; ++i) {
int j = random.nextInt(i + 1);
char t = s[i];
s[i] = s[j];
s[j] = t;
}
}
public static <T> void shuffle(T[] s) {
for (int i = 0; i < s.length; ++i) {
int j = random.nextInt(i + 1);
T t = s[i];
s[i] = s[j];
s[j] = t;
}
}
public static void sortArray(int[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void sortArray(long[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void sortArray(float[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void sortArray(double[] a) {
shuffle(a);
Arrays.sort(a);
}
public static void sortArray(char[] a) {
shuffle(a);
Arrays.sort(a);
}
public static <T extends Comparable<T>> void sortArray(T[] a) {
Arrays.sort(a);
}
}
static class Writer {
private PrintWriter pw;
public Writer(){
pw = new PrintWriter(System.out);
}
public Writer(String f){
try {
pw = new PrintWriter(new FileWriter(f));
} catch (IOException e) {
e.printStackTrace();
}
}
public void yesNo(boolean condition) {
println(condition?"YES":"NO");
}
public void printArray(int[] a) {
for(int i = 0; i<a.length; i++) print(a[i]+" ");
}
public void printlnArray(int[] a) {
for(int i = 0; i<a.length; i++) print(a[i]+" ");
pw.println();
}
public void printArray(long[] a) {
for(int i = 0; i<a.length; i++) print(a[i]+" ");
}
public void printlnArray(long[] a) {
for(int i = 0; i<a.length; i++) print(a[i]+" ");
pw.println();
}
public void print(Object o) {
pw.print(o.toString());
}
public void println(Object o) {
pw.println(o.toString());
}
public void println() {
pw.println();
}
public void flush() {
pw.flush();
}
public void exit() {
pw.close();
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 950689920192c22e3a1ff0e20d7fdd18 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class C_Complex_Market_Analysis {
static long mod = Long.MAX_VALUE;
public static void main(String[] args) {
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
FastReader f = new FastReader();
int t = f.nextInt();
while(t-- > 0){
solve(f, out);
}
out.close();
}
public static void solve(FastReader f, PrintWriter out) {
int n = f.nextInt();
int e = f.nextInt();
int arr[] = f.nextArray(n);
long ans = 0;
for(int i = 0; i < e; i++) {
ArrayList<Integer> arrli = new ArrayList<>();
for(int j = i; j < n; j += e) {
if(arr[j] == 1 || isPrime(arr[j])) {
arrli.add(arr[j]);
} else {
arrli.add(0);
}
}
for(int j = 0; j < arrli.size(); j++) {
if(arrli.get(j) == 1 || arrli.get(j) == 0) {
continue;
}
long prev = 0;
long next = 0;
int ind = j-1;
while(ind >= 0 && arrli.get(ind) == 1) {
prev++;
ind--;
}
ind = j+1;
while(ind < arrli.size() && arrli.get(ind) == 1) {
next++;
ind++;
}
ans += prev * (next+1);
ans += next;
}
}
out.println(ans);
}
// Sort an array
public static void sort(int arr[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i: arr) {
al.add(i);
}
Collections.sort(al);
for(int i = 0; i < arr.length; i++) {
arr[i] = al.get(i);
}
}
// Find all divisors of n
public static void allDivisors(int n) {
for(int i = 1; i*i <= n; i++) {
if(n%i == 0) {
System.out.println(i + " ");
if(i != n/i) {
System.out.println(n/i + " ");
}
}
}
}
// Check if n is prime or not
public static boolean isPrime(int n) {
if(n < 1) return false;
if(n == 2 || n == 3) return true;
if(n % 2 == 0 || n % 3 == 0) return false;
for(int i = 5; i*i <= n; i += 6) {
if(n % i == 0 || n % (i+2) == 0) {
return false;
}
}
return true;
}
// Find gcd of a and b
public static long gcd(long a, long b) {
long dividend = a > b ? a : b;
long divisor = a < b ? a : b;
while(divisor > 0) {
long reminder = dividend % divisor;
dividend = divisor;
divisor = reminder;
}
return dividend;
}
// Find lcm of a and b
public static long lcm(long a, long b) {
long lcm = gcd(a, b);
long hcf = (a * b) / lcm;
return hcf;
}
// Find factorial in O(n) time
public static long fact(int n) {
long res = 1;
for(int i = 2; i <= n; i++) {
res *= res * i;
}
return res;
}
// Find power in O(logb) time
public static long power(long a, long b) {
long res = 1;
while(b > 0) {
if((b&1) == 1) {
res = (res * a)%mod;
}
a = (a * a)%mod;
b >>= 1;
}
return res;
}
// Find nCr
public static long nCr(int n, int r) {
if(r < 0 || r > n) {
return 0;
}
long ans = fact(n) / (fact(r) * fact(n-r));
return ans;
}
// Find nPr
public static long nPr(int n, int r) {
if(r < 0 || r > n) {
return 0;
}
long ans = fact(n) / fact(r);
return ans;
}
// sort all characters of a string
public static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
// User defined class for fast I/O
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
boolean hasNext() {
if (st != null && st.hasMoreTokens()) {
return true;
}
String tmp;
try {
br.mark(1000);
tmp = br.readLine();
if (tmp == null) {
return false;
}
br.reset();
} catch (IOException e) {
return false;
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
float nextFloat() {
return Float.parseFloat(next());
}
boolean nextBoolean() {
return Boolean.parseBoolean(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] nextArray(int n) {
int[] a = new int[n];
for(int i=0; i<n; i++) {
a[i] = nextInt();
}
return a;
}
}
}
/**
Dec Char Dec Char Dec Char Dec Char
--------- --------- --------- ----------
0 NUL (null) 32 SPACE 64 @ 96 `
1 SOH (start of heading) 33 ! 65 A 97 a
2 STX (start of text) 34 " 66 B 98 b
3 ETX (end of text) 35 # 67 C 99 c
4 EOT (end of transmission) 36 $ 68 D 100 d
5 ENQ (enquiry) 37 % 69 E 101 e
6 ACK (acknowledge) 38 & 70 F 102 f
7 BEL (bell) 39 ' 71 G 103 g
8 BS (backspace) 40 ( 72 H 104 h
9 TAB (horizontal tab) 41 ) 73 I 105 i
10 LF (NL line feed, new line) 42 * 74 J 106 j
11 VT (vertical tab) 43 + 75 K 107 k
12 FF (NP form feed, new page) 44 , 76 L 108 l
13 CR (carriage return) 45 - 77 M 109 m
14 SO (shift out) 46 . 78 N 110 n
15 SI (shift in) 47 / 79 O 111 o
16 DLE (data link escape) 48 0 80 P 112 p
17 DC1 (device control 1) 49 1 81 Q 113 q
18 DC2 (device control 2) 50 2 82 R 114 r
19 DC3 (device control 3) 51 3 83 S 115 s
20 DC4 (device control 4) 52 4 84 T 116 t
21 NAK (negative acknowledge) 53 5 85 U 117 u
22 SYN (synchronous idle) 54 6 86 V 118 v
23 ETB (end of trans. block) 55 7 87 W 119 w
24 CAN (cancel) 56 8 88 X 120 x
25 EM (end of medium) 57 9 89 Y 121 y
26 SUB (substitute) 58 : 90 Z 122 z
27 ESC (escape) 59 ; 91 [ 123 {
28 FS (file separator) 60 < 92 \ 124 |
29 GS (group separator) 61 = 93 ] 125 }
30 RS (record separator) 62 > 94 ^ 126 ~
31 US (unit separator) 63 ? 95 _ 127 DEL
*/
// (a/b)%mod == (a * moduloInverse(b)) % mod;
// moduloInverse(b) = power(b, mod-2);
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7750691513ca3a26a77b286e6146eee7 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | // JAI SHREE RAM, HAR HAR MAHADEV, HARE KRISHNA
import java.util.*;
import java.util.Map.Entry;
import java.util.stream.*;
import java.lang.*;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.io.*;
import java.io.ObjectInputFilter.Status;
public class CodeForces {
static private final String INPUT = "input.txt";
static private final String OUTPUT = "output.txt";
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st;
static PrintWriter out = new PrintWriter(System.out);
static DecimalFormat df = new DecimalFormat("0.00000");
final static int MAX = Integer.MAX_VALUE, MIN = Integer.MIN_VALUE, mod = (int) (1e9 + 7);
final static long LMAX = Long.MAX_VALUE, LMIN = Long.MIN_VALUE;
final static long INF = (long) 1e18, Neg_INF = (long) -1e18;
static Random rand = new Random();
// ======================= MAIN ==================================
public static void main(String[] args) throws IOException {
long time = System.currentTimeMillis();
boolean oj = System.getProperty("ONLINE_JUDGE") != null;
// ==== start ====
input();
preprocess();
int t = 1;
t = readInt();
while (t-- > 0) {
solve();
}
out.flush();
// ==== end ====
if (!oj)
System.out.println(Arrays.deepToString(new Object[] { System.currentTimeMillis() - time + " ms" }));
}
private static void solve() throws IOException {
int n = readInt(), k = readInt();
int[] arr = readIntArray(n);
boolean[] vis = new boolean[n];
long ans = 0L;
for (int i = 0; i < n; i++) {
if (vis[i])
break;
boolean flag = false;
int t = i, one = 0, prev = 0;
while (t < n) {
vis[t] = true;
if (arr[t] == 1)
one++;
else {
if (flag)
ans += 1L * (prev + 1) * (one + 1) - 1;
if (isPrime[arr[t]])
flag = true;
else
flag = false;
prev = one;
one = 0;
}
t += k;
}
if (flag)
ans += 1L * (prev + 1) * (one + 1) - 1;
}
out.println(ans);
}
static boolean[] isPrime;
private static void preprocess() throws IOException {
isPrime = getPrimes((int) 1e6);
}
// cd C:\Users\Eshan Bhatt\Visual Studio Code\Competitive Programming\CodeForces
// javac CodeForces.java
// java CodeForces
// javac CodeForces.java && java CodeForces
// change Stack size -> java -Xss16M CodeForces.java
// ==================== CUSTOM CLASSES ================================
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
first = f;
second = s;
}
public int compareTo(Pair o) {
if (this.first == o.first)
return this.second - o.second;
return this.first - o.first;
}
@Override
public boolean equals(Object obj) {
if (obj == this)
return true;
if (obj == null)
return false;
if (this.getClass() != obj.getClass())
return false;
Pair other = (Pair) (obj);
if (this.first != other.first)
return false;
if (this.second != other.second)
return false;
return true;
}
@Override
public int hashCode() {
return this.first ^ this.second;
}
@Override
public String toString() {
return this.first + " " + this.second;
}
}
static class DequeNode {
DequeNode prev, next;
int val;
DequeNode(int val) {
this.val = val;
}
DequeNode(int val, DequeNode prev, DequeNode next) {
this.val = val;
this.prev = prev;
this.next = next;
}
}
// ======================= FOR INPUT ==================================
private static void input() {
FileInputStream instream = null;
PrintStream outstream = null;
try {
instream = new FileInputStream(INPUT);
outstream = new PrintStream(new FileOutputStream(OUTPUT));
System.setIn(instream);
System.setOut(outstream);
} catch (Exception e) {
System.err.println("Error Occurred.");
}
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}
static String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(readLine());
return st.nextToken();
}
static long readLong() throws IOException {
return Long.parseLong(next());
}
static int readInt() throws IOException {
return Integer.parseInt(next());
}
static double readDouble() throws IOException {
return Double.parseDouble(next());
}
static char readCharacter() throws IOException {
return next().charAt(0);
}
static String readString() throws IOException {
return next();
}
static String readLine() throws IOException {
return br.readLine().trim();
}
static int[] readIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = readInt();
return arr;
}
static int[][] read2DIntArray(int n, int m) throws IOException {
int[][] arr = new int[n][m];
for (int i = 0; i < n; i++)
arr[i] = readIntArray(m);
return arr;
}
static List<Integer> readIntList(int n) throws IOException {
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(readInt());
return list;
}
static long[] readLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = readLong();
return arr;
}
static long[][] read2DLongArray(int n, int m) throws IOException {
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++)
arr[i] = readLongArray(m);
return arr;
}
static List<Long> readLongList(int n) throws IOException {
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(readLong());
return list;
}
static char[] readCharArray() throws IOException {
return readString().toCharArray();
}
static char[][] readMatrix(int n, int m) throws IOException {
char[][] mat = new char[n][m];
for (int i = 0; i < n; i++)
mat[i] = readCharArray();
return mat;
}
// ========================= FOR OUTPUT ==================================
private static void printIList(List<Integer> list) {
for (int i = 0; i < list.size(); i++)
out.print(list.get(i) + " ");
out.println(" ");
}
private static void printLList(List<Long> list) {
for (int i = 0; i < list.size(); i++)
out.print(list.get(i) + " ");
out.println(" ");
}
private static void printIArray(int[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println(" ");
}
private static void print2DIArray(int[][] arr) {
for (int i = 0; i < arr.length; i++)
printIArray(arr[i]);
}
private static void printLArray(long[] arr) {
for (int i = 0; i < arr.length; i++)
out.print(arr[i] + " ");
out.println(" ");
}
private static void print2DLArray(long[][] arr) {
for (int i = 0; i < arr.length; i++)
printLArray(arr[i]);
}
private static void yes() {
out.println("YES");
}
private static void no() {
out.println("NO");
}
// ====================== TO CHECK IF STRING IS NUMBER ========================
private static boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
private static boolean isLong(String s) {
try {
Long.parseLong(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
// ==================== FASTER SORT ================================
private static void sort(int[] arr) {
int n = arr.length;
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void reverseSort(int[] arr) {
int n = arr.length;
List<Integer> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void sort(long[] arr) {
int n = arr.length;
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
private static void reverseSort(long[] arr) {
int n = arr.length;
List<Long> list = new ArrayList<>(n);
for (int i = 0; i < n; i++)
list.add(arr[i]);
Collections.sort(list, Collections.reverseOrder());
for (int i = 0; i < n; i++)
arr[i] = list.get(i);
}
// ==================== MATHEMATICAL FUNCTIONS ===========================
private static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
private static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
private static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static int mod_pow(long a, long b, int mod) {
if (b == 0)
return 1;
int temp = mod_pow(a, b >> 1, mod);
temp %= mod;
temp = (int) ((1L * temp * temp) % mod);
if ((b & 1) == 1)
temp = (int) ((1L * temp * a) % mod);
return temp;
}
private static int multiply(int a, int b) {
return (int) ((((1L * a) % mod) * ((1L * b) % mod)) % mod);
}
private static int divide(int a, int b) {
return multiply(a, mod_pow(b, mod - 2, mod));
}
private static boolean isPrime(long n) {
for (long i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return true;
}
private static long nCr(long n, long r) {
if (n - r > r)
r = n - r;
long ans = 1L;
for (long i = r + 1; i <= n; i++)
ans *= i;
for (long i = 2; i <= n - r; i++)
ans /= i;
return ans;
}
private static List<Integer> factors(int n) {
List<Integer> list = new ArrayList<>();
for (int i = 1; 1L * i * i <= n; i++)
if (n % i == 0) {
list.add(i);
if (i != n / i)
list.add(n / i);
}
return list;
}
private static List<Long> factors(long n) {
List<Long> list = new ArrayList<>();
for (long i = 1; i * i <= n; i++)
if (n % i == 0) {
list.add(i);
if (i != n / i)
list.add(n / i);
}
return list;
}
// ==================== Primes using Seive =====================
private static boolean[] getPrimes(int n) {
boolean[] prime = new boolean[n + 1];
Arrays.fill(prime, true);
for (int i = 2; 1L * i * i <= n; i++)
if (prime[i])
for (int j = i * i; j <= n; j += i)
prime[j] = false;
return prime;
// List<Integer> list = new ArrayList<>();
// for (int i = 2; i <= n; i++)
// if (prime[i])
// list.add(i);
// return list;
}
private static int[] SeivePrime(int n) {
int[] primes = new int[n];
for (int i = 0; i < n; i++)
primes[i] = i;
for (int i = 2; 1L * i * i < n; i++) {
if (primes[i] != i)
continue;
for (int j = i * i; j < n; j += i)
if (primes[j] == j)
primes[j] = i;
}
return primes;
}
// ==================== STRING FUNCTIONS ================================
private static boolean isPalindrome(String str) {
int i = 0, j = str.length() - 1;
while (i < j)
if (str.charAt(i++) != str.charAt(j--))
return false;
return true;
}
// check if a is subsequence of b
private static boolean isSubsequence(String a, String b) {
int idx = 0;
for (int i = 0; i < b.length() && idx < a.length(); i++)
if (a.charAt(idx) == b.charAt(i))
idx++;
return idx == a.length();
}
private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
private static String sortString(String str) {
int[] arr = new int[256];
for (char ch : str.toCharArray())
arr[ch]++;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < 256; i++)
while (arr[i]-- > 0)
sb.append((char) i);
return sb.toString();
}
// ==================== LIS & LNDS ================================
private static int LIS(int arr[], int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find1(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find1(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) >= val) {
ret = mid;
j = mid - 1;
} else {
i = mid + 1;
}
}
return ret;
}
private static int LNDS(int[] arr, int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
int idx = find2(list, arr[i]);
if (idx < list.size())
list.set(idx, arr[i]);
else
list.add(arr[i]);
}
return list.size();
}
private static int find2(List<Integer> list, int val) {
int ret = list.size(), i = 0, j = list.size() - 1;
while (i <= j) {
int mid = (i + j) / 2;
if (list.get(mid) <= val) {
i = mid + 1;
} else {
ret = mid;
j = mid - 1;
}
}
return ret;
}
// =============== Lower Bound & Upper Bound ===========
// less than or equal
private static int lower_bound(List<Integer> list, int val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(List<Long> list, long val) {
int ans = -1, lo = 0, hi = list.size() - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(int[] arr, int val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
private static int lower_bound(long[] arr, long val) {
int ans = -1, lo = 0, hi = arr.length - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] <= val) {
ans = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return ans;
}
// greater than or equal
private static int upper_bound(List<Integer> list, int val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(List<Long> list, long val) {
int ans = list.size(), lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (list.get(mid) >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(int[] arr, int val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
private static int upper_bound(long[] arr, long val) {
int ans = arr.length, lo = 0, hi = ans - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (arr[mid] >= val) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
// ==================== UNION FIND =====================
private static int find(int x, int[] parent) {
if (parent[x] == x)
return x;
return parent[x] = find(parent[x], parent);
}
private static boolean union(int x, int y, int[] parent, int[] rank) {
int lx = find(x, parent), ly = find(y, parent);
if (lx == ly)
return false;
if (rank[lx] > rank[ly])
parent[ly] = lx;
else if (rank[lx] < rank[ly])
parent[lx] = ly;
else {
parent[lx] = ly;
rank[ly]++;
}
return true;
}
// ==================== TRIE ================================
static class Trie {
class Node {
Node[] children;
boolean isEnd;
Node() {
children = new Node[26];
}
}
Node root;
Trie() {
root = new Node();
}
void insert(String word) {
Node curr = root;
for (char ch : word.toCharArray()) {
if (curr.children[ch - 'a'] == null)
curr.children[ch - 'a'] = new Node();
curr = curr.children[ch - 'a'];
}
curr.isEnd = true;
}
boolean find(String word) {
Node curr = root;
for (char ch : word.toCharArray()) {
if (curr.children[ch - 'a'] == null)
return false;
curr = curr.children[ch - 'a'];
}
return curr.isEnd;
}
}
// ================== SEGMENT TREE (RANGE SUM & RANGE UPDATE) ==================
public static class SegmentTree {
int n;
long[] arr, tree, lazy;
SegmentTree(long arr[]) {
this.arr = arr;
this.n = arr.length;
this.tree = new long[(n << 2)];
this.lazy = new long[(n << 2)];
build(1, 0, n - 1);
}
void build(int id, int start, int end) {
if (start == end)
tree[id] = arr[start];
else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
build(left, start, mid);
build(right, mid + 1, end);
tree[id] = tree[left] + tree[right];
}
}
void update(int l, int r, long val) {
update(1, 0, n - 1, l, r, val);
}
void update(int id, int start, int end, int l, int r, long val) {
distribute(id, start, end);
if (end < l || r < start)
return;
if (start == end)
tree[id] += val;
else if (l <= start && end <= r) {
lazy[id] += val;
distribute(id, start, end);
} else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
update(left, start, mid, l, r, val);
update(right, mid + 1, end, l, r, val);
tree[id] = tree[left] + tree[right];
}
}
long query(int l, int r) {
return query(1, 0, n - 1, l, r);
}
long query(int id, int start, int end, int l, int r) {
if (end < l || r < start)
return 0L;
distribute(id, start, end);
if (start == end)
return tree[id];
else if (l <= start && end <= r)
return tree[id];
else {
int mid = (start + end) / 2, left = (id << 1), right = left + 1;
return query(left, start, mid, l, r) + query(right, mid + 1, end, l, r);
}
}
void distribute(int id, int start, int end) {
if (start == end)
tree[id] += lazy[id];
else {
tree[id] += lazy[id] * (end - start + 1);
lazy[(id << 1)] += lazy[id];
lazy[(id << 1) + 1] += lazy[id];
}
lazy[id] = 0;
}
}
// ==================== FENWICK TREE ================================
static class FT {
int n;
int[] arr;
int[] tree;
FT(int[] arr, int n) {
this.arr = arr;
this.n = n;
this.tree = new int[n + 1];
for (int i = 1; i <= n; i++) {
update(i, arr[i - 1]);
}
}
FT(int n) {
this.n = n;
this.tree = new int[n + 1];
}
// 1 based indexing
void update(int idx, int val) {
while (idx <= n) {
tree[idx] += val;
idx += idx & -idx;
}
}
// 1 based indexing
long query(int l, int r) {
return getSum(r) - getSum(l - 1);
}
long getSum(int idx) {
long ans = 0L;
while (idx > 0) {
ans += tree[idx];
idx -= idx & -idx;
}
return ans;
}
}
// ==================== BINARY INDEX TREE ================================
static class BIT {
long[][] tree;
int n, m;
BIT(int[][] mat, int n, int m) {
this.n = n;
this.m = m;
tree = new long[n + 1][m + 1];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
update(i, j, mat[i - 1][j - 1]);
}
}
}
void update(int x, int y, int val) {
while (x <= n) {
int t = y;
while (t <= m) {
tree[x][t] += val;
t += t & -t;
}
x += x & -x;
}
}
long query(int x1, int y1, int x2, int y2) {
return getSum(x2, y2) - getSum(x1 - 1, y2) - getSum(x2, y1 - 1) + getSum(x1 - 1, y1 - 1);
}
long getSum(int x, int y) {
long ans = 0L;
while (x > 0) {
int t = y;
while (t > 0) {
ans += tree[x][t];
t -= t & -t;
}
x -= x & -x;
}
return ans;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | cb9161365cdbce098011cdef1e939bbc | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.lang.Math;
import java.lang.reflect.Array;
import java.util.*;
import javax.swing.text.DefaultStyledDocument.ElementSpec;
public final class Solution {
static BufferedReader br = new BufferedReader(
new InputStreamReader(System.in)
);
static BufferedWriter bw = new BufferedWriter(
new OutputStreamWriter(System.out)
);
static StringTokenizer st;
/*write your constructor and global variables here*/
static class sortCond implements Comparator<Pair<Integer, Integer>> {
@Override
public int compare(Pair<Integer, Integer> p1, Pair<Integer, Integer> p2) {
if (p1.a <= p2.a) {
return -1;
} else {
return 1;
}
}
}
static class Rec {
int a;
int b;
long c;
Rec(int a, int b, long c) {
this.a = a;
this.b = b;
this.c = c;
}
}
static class Pair<f, s> {
f a;
s b;
Pair(f a, s b) {
this.a = a;
this.b = b;
}
}
interface modOperations {
int mod(int a, int b, int mod);
}
static int findBinaryExponentian(int a, int pow, int mod) {
if (pow == 1) {
return a;
} else if (pow == 0) {
return 1;
} else {
int retVal = findBinaryExponentian(a, (int) pow / 2, mod);
int val = (pow % 2 == 0) ? 1 : a;
return modMul.mod(modMul.mod(retVal, retVal, mod), val, mod);
}
}
static int findPow(int a, int b, int mod) {
if (b == 1) {
return a % mod;
} else if (b == 0) {
return 1;
} else {
int res = findPow(a, (int) b / 2, mod);
return modMul.mod(res, modMul.mod(res, (b % 2 == 1 ? a : 1), mod), mod);
}
}
static int bleft(long ele, ArrayList<Long> sortedArr) {
int l = 0;
int h = sortedArr.size() - 1;
int ans = -1;
while (l <= h) {
int mid = l + (int) (h - l) / 2;
if (sortedArr.get(mid) < ele) {
l = mid + 1;
} else if (sortedArr.get(mid) >= ele) {
ans = mid;
h = mid - 1;
}
}
return ans;
}
static long gcd(long a, long b) {
long div = b;
long rem = a % b;
while (rem != 0) {
long temp = rem;
rem = div % rem;
div = temp;
}
return div;
}
static int log(int no) {
int i = 0;
while ((1 << i) <= no) {
i++;
}
if ((1 << (i - 1)) == no) {
return i - 1;
} else {
return i;
}
}
static modOperations modAdd = (int a, int b, int mod) -> {
return (a % mod + b % mod) % mod;
};
static modOperations modSub = (int a, int b, int mod) -> {
return (int) ((1l * a % mod - 1l * b % mod + 1l * mod) % mod);
};
static modOperations modMul = (int a, int b, int mod) -> {
return (int) ((1l * (a % mod) * 1l * (b % mod)) % (1l * mod));
};
static modOperations modDiv = (int a, int b, int mod) -> {
return modMul.mod(a, findBinaryExponentian(b, mod - 1, mod), mod);
};
static HashSet<Integer> primeList(int MAXI) {
int[] prime = new int[MAXI + 1];
HashSet<Integer> obj = new HashSet<>();
for (int i = 2; i <= (int) Math.sqrt(MAXI) + 1; i++) {
if (prime[i] == 0) {
obj.add(i);
for (int j = i * i; j <= MAXI; j += i) {
prime[j] = 1;
}
}
}
for (int i = (int) Math.sqrt(MAXI) + 1; i <= MAXI; i++) {
if (prime[i] == 0) {
obj.add(i);
}
}
return obj;
}
static int[] factorialList(int MAXI, int mod) {
int[] factorial = new int[MAXI + 1];
factorial[2] = 1;
for (int i = 3; i < MAXI + 1; i++) {
factorial[i] = modMul.mod(factorial[i - 1], i, mod);
}
return factorial;
}
static void put(HashMap<Integer, Integer> cnt, int key) {
if (cnt.containsKey(key)) {
cnt.replace(key, cnt.get(key) + 1);
} else {
cnt.put(key, 1);
}
}
static long arrSum(ArrayList<Long> arr) {
long tot = 0;
for (int i = 0; i < arr.size(); i++) {
tot += arr.get(i);
}
return tot;
}
static int ord(char b) {
return (int) b - (int) 'a';
}
static int optimSearch(int[] cnt, int lower_bound, int pow, int n) {
int l = lower_bound + 1;
int h = n;
int ans = 0;
while (l <= h) {
int mid = l + (h - l) / 2;
if (cnt[mid] - cnt[lower_bound] == pow) {
return mid;
} else if (cnt[mid] - cnt[lower_bound] < pow) {
ans = mid;
l = mid + 1;
} else {
h = mid - 1;
}
}
return ans;
}
static Pair<Long, Integer> ret_ans(ArrayList<Integer> ans) {
int size = ans.size();
int mini = 1000000000 + 1;
long tit = 0l;
for (int i = 0; i < size; i++) {
tit += 1l * ans.get(i);
mini = Math.min(mini, ans.get(i));
}
return new Pair<>(tit - mini, mini);
}
static int factorList(
HashMap<Integer, Integer> maps,
int no,
int maxK,
int req
) {
int i = 1;
while (i * i <= no) {
if (no % i == 0) {
if (i != no / i) {
put(maps, no / i);
}
put(maps, i);
if (maps.get(i) == req) {
maxK = Math.max(maxK, i);
}
if (maps.get(no / i) == req) {
maxK = Math.max(maxK, no / i);
}
}
i++;
}
return maxK;
}
static ArrayList<Integer> getKeys(HashMap<Integer, Integer> maps) {
ArrayList<Integer> vals = new ArrayList<>();
for (Map.Entry<Integer, Integer> map : maps.entrySet()) {
vals.add(map.getKey());
}
return vals;
}
static ArrayList<Integer> getValues(HashMap<Integer, Integer> maps) {
ArrayList<Integer> vals = new ArrayList<>();
for (Map.Entry<Integer, Integer> map : maps.entrySet()) {
vals.add(map.getValue());
}
return vals;
}
/*write your methods here*/
static int getMax(ArrayList<Integer> arr) {
int max = arr.get(0);
for (int i = 1; i < arr.size(); i++) {
if (arr.get(i) > max) {
max = arr.get(i);
}
}
return max;
}
static int getMin(ArrayList<Integer> arr) {
int max = arr.get(0);
for (int i = 1; i < arr.size(); i++) {
if (arr.get(i) < max) {
max = arr.get(i);
}
}
return max;
}
public static void main(String[] args) throws IOException {
int cases = Integer.parseInt(br.readLine()), n, i, j, e;
HashSet<Integer> prime = primeList(1000000);
while (cases-- != 0) {
st = new StringTokenizer(br.readLine());
n = Integer.parseInt(st.nextToken());
e = Integer.parseInt(st.nextToken());
st = new StringTokenizer(br.readLine());
int arr[] = new int[n];
boolean vis[] = new boolean[n];
for (i = 0; i < n; i++) {
arr[i] = Integer.parseInt(st.nextToken());
}
long tot = 0l;
for (i = 0; i < n; i++) {
if (!vis[i]) {
j = i;
ArrayList<Integer> rec = new ArrayList<>();
//System.out.println(j + " " + tot);
int start = j;
while (j < n) {
vis[j] = true;
if (arr[j] != 1 && !prime.contains(arr[j])) {
break;
} else {
if (prime.contains(arr[j])) {
rec.add(j);
}
j += e;
}
}
if (rec.isEmpty()) {
continue;
} else {
int end = j;
boolean ok = true;
if (rec.get(0) != start) {
ok = false;
}
rec.add(0, start - e);
rec.add(end);
//System.out.println(rec.toString());
for (int k = 0; k < rec.size() - 1; k++) {
int no = (rec.get(k + 1) - rec.get(k)) / e - 1;
int frontno = 0;
if (k + 2 < rec.size()) {
frontno += (rec.get(k + 2) - rec.get(k + 1)) / e;
}
tot += 1l * no * frontno;
if (k != 0 || ok) {
tot += 1l * (rec.get(k + 1) - rec.get(k)) / e - 1;
}
}
}
}
}
bw.write(Long.toString(tot) + "\n");
}
bw.flush();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | b2e9e29a9b33bd24cd4da3937bdf8185 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | //Some of the methods are copied from GeeksforGeeks Website
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
//static Scanner sc=new Scanner(System.in);
static Reader sc=new Reader();
// static FastReader sc=new FastReader(System.in);
static long mod = (long)(1e9)+ 7;
static int max_num=(int)1e6+5;
static boolean prime[];
static void sieveOfEratosthenes(int n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
prime = new boolean[n+1];
for(int i=0;i<=n;i++)
prime[i] = true;
for(int p = 2; p*p <=n; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p] == true)
{
// Update all multiples of p
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
}
public static void main (String[] args) throws java.lang.Exception
{
try{
/*
Collections.sort(al,(a,b)->a.x-b.x);
Collections.sort(al,Collections.reverseOrder());
long n=sc.nextLong();
String s=sc.next();
char a[]=s.toCharArray();
StringBuilder sb=new StringBuilder();
map.put(a[i],map.getOrDefault(a[i],0)+1);
*/
int t = sc.nextInt();
sieveOfEratosthenes(max_num);
while(t-->0)
{
int n=sc.nextInt();
int e=sc.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
{
int x=sc.nextInt();
if(x==1)
a[i]=1;
else if(prime[x])
a[i]=-1;
}
//print(a);
long ans=0;
for(int i=0;i<n;i++)
{
if(a[i]==-1)
{
long p=0,q=0;
for(int j=i-e;j>=0;j-=e)
{
if(a[j]!=1)
break;
p++;
}
for(int j=i+e;j<n;j+=e)
{
if(a[j]!=1)
break;
q++;
}
// actually , (p+1) * (q+1) -1
ans+=p*(q+1l)+q;
}
}
out.println(ans);
}
out.flush();
out.close();
}
catch(Exception e)
{}
}
/*
...SOLUTION ENDS HERE...........SOLUTION ENDS HERE...
*/
static void flag(boolean flag)
{
out.println(flag ? "YES" : "NO");
out.flush();
}
/*
Map<Long,Long> map=new HashMap<>();
for(int i=0;i<n;i++)
{
if(!map.containsKey(a[i]))
map.put(a[i],1);
else
map.replace(a[i],map.get(a[i])+1);
}
Set<Map.Entry<Long,Long>> hmap=map.entrySet();
for(Map.Entry<Long,Long> data : hmap)
{
}
Iterator<Integer> itr = set.iterator();
while(itr.hasNext())
{
int val=itr.next();
}
*/
// static class Pair
// {
// int x,y;
// Pair(int x,int y)
// {
// this.x=x;
// this.y=y;
// }
// }
// Arrays.sort(p, new Comparator<Pair>()
// {
// @Override
// public int compare(Pair o1,Pair o2)
// {
// if(o1.x>o2.x) return 1;
// else if(o1.x==o2.x)
// {
// if(o1.y>o2.y) return 1;
// else return -1;
// }
// else return -1;
// }});
static void print(int a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print(long a[])
{
int n=a.length;
for(int i=0;i<n;i++)
{
out.print(a[i]+" ");
}
out.println();
out.flush();
}
static void print_int(ArrayList<Integer> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static void print_long(ArrayList<Long> al)
{
int si=al.size();
for(int i=0;i<si;i++)
{
out.print(al.get(i)+" ");
}
out.println();
out.flush();
}
static int LowerBound(int a[], int x)
{ // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
static int UpperBound(int a[], int x)
{// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
static void DFS(ArrayList<Integer> graph[],boolean[] visited, int u)
{
visited[u]=true;
int v=0;
for(int i=0;i<graph[u].size();i++)
{
v=graph[u].get(i);
if(!visited[v])
DFS(graph,visited,v);
}
}
static boolean[] prime(int num)
{
boolean[] bool = new boolean[num];
for (int i = 0; i< bool.length; i++) {
bool[i] = true;
}
for (int i = 2; i< Math.sqrt(num); i++) {
if(bool[i] == true) {
for(int j = (i*i); j<num; j = j+i) {
bool[j] = false;
}
}
}
if(num >= 0) {
bool[0] = false;
bool[1] = false;
}
return bool;
}
static long nCr(long a,long b,long mod)
{
return (((fact[(int)a] * modInverse(fact[(int)b],mod))%mod * modInverse(fact[(int)(a - b)],mod))%mod + mod)%mod;
}
static long fact[]=new long[max_num];
static void fact_fill()
{
fact[0]=1l;
for(int i=1;i<max_num;i++)
{
fact[i]=(fact[i-1]*(long)i);
if(fact[i]>=mod)
fact[i]%=mod;
}
}
static long modInverse(long a, long m)
{
return power(a, m - 2, m);
}
static long power(long x, long y, long m)
{
if (y == 0)
return 1;
long p = power(x, y / 2, m) % m;
p = (long)((p * (long)p) % m);
if (y % 2 == 0)
return p;
else
return (long)((x * (long)p) % m);
}
static long sum_array(int a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static long sum_array(long a[])
{
int n=a.length;
long sum=0;
for(int i=0;i<n;i++)
sum+=a[i];
return sum;
}
static void sort(int[] a)
{
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void sort(long[] a)
{
ArrayList<Long> l=new ArrayList<>();
for (long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static void reverse_array(int a[])
{
int n=a.length;
int i,t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static void reverse_array(long a[])
{
int n=a.length;
int i; long t;
for (i = 0; i < n / 2; i++) {
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static class FastReader{
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan());
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan());
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static PrintWriter out=new PrintWriter(System.out);
static int int_max=Integer.MAX_VALUE;
static int int_min=Integer.MIN_VALUE;
static long long_max=Long.MAX_VALUE;
static long long_min=Long.MIN_VALUE;
}
// Thank You ! | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7e59dd1250c803d9f31820ca2acbf1f0 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | /*----------- ---------------*
Author : Ryan Ranaut
__Hope is a big word, never lose it__
------------- --------------*/
import java.io.*;
import java.util.*;
public class Codeforces1 {
static PrintWriter out = new PrintWriter(System.out);
static final int mod = 1_000_000_007;
static int max = (int)(1e6);
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
/*--------------------------------------------------------------------------*/
//Try seeing general case
//Minimization Maximization - BS..... Connections - Graphs.....
//Greedy not worthy - Try DP
public static void main(String[] args)
{
FastReader s = new FastReader();
sieve();
int t = s.nextInt();
while(t-->0)
{
int n = s.nextInt();
int e = s.nextInt();
int[] a = s.readIntArray(n);
out.println(find(n, e, a));
}
out.close();
}
public static long find(int n, int e, int[] a)
{
long res = 0;
for(int i=0;i<n;i++)
{
if(isPrime[a[i]])
{
long left = 0, right = 0;
for(int j=i-e;j>=0;j-=e)
{
if(a[j] == 1)
left++;
else
break;
}
for(int j=i+e;j<n;j+=e)
{
if(a[j] == 1)
right++;
else
break;
}
res += left + right + left*right;
}
}
return res;
}
static boolean[] isPrime = new boolean[max+1];
public static void sieve()
{
Arrays.fill(isPrime, true);
isPrime[0] = isPrime[1] = false;//0 and 1 are not prime
for(int i=2;i*i<=max;i++)//Go till the square root of the number
{
if(isPrime[i])
{
for(int j=i*i;j<=max;j+=i)//Start from the square of the number
isPrime[j] = false;
}
}
}
/*----------------------------------End of the road--------------------------------------*/
static class DSU {
int[] parent;
int[] ranks;
int[] groupSize;
int size;
public DSU(int n) {
size = n;
parent = new int[n];//0 based
ranks = new int[n];
groupSize = new int[n];//Size of each component
for (int i = 0; i < n; i++) {
parent[i] = i;
ranks[i] = 1; groupSize[i] = 1;
}
}
public int find(int x)//Path Compression
{
if (parent[x] == x)
return x;
else
return parent[x] = find(parent[x]);
}
public void union(int x, int y)//Union by rank
{
int x_rep = find(x);
int y_rep = find(y);
if (x_rep == y_rep)
return;
if (ranks[x_rep] < ranks[y_rep])
{
parent[x_rep] = y_rep;
groupSize[y_rep] += groupSize[x_rep];
}
else if (ranks[x_rep] > ranks[y_rep])
{
parent[y_rep] = x_rep;
groupSize[x_rep] += groupSize[y_rep];
}
else {
parent[y_rep] = x_rep;
ranks[x_rep]++;
groupSize[x_rep] += groupSize[y_rep];
}
size--;//Total connected components
}
}
public static int gcd(int x,int y)
{
return y==0?x:gcd(y,x%y);
}
public static long gcd(long x,long y)
{
return y==0L?x:gcd(y,x%y);
}
public static int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static long pow(long a,long b)
{
if(b==0L)
return 1L;
long tmp=1;
while(b>1L)
{
if((b&1L)==1)
tmp*=a;
a*=a;
b>>=1;
}
return (tmp*a);
}
public static long modPow(long a,long b,long mod)
{
if(b==0L)
return 1L;
long tmp=1;
while(b>1L)
{
if((b&1L)==1L)
tmp*=a;
a*=a;
a%=mod;
tmp%=mod;
b>>=1;
}
return (tmp*a)%mod;
}
static long mul(long a, long b) {
return a*b%mod;
}
static long fact(int n) {
long ans=1;
for (int i=2; i<=n; i++) ans=mul(ans, i);
return ans;
}
static long fastPow(long base, long exp) {
if (exp==0) return 1;
long half=fastPow(base, exp/2);
if (exp%2==0) return mul(half, half);
return mul(half, mul(half, base));
}
static void debug(int ...a)
{
for(int x: a)
out.print(x+" ");
out.println();
}
static void debug(long ...a)
{
for(long x: a)
out.print(x+" ");
out.println();
}
static void debugMatrix(int[][] a)
{
for(int[] x:a)
out.println(Arrays.toString(x));
}
static void debugMatrix(long[][] a)
{
for(long[] x:a)
out.println(Arrays.toString(x));
}
static void reverseArray(int[] a) {
int n = a.length;
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void sort(int[] a)
{
ArrayList<Integer> ls = new ArrayList<>();
for(int x: a) ls.add(x);
Collections.sort(ls);
for(int i=0;i<a.length;i++)
a[i] = ls.get(i);
}
static void sort(long[] a)
{
ArrayList<Long> ls = new ArrayList<>();
for(long x: a) ls.add(x);
Collections.sort(ls);
for(int i=0;i<a.length;i++)
a[i] = ls.get(i);
}
// static class Pair{
// int x, y;
// Pair(int x, int y)
// {
// this.x = x;
// this.y = y;
// }
//
// @Override
// public boolean equals(Object o) {
// if (this == o) return true;
// if (o == null || getClass() != o.getClass()) return false;
// Pair pair = (Pair) o;
// return x == pair.x && y == pair.y;
// }
//
// @Override
// public int hashCode() {
// return Objects.hash(x, y);
// }
// }
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7edb31ea289cdf5c32bbc4331990a201 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author El Mehdi ASSALI
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
CComplexMarketAnalysis solver = new CComplexMarketAnalysis();
solver.solve(1, in, out);
out.close();
}
static class CComplexMarketAnalysis {
boolean[] isPrime = new boolean[(int) 1e6 + 7];
public void solve(int testNumber, InputReader in, PrintWriter out) {
Arrays.fill(isPrime, true);
isPrime[0] = isPrime[1] = false;
for (int i = 2; i < isPrime.length; i++) {
if (isPrime[i]) {
for (int j = i + i; j < isPrime.length; j += i) {
isPrime[j] = false;
}
}
}
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt(), e = in.nextInt();
int[] arr = new int[n];
int[] groups = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = in.nextInt();
}
for (int i = 0; i < n; i++) {
if (arr[i] == 1 && groups[i] == 0) {
int count = 0;
for (int j = i; j < n && arr[j] == 1; j += e) {
count++;
}
for (int j = i; j < n && arr[j] == 1; j += e) {
groups[j] = count;
}
}
}
long answer = 0;
for (int i = 0; i < n; i++) {
if (isPrime[arr[i]]) {
answer += (long) (1 + (i - e >= 0 ? groups[i - e] : 0)) * (1 + (i + e < n ? groups[i + e] : 0)) - 1;
}
}
out.println(answer);
}
}
}
static class InputReader {
private final BufferedReader reader;
private StringTokenizer tokenizer;
public InputReader(InputStream inputStream) {
reader = new BufferedReader(new InputStreamReader(inputStream));
}
public String nextLine() {
String line = null;
try {
line = reader.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return line;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(nextLine());
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | b71b4dd5f5439e003fa9342e11c972a5 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
import java.time.*;
import static java.lang.Math.*;
@SuppressWarnings("unused")
public class B {
static boolean DEBUG = false;
static Reader fs;
static PrintWriter pw;
public static ArrayList<Integer> sieve(int n) {
int prime[] = new int[n + 1];
ArrayList<Integer> ans = new ArrayList<Integer>();
for (int i = 2; i <= n; i++) {
if (prime[i] == 0) {
prime[i] = i;
ans.add(i);
}
for (int j = 0; j < (int) ans.size() && ans.get(j) <= prime[i] && i * ans.get(j) <= n; ++j) {
prime[i * ans.get(j)] = ans.get(j);
}
}
return ans;
}
static void solve() {
int n = fs.nextInt(), e = fs.nextInt();
int a[] = fs.readArray(n);
boolean isones[] = new boolean[n + 1];
for (int i = 0; i < n; i++) {
if (a[i] == 1)
isones[i+1] = true;
}
boolean isprime[] = new boolean[n + 1];
for (int i = 0; i < n; i++) {
if (conatainer.contains(a[i])) {
isprime[i+1] = true;
}
}
pw.println(go(n, e, isones, isprime));
}
static long go(int n, int k, boolean isone[], boolean isprime[]) {
long res = 0;
ArrayList<Integer> ones = new ArrayList<>();
boolean chained[] = new boolean[n + 1];
for (int i = 1; i <= n; i++)
if (!chained[i]) {
ones.clear();
int cur_ones = 0;
for (int j = i; j <= n; j += k) {
if ((!isone[j] && !isprime[j]) || chained[j])
break;
chained[j] = true;
if (isone[j])
cur_ones++;
else {
ones.add(cur_ones);
cur_ones = 0;
}
}
if (ones.size() == 0)
continue;
ones.add(cur_ones);
// System.err.p/rintln(ones);
for (int j = 0; j < ones.size(); ++j) {
res += ones.get(j);
if (j > 0 && j < ones.size() - 1) {
res += ones.get(j);
}
if (j < ones.size() - 1) {
res += (long )(ones.get(j)) * (long )(ones.get(j + 1));
}
}
}
return res;
}
static ArrayList<Integer> primes = sieve(1_000_000 + 1);
static TreeSet<Integer > conatainer = new TreeSet<Integer>();
public static void main(String[] args) throws IOException {
Instant start = Instant.now();
if (args.length == 2) {
System.setIn(new FileInputStream(new File("D:\\program\\javaCPEclipse\\CodeForces\\src\\input.txt")));
// System.setOut(new PrintStream(new File("output.txt")));
System.setErr(new PrintStream(new File("D:\\program\\javaCPEclipse\\CodeForces\\src\\error.txt")));
DEBUG = true;
}
for(int x : primes)conatainer.add(x);
fs = new Reader();
pw = new PrintWriter(System.out);
int t = fs.nextInt();
while (t-- > 0) {
solve();
}
Instant end = Instant.now();
if (DEBUG) {
pw.println(Duration.between(start, end));
}
pw.close();
}
static void sort(int a[]) {
ArrayList<Integer> l = new ArrayList<Integer>();
for (int x : a)
l.add(x);
Collections.sort(l);
for (int i = 0; i < a.length; i++) {
a[i] = l.get(i);
}
}
public static void print(long a, long b, long c, PrintWriter pw) {
pw.println(a + " " + b + " " + c);
return;
}
static class Reader {
BufferedReader br;
StringTokenizer st;
public Reader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
int[][] read2Array(int n, int m) {
int a[][] = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
a[i][j] = nextInt();
}
}
return a;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 8ceade9dfd2b3682b7b1b736e3f5fedb | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class a {
static FastReader sc = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static boolean prime[];
public static void main (String[] args) throws java.lang.Exception {
int t = sc.nextInt();
int max = (int)1e6;
sieveOfEratosthenes(max);
while(t-->0){
int n = sc.nextInt();
int k = sc.nextInt();
int arr[] = new int[n];
for(int i = 0;i < n; i++){
arr[i] = sc.nextInt();
}
long pairs = 0;
for(int i = 0 ; i < n; i++){
if(prime[arr[i]]){
int index = i-k;
long lcnt = 0;
while(index >= 0){
if(arr[index] == 1) {
lcnt++;
index -= k;
}else{
break;
}
}
index = i+k;
long rcnt = 0;
while(index < n){
if(arr[index] == 1) {
rcnt++;
index += k;
}else{
break;
}
}
// out.println(lcnt+" "+rcnt);
pairs += lcnt+lcnt*rcnt+rcnt;
}
}
out.println(pairs);
}
// Discipline is doing what Needs to be done even if you don't want to Do it.
out.flush();
}
static void sieveOfEratosthenes(int n)
{
prime = new boolean[n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
// for (int i = 2; i <= 20; i++)
// {
// if (prime[i] == true)
// System.out.print(i + " ");
// }
}
}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | ce5c9f91300ada69946b7c5531aa8b97 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import static java.lang.Math.max;
public class A {
static PrintWriter out;
static boolean[] primes;
public static void main(String[] args) {
FastReader in = new FastReader();
out = new PrintWriter(System.out);
primes = sieveOfEratosthenes(1000000);
long t = in.nextLong();
long test = 1;
while (test <= t) {
solve(in);
test++;
}
out.close();
}
/*--------------------------------------------------------------------------------------------------------------------------------*/
/*--------------------------------------------------------------------------------------------------------------------------------*/
private static void solve(FastReader in) {
int n = in.nextInt();
int e = in.nextInt();
int[] a = in.readArray(n);
long ans = 0;
for (int i = 0; i < n; i++) {
if (primes[a[i]]) {
int r = 1, l = 1;
while (r * e + i < n && a[r * e + i] == 1) r++;
while (i - (l * e) >= 0 && a[i - (l * e)] == 1) l++;
l--;
r--;
ans += r + l + (long) l * r;
}
}
out.println(ans);
}
/*--------------------------------------------------------------------------------------------------------------------------------*/
/*--------------------------------------------------------------------------------------------------------------------------------*/
private static int gcd(int a, int b) {
return b == 0 ? a : (gcd(b, a % b));
}
static boolean[] sieveOfEratosthenes(int n) {
boolean[] prime = new boolean[n + 1];
Arrays.fill(prime, true);
prime[0] = prime[1] = false;
for (int p = 2; p * p <= n; p++) {
if (prime[p]) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static long modPow(long a, long b, long m) {
long res = 1;
a %= m;
while (b > 0) {
if ((b & 1) != 0) {
res = res * a;
res %= m;
}
b >>= 1;
a *= a;
a %= m;
}
return res;
}
private static class Pair implements Comparable<Pair> {
int ff, ss;
Pair(int x, int y) {
this.ff = x;
this.ss = y;
}
public int compareTo(Pair o) {
return this.ff == o.ff ? this.ss - o.ss : this.ff - o.ff;
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 9c5adcefaf65fef5ed1f032319e9d80f | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.lang.reflect.Array;
import java.util.*;
public class Main {
public static void main(String[] args) {
// write your code here
Scanner sc = new Scanner(System.in);
boolean[] isPrime = seive();
int t = sc.nextInt();
while (t-->0){
int n = sc.nextInt();
int e = sc.nextInt();
int[] arr = new int[n];
for (int i =0; i<n ; i++){
arr[i] = sc.nextInt();
}
long count = 0;
for (int i =0 ; i<n ; i++){
if (isPrime[arr[i]] && arr[i]>1){
long l =0;
long r =0;
for (int j = i+e; j<n ; j+=e){
if (arr[j]==1){
r++;
}else {
break;
}
}
for (int j = i-e ; j>=0 ; j-=e){
if (arr[j]==1){
l++;
}else {
break;
}
}
count+=(l+r+(l*r));
}
}
System.out.println(count);
}
}
// https://www.geeksforgeeks.org/sieve-of-eratosthenes/
public static boolean[] seive(){
boolean[] primes = new boolean[(int)Math.pow(10,6)+1];
Arrays.fill(primes , true);
for (int p = 2 ; p*p<= primes.length ; p++){
if (primes[p]){
for (int i = p*p ; i<primes.length ; i+=p){
primes[i] = false;
}
}
}
return primes;
}
public boolean isPrime(int n){
for (int i = 2 ; i<n ; i++){
if (n%i == 0){
return false;
}
}
return true;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 87ebfcb879b1f3275f8d886daa61d084 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | /**
*
* YUVRAJ PARASHAR
*
* */
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Cp{
public static void main(String[] args) {
/*==============Sieve of Eratosthenes==============*/
boolean isPrime[] = new boolean[1000005];
Arrays.fill(isPrime, true);
isPrime[0] = isPrime[1] = false;
for(int i = 2; i < 1000005; i++){
if(isPrime[i] == true){
for(int j = i*2; j < 1000005; j += i){
isPrime[j] = false;
}
}
}
/*================================================*/
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
long ans = 0;
for(int i = 0; i < n; i++){
if(isPrime[arr[i]]){
long before = 0;
long after = 0;
for(int j = i-e; j >= 0; j -= e){
if(arr[j]!=1)break;
before++;
}
for(int j = i+e; j < n; j += e){
if(arr[j]!=1)break;
after++;
}
ans += (before+1) * (after+1) -1;
}
}
System.out.println(ans);
}
sc.close();
}
// print array [for debugging purpose only]
public static void print(int arr[]){
for(int x : arr){
System.out.println(x+" ");
}System.out.println();
}
// GCD - Greatest Common Divisor
public static long gcd(long a, long b){
if(b == 0L){
return a;
}
return gcd(b, a%b);
}
// sort
public static void sort(int[] arr){
// Arrays.sort is Quick Sort which can take upto O(N^2) if array is in reverse order
ArrayList<Integer> ls = new ArrayList<>();
for(int x : arr){
ls.add(x);
}
Collections.sort(ls); // Merge sort - Always O(N*log N)
for(int i = 0; i < arr.length; i++){
arr[i] = ls.get(i);
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 58c458f0fb7a9974f430ff2bb17f683e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC {
public void solve(int testNumber, Scanner in, PrintWriter out) {
int T = in.nextInt();
while (T-- > 0) {
solveOne(in, out);
}
}
private void solveOne(Scanner in, PrintWriter out) {
int N = in.nextInt();
int E = in.nextInt();
int A[] = L.readIntArray(N, in);
int[] rightOnes = new int[N];
int[] leftOnes = new int[N];
for (int i = N - 1; i >= 0; i--) {
if (A[i] != 1) continue;
rightOnes[i]++;
if (i + E < N) rightOnes[i] += rightOnes[i + E];
}
for (int i = 0; i < N; i++) {
if (A[i] != 1) continue;
leftOnes[i]++;
if (i - E >= 0) leftOnes[i] += leftOnes[i - E];
}
long pairs = 0;
for (int i = 0; i < N; i++) {
if (A[i] != 1 && L.isPrime(A[i])) {
if (i + E < N) pairs += rightOnes[i + E];
if (i - E >= 0) pairs += leftOnes[i - E];
if (i - E >= 0 && E + i < N) pairs += leftOnes[i - E] * (long) rightOnes[i + E];
}
}
out.println(pairs);
}
}
static class L {
public static int[] readIntArray(int size, Scanner in) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = in.nextInt();
}
return array;
}
public static boolean isPrime(long n) {
int s = (int) Math.sqrt(n);
for (int i = 2; i <= s; i++)
if (n % i == 0)
return false;
return true;
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 7a0c1b4f4f1e686336fa9db261ed9080 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Random;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
static int mod = (int) (1e9) + 7;
/* ======================DSU===================== */
static class dsu {
static int parent[], n;// min[],value[];
static long size[];
dsu(int n) {
parent = new int[n + 1];
size = new long[n + 1];
// min=new int[n+1];
// value=new int[n+1];
Main.dsu.n = n;
makeSet();
}
static void makeSet() {
for (int i = 1; i <= n; i++) {
parent[i] = i;
size[i] = 1;
// min[i]=i;
}
}
static int find(int a) {
if (parent[a] == a)
return a;
else {
return parent[a] = find(parent[a]);// Path Compression
}
}
static void union(int a, int b) {
int setA = find(a);
int setB = find(b);
if (setA == setB)
return;
if (size[setA] >= size[setB]) {
parent[setB] = setA;
size[setA] += size[setB];
} else {
parent[setA] = setB;
size[setB] += size[setA];
}
}
}
/* ======================================================== */
static class Pair<X extends Number, Y extends Number> implements Comparator<Pair> {
X x;
Y y;
// Constructor
public Pair(X x, Y y) {
this.x = x;
this.y = y;
}
public Pair() {
}
@Override
public int compare(Main.Pair o1, Main.Pair o2) {
return ((int) (o1.y.intValue() - o2.y.intValue()));// Ascending Order based on 'y'
}
}
/* ===============================Tries================================= */
static class TrieNode {
private HashMap<Character, TrieNode> children = new HashMap<>();
public int size;
boolean endOfWord;
public void putChildIfAbsent(char ch) {
children.putIfAbsent(ch, new TrieNode());
}
public TrieNode getChild(char ch) {
return children.get(ch);
}
}
static private TrieNode root;
public static void insert(String str) {
TrieNode curr = root;
for (char ch : str.toCharArray()) {
curr.putChildIfAbsent(ch);
curr = curr.getChild(ch);
curr.size++;
}
// mark the current nodes endOfWord as true
curr.endOfWord = true;
}
public static int search(String word) {
TrieNode curr = root;
for (char ch : word.toCharArray()) {
curr = curr.getChild(ch);
if (curr == null) {
return 0;
}
}
// size contains words starting with prefix- word
return curr.size;
}
public boolean delete(TrieNode current, String word, int index) {
if (index == word.length()) {
// when end of word is reached only delete if currrent.endOfWord is true.
if (!current.endOfWord) {
return false;
}
current.endOfWord = false;
// if current has no other mapping then return true
return current.children.size() == 0;
}
char ch = word.charAt(index);
TrieNode node = current.children.get(ch);
if (node == null) {
return false;
}
boolean shouldDeleteCurrentNode = delete(node, word, index + 1);
// if true is returned then delete the mapping of character and trienode
// reference from map.
if (shouldDeleteCurrentNode) {
current.children.remove(ch);
// return true if no mappings are left in the map.
return current.children.size() == 0;
}
return false;
}
/* ================================================================= */
static boolean isPrime(long n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static Pair<Integer, Integer> lowerBound(long[] a, long x) { // x is the target, returns lowerBound. If not found
// return -1
int l = -1, r = a.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (a[m] == x) {
return new Pair<>(m, 1);
}
if (a[m] >= x)
r = m;
else
l = m;
}
return new Pair<>(r, 0);
}
static Pair<Integer, Integer> upperBound(long[] a, long x) {// x is the target, returns upperBound. If not found
int l = -1, r = a.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (a[m] == x) {
return new Pair<>(m, 1);
}
if (a[m] <= x)
l = m;
else
r = m;
}
return new Pair<>(l + 1, 0);
}
static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long power(long x, long y) {
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x);
y >>= 1;
x = (x * x);
}
return res;
}
public double binPow(double x, int n) {// binary exponentiation with negative power as well
if (n == 0)
return 1.0;
double binPow = binPow(x, n / 2);
if (n % 2 == 0) {
return binPow * binPow;
} else {
return n > 0 ? (binPow * binPow * x) : (binPow * binPow / x);
}
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
/*
* ===========Modular Operations==================
*/
static long modPower(long x, long y, long p) {
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
static long modInverse(long n, long p) {
return modPower(n, p - 2, p);
}
static long modAdd(long a, long b) {
return ((a + b + mod) % mod);
}
static long modMul(long a, long b) {
return ((a % mod) * (b % mod)) % mod;
}
static long[] fac = new long[200000 + 5];
static long[] invFac = new long[200000 + 5];
static long nCrModPFermat(int n, int r) {
if (r == 0)
return 1;
// return (fac[n] * modInverse(fac[r], mod) % mod * modInverse(fac[n - r], mod)
// % mod) % mod;
return (fac[n] * invFac[r] % mod * invFac[n - r] % mod) % mod;
}
/*
* ===============================================
*/
static void ruffleSort(long[] a) {
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n);
long temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(int[] a) {
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n);
int temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
static String binString32(int n) {
StringBuilder str = new StringBuilder("");
String bin = Integer.toBinaryString(n);
if (bin.length() != 32) {
for (int k = 0; k < 32 - bin.length(); k++) {
str.append("0");
}
str.append(bin);
}
return str.toString();
}
static class sparseTable {
public static int st[][];
public static int log = 4;
static int func(int a, int b) {// make func as per question(here min range query)
return (int) gcd(a, b);
}
void makeTable(int n, int a[]) {
st = new int[n][log];
for (int i = 0; i < n; i++) {
st[i][0] = a[i];
}
for (int j = 1; j < log; j++) {
for (int i = 0; i + (1 << j) <= n; i++) {
st[i][j] = func(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
}
}
static int query(int l, int r) {
int length = r - l + 1;
int k = 0;
while ((1 << (k + 1)) <= length) {
k++;
}
return func(st[l][k], st[r - (1 << k) + 1][k]);
}
static void printTable(int n) {
for (int j = 0; j < log; j++) {
for (int i = 0; i < n; i++) {
System.out.print(st[i][j] + " ");
}
System.out.println();
}
}
}
/*
* ====================================Main=================================
*/
// static int st[], a[];
// static void buildTree(int treeIndex, int lo, int hi) {
// if (hi == lo) {
// st[treeIndex] = a[lo];
// return;
// }
// int mid = (lo + hi) / 2;
// buildTree(treeIndex * 2 + 1, lo, mid);
// buildTree(treeIndex * 2 + 2, mid + 1, hi);
// st[treeIndex] = merge(st[treeIndex * 2 + 1], st[treeIndex * 2 + 2]);
// }
// static void update(int treeIndex, int lo, int hi, int arrIndex, int val) {
// if (hi == lo) {
// st[treeIndex] = val;
// a[arrIndex] = val;
// return;
// }
// int mid = (hi + lo) / 2;
// if (mid < arrIndex) {
// update(treeIndex * 2 + 2, mid + 1, hi, arrIndex, val);
// } else {
// update(treeIndex * 2 + 1, lo, mid, arrIndex, val);
// }
// st[treeIndex] = merge(st[treeIndex * 2 + 1], st[treeIndex * 2 + 2]);
// }
// static int query(int treeIndex, int lo, int hi, int l, int r) {
// if (l <= lo && r >= hi) {
// return st[treeIndex];
// }
// if (l > hi || r < lo) {
// return 0;
// }
// int mid = (hi + lo) / 2;
// return query(treeIndex * 2 + 1, lo, mid, l, Math.min(mid, r));
// }
// static int merge(int a, int b) {
// return a + b;
// }
public static long findKthPositive(long[] A, long k) {
int l = 0, r = A.length, m;
while (l < r) {
m = (l + r) / 2;
if (A[m] - 1 - m < k)
l = m + 1;
else
r = m;
}
return l + k;
}
static int[] z_function(char ar[]) {
int[] z = new int[ar.length];
z[0] = ar.length;
int l = 0;
int r = 0;
for (int i = 1; i < ar.length; i++) {
if (r < i) {
l = i;
r = i;
while (r < ar.length && ar[r - l] == ar[r])
r++;
z[i] = r - l;
r--;
} else {
int k = i - l;
if (z[k] < r - i + 1) {
z[i] = z[k];
} else {
l = i;
while (r < ar.length && ar[r - l] == ar[r])
r++;
z[i] = r - l;
r--;
}
}
}
return z;
}
static void mergeSort(int a[]) {
int n = a.length;
if (n >= 2) {
int mid = n / 2;
int left[] = new int[mid];
int right[] = new int[n - mid];
for (int i = 0; i < mid; i++) {
left[i] = a[i];
}
for (int i = mid; i < n; i++) {
right[i - mid] = a[i];
}
mergeSort(left);
mergeSort(right);
mergeSortedArray(left, right, a);
}
}
static void mergeSortedArray(int left[], int right[], int a[]) {
int i = 0, j = 0, k = 0, n = left.length, m = right.length;
while (i != n && j != m) {
if (left[i] < right[j]) {
a[k++] = left[i++];
} else {
a[k++] = right[j++];
}
}
while (i != n) {
a[k++] = left[i++];
}
while (j != m) {
a[k++] = right[j++];
}
}
// BINARY SEARCH
// count of set bits in a particular position
// suffix max array/ suffix sum array/ prefix
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] intArr(int n) {
int res[] = new int[n];
for (int i = 0; i < n; i++)
res[i] = nextInt();
return res;
}
long[] longArr(int n) {
long res[] = new long[n];
for (int i = 0; i < n; i++)
res[i] = nextLong();
return res;
}
}
static FastReader f = new FastReader();
static BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
static ArrayList<String> strArr = new ArrayList<>();
public static void main(String args[]) throws Exception {
int t = 1;
t = f.nextInt();
int tc = 1;
// fac[0] = 1;
// for (int i = 1; i <= 200000; i++) {
// fac[i] = fac[i - 1] * i % mod;
// invFac[i] = modInverse(fac[i], mod);
// }
while (t-- != 0) {
int n=f.nextInt();
int e=f.nextInt();
long ans=0;
int a[]=new int[n];
a=f.intArr(n);
for(int i=0;i<n;i++){
if(isPrime(a[i])){
long left=i,right=i,l=1,r=1;
left-=e;
right+=e;
while(left>=0 && a[(int)left]==1){
left-=e;
l++;
}
while(right<n && a[(int)right]==1){
right+=e;
r++;
}
ans+=l*r-1;
}
}
w.write(ans+"\n");
}
w.flush();
}
}
/*
*/ | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | a298fd6fd569b6e41cbaafd0ad60110e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.File;import java.io.FileInputStream;import java.io.FileNotFoundException;
import java.io.IOException;import java.io.InputStream;import java.io.PrintWriter;
import java.security.AccessControlException;import java.util.ArrayList;import java.util.Arrays;
import java.util.List;import java.util.TreeSet;public class _C {static public void main(final String[] args)
throws IOException{C._main(args);}
static class C extends Solver{public C(){}TreeSet<Integer>herato=new TreeSet<>();
@Override public void solve()throws IOException{int n=sc.nextInt();int e=sc.nextInt();sc.nextLine();
int[]a=Arrays.stream(sc.nextLine().trim().split("\\s+")).filter($s2->!$s2.isEmpty()).mapToInt(Integer::parseInt).toArray();
if(herato.isEmpty()){herato.addAll(Herato.apply(1000000));}ArrayList<int[]>list=
new ArrayList<>();for(int i=0;i<e;i++){int[]cur=null;for(int j=i;;j+=e){if(j>=n)
{break;}if(a[j]!=1){if(herato.contains(a[j])){if(cur==null){cur=new int[]{j,-1,-1};
}if(cur[1]!=-1){if(cur[2]!=cur[0]&& cur[1]!=-1){list.add(cur);}cur=new int[]{cur[1]
+e,-1,-1};}cur[1]=j;}else if(cur!=null){if(cur[2]!=cur[0]&& cur[1]!=-1){list.add(cur);
}cur=null;}}else{if(cur==null){cur=new int[]{j,-1,-1};}}if(cur!=null){cur[2]=j;}
}if(cur!=null && cur[2]!=cur[0]&& cur[1]!=-1){list.add(cur);}}long res=0;for(int[]
v:list){res+=(((long)v[1]-v[0])/e+1)*(((long)v[2]-v[1])/e+1)-1;}pw.println(res);
}static public void _main(String[]args)throws IOException{new C().run();}}static class
Herato{public static void apply(final int a[]){for(int i=2;i<a.length;){for(int
j=i*2;j<a.length;j+=i){a[j]=1;}for(i++;i<a.length && a[i]==1;i++){}}}public static
List<Integer>apply(final int n){int[]a=new int[n+1];ArrayList<Integer>res=new ArrayList<>();
for(int i=2;i<a.length;){res.add(i);for(int j=i*2;j<a.length;j+=i){a[j]=1;}for(i++;
i<a.length && a[i]==1;i++){}}return res;}}static class MyScanner{private StringBuilder
cache=new StringBuilder();int cache_pos=0;private int first_linebreak=-1;private
int second_linebreak=-1;private StringBuilder sb=new StringBuilder();private InputStream
is=null;public MyScanner(final InputStream is){this.is=is;}public String charToString(final
int c){return String.format("'%s'",c=='\n'?"\\n":(c=='\r'?"\\r":String.valueOf((char)c)));
}public int get(){int res=-1;if(cache_pos<cache.length()){res=cache.charAt(cache_pos);
cache_pos++;if(cache_pos==cache.length()){cache.delete(0,cache_pos);cache_pos=0;
}}else{try{res=is.read();}catch(IOException ex){throw new RuntimeException(ex);}
}return res;}private void unget(final int c){if(cache_pos==0){cache.insert(0,(char)c);
}else{cache_pos--;}}public String nextLine(){sb.delete(0,sb.length());int c;boolean
done=false;boolean end=false;while((c=get())!=-1){if(check_linebreak(c)){done=true;
if(c==first_linebreak){if(!end){end=true;}else{cache.append((char)c);break;}}else
if(second_linebreak!=-1 && c==second_linebreak){break;}}if(end && c!=first_linebreak
&& c!=second_linebreak){cache.append((char)c);break;}if(!done){sb.append((char)c);
}}return!done && sb.length()==0?null:sb.toString();}private boolean check_linebreak(int
c){if(c=='\n'|| c=='\r'){if(first_linebreak==-1){first_linebreak=c;}else if(c!=first_linebreak
&& second_linebreak==-1){second_linebreak=c;}return true;}return false;}public int
nextInt(){return Integer.parseInt(next());}public long nextLong(){return Long.parseLong(next());
}public boolean hasNext(){boolean res=false;int c;while((c=get())!=-1){if(!check_linebreak(c)
&& c!=' '&& c!='\t'){res=true;unget(c);break;}}return res;}public String next(){
sb.delete(0,sb.length());boolean started=false;int c;while((c=get())!=-1){if(check_linebreak(c)
|| c==' '|| c=='\t'){if(started){unget(c);break;}}else{started=true;sb.append((char)c);
}}return sb.toString();}public int nextChar(){return get();}public boolean eof()
{int c=get();boolean res=false;if(c!=-1){unget(c);}else{res=true;}return res;}public
double nextDouble(){return Double.parseDouble(next());}}static abstract class Solver
{protected String nameIn=null;protected String nameOut=null;protected boolean singleTest
=false;protected MyScanner sc=null;protected PrintWriter pw=null;private int current_test
=0;private int count_tests=0;protected int currentTest(){return current_test;}protected
int countTests(){return count_tests;}private void process()throws IOException{if(!singleTest)
{count_tests=sc.nextInt();sc.nextLine();for(current_test=1;current_test<=count_tests;
current_test++){solve();pw.flush();}}else{count_tests=1;current_test=1;solve();}
}abstract protected void solve()throws IOException;public void run()throws IOException
{boolean done=false;try{if(nameIn!=null){if(new File(nameIn).exists()){try(FileInputStream
fis=new FileInputStream(nameIn);PrintWriter pw0=select_output();){select_output();
done=true;sc=new MyScanner(fis);pw=pw0;process();}}else{throw new RuntimeException("File "
+new File(nameIn).getAbsolutePath()+" does not exist!");}}}catch(IOException | AccessControlException
ex){}if(!done){try(PrintWriter pw0=select_output();){sc=new MyScanner(System.in);
pw=pw0;process();}}}private PrintWriter select_output()throws FileNotFoundException
{if(nameOut!=null){return new PrintWriter(nameOut);}return new PrintWriter(System.out);
}}static abstract class Tester{static public int getRandomInt(final int min,final
int max){return(min+(int)Math.floor(Math.random()*(max-min+1)));}static public long
getRandomLong(final long min,final long max){return(min+(long)Math.floor(Math.random()
*(max-min+1)));}static public double getRandomDouble(final double min,final double
maxExclusive){return(min+Math.random()*(maxExclusive-min));}abstract protected void
testOutput(final List<String>output_data);abstract protected void generateInput();
abstract protected String inputDataToString();private boolean break_tester=false;
protected void beforeTesting(){}protected void breakTester(){break_tester=true;}
public boolean broken(){return break_tester;}}}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | b643ea5380d1412588a22a626d099c70 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Solution{
static long mod=(long)1e9+7;
static boolean[] prime=sieveOfEratosthenes(((int)1e6+7));
static FastScanner sc = new FastScanner();
public static void solve(){
int n=sc.nextInt();
int e=sc.nextInt();
int[] a=sc.readArray(n);
int[] pre=new int[n];
int[] suff=new int[n];
ArrayList<Integer> lst=new ArrayList<>();
for (int i=0;i<n;i++ ) {
if (a[i]==1) {
lst.add(i);
}
}
for (int i:lst) {
pre[i]=1;
if (i-e>=0) {
pre[i]+=pre[i-e];
}
}
for (int i=lst.size()-1;i>=0;i--) {
int j=lst.get(i);
suff[j]=1;
if (j+e<n){
suff[j]+=suff[j+e];
}
}
long ans=0l;
for (int i=0;i<n;i++) {
if (prime[a[i]]) {
int p=0;
int s=0;
if (i-e>=0)
p=pre[i-e];
if (i+e<n) {
s=suff[i+e];
}
ans+=((long)(p+1)*(s+1))-1l;
}
}
System.out.println(ans);
}
public static void main(String[] args) {
int t = sc.nextInt();
// int t=1;
outer: for (int tt = 0; tt < t; tt++) {
solve();
}
}
static long log2(long n){
return (long)((double)Math.log((double)n)/Math.log((double)2));
}
static boolean isPrime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static int[] LPS(String s){
int[] lps=new int[s.length()];
int i=0,j=1;
while (j<s.length()) {
if(s.charAt(i)==s.charAt(j)){
lps[j]=i+1;
i++;
j++;
continue;
}else{
if (i==0) {
j++;
continue;
}
i=lps[i-1];
while(s.charAt(i)!=s.charAt(j) && i!=0) {
i=lps[i-1];
}
if(s.charAt(i)==s.charAt(j)){
lps[j]=i+1;
i++;
}
j++;
}
}
return lps;
}
static long getPairsCount(int n, double sum,int[] arr)
{
HashMap<Double, Integer> hm = new HashMap<>();
for (int i = 0; i < n; i++) {
if (!hm.containsKey((double)arr[i]))
hm.put((double)arr[i], 0);
hm.put((double)arr[i], hm.get((double)arr[i]) + 1);
}
long twice_count = 0;
for (int i = 0; i < n; i++) {
if (hm.get(sum - arr[i]) != null)
twice_count += hm.get(sum - arr[i]);
if (sum - (double)arr[i] == (double)arr[i])
twice_count--;
}
return twice_count / 2l;
}
static boolean[] sieveOfEratosthenes(int n)
{
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
prime[1]=false;
return prime;
}
static long power(long x, long y, long p)
{
long res = 1l;
x = x % p;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % p;
y>>=1;
x = (x * x) % p;
}
return res;
}
public static int log2(int N)
{
int result = (int)(Math.log(N) / Math.log(2));
return result;
}
////////////////////////////////////////////////////////////////////////////////////
////////////////////DO NOT READ AFTER THIS LINE //////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////
static long modFact(int n,
int p)
{
if (n >= p)
return 0;
long result = 1l;
for (int i = 3; i <= n; i++)
result = (result * i) % p;
return result;
}
static boolean isPalindrom(char[] arr, int i, int j) {
boolean ok = true;
while (i <= j) {
if (arr[i] != arr[j]) {
ok = false;
break;
}
i++;
j--;
}
return ok;
}
static int max(int a, int b) {
return Math.max(a, b);
}
static int min(int a, int b) {
return Math.min(a, b);
}
static long max(long a, long b) {
return Math.max(a, b);
}
static long min(long a, long b) {
return Math.min(a, b);
}
static int abs(int a) {
return Math.abs(a);
}
static long abs(long a) {
return Math.abs(a);
}
static void swap(long arr[], int i, int j) {
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static void swap(int arr[], int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static int maxArr(int arr[]) {
int maxi = Integer.MIN_VALUE;
for (int x : arr)
maxi = max(maxi, x);
return maxi;
}
static int minArr(int arr[]) {
int mini = Integer.MAX_VALUE;
for (int x : arr)
mini = min(mini, x);
return mini;
}
static long maxArr(long arr[]) {
long maxi = Long.MIN_VALUE;
for (long x : arr)
maxi = max(maxi, x);
return maxi;
}
static long minArr(long arr[]) {
long mini = Long.MAX_VALUE;
for (long x : arr)
mini = min(mini, x);
return mini;
}
static int lcm(int a,int b){
return (int)(((long)a*b)/(long)gcd(a,b));
}
static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static void ruffleSort(int[] a) {
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n);
int temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
public static int binarySearch(int a[], int target) {
int left = 0;
int right = a.length - 1;
int mid = (left + right) / 2;
int i = 0;
while (left <= right) {
if (a[mid] <= target) {
i = mid + 1;
left = mid + 1;
} else {
right = mid - 1;
}
mid = (left + right) / 2;
}
return i-1;
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
int[][] read2dArray(int n, int m) {
int arr[][] = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr[i][j] = nextInt();
}
}
return arr;
}
ArrayList<Integer> readArrayList(int n) {
ArrayList<Integer> arr = new ArrayList<Integer>();
for (int i = 0; i < n; i++) {
int a = nextInt();
arr.add(a);
}
return arr;
}
long nextLong() {
return Long.parseLong(next());
}
}
static class Pair {
int val, ind;
Pair(int fr, int sc) {
this.val = fr;
this.ind = sc;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | d0864927d40265243bed3ec252bf4140 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.util.*;
import java.io.*;
public class codeforceb
{
//ArrayList<Integer> arr2=new ArrayList<Integer>();
//ArrayList<Integer> arr1=new ArrayList<Integer>();
//Map<Integer,Integer> hm=new HashMap<Integer,Integer>();
//Set<Integer> st1=new HashSet<Integer>();
static Set<Integer> prime=new HashSet<Integer>();
static int inf = Integer.MAX_VALUE ;
static long infL = Long.MAX_VALUE ;
static int mod=1000000007;
static FastReader sc=new FastReader();
static PrintWriter out=new PrintWriter(System.out);
public static void main (String[] args)
{
/*Arrays.sort(time,new Comparator<Pair>(){
public int compare(Pair p1,Pair p2){
if(p1.a==p2.a)
return p1.b-p2.b;
return p1.a-p2.a;
}
});*/
//int r[][]= {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}
// };
int t=sc.nextInt();
seive(1000000);
while(t-->0)
{
solve();
}
out.close();
}
public static void solve() {
int n=sc.nextInt();
int e=sc.nextInt();
int arr[]=sc.readArray(n);
int left[]=new int[n];
int right[]=new int[n];
boolean f;
int c=0,c1=0;
for(int i=e;i<n;i++) {
if(arr[i-e]==1) left[i]=left[i-e]+1;
}
for(int i=n-e-1;i>=0;i--) {
if(arr[i+e]==1) right[i]=right[i+e]+1;
}
long sum=0;
for(int i=0;i<n;i++) {
if(prime.contains(arr[i])) {
long x=((long)left[i]+1)*(1+(long)right[i])-1;
sum+=x;
}
}
out.println(sum);
}
public static boolean check(int r1,int c1,int n,int m) {
if(r1<0 || c1<0 || r1>=n || c1>=m) return false;
return true;
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static class Pair{
int a;
int b;
Pair(int x, int y)
{
this.a = x;
this.b = y;
}
}
static boolean isPrime(long x)
{
if(x==1)
return false;
if(x<=3)
return true;
if(x%2==0 || x%3==0)
return false;
for(int i=5;i<=Math.sqrt(x);i+=2)
if(x%i==0)
return false;
return true;
}
static long gcd(long a,long b)
{
return (b==0)?a:gcd(b,a%b);
}
static int gcd(int a,int b)
{
return (b==0)?a:gcd(b,a%b);
}
public static long myPow(long x, long n) {
if(x==0)
return 0;
if(n==0)
return 1;
if(n==1)
return x;
long calc=myPow((long)x,n/2);
//System.out.println(calc+" "+n/2);
if(Math.abs(n)%2==0)
return (calc*calc);
else
return (x*calc*calc);
}
public static long lcm(long a , long b){
return a * (b/gcd(a,b));
}
public static int lcm(int a , int b){
return (a * b)/gcd(a,b);
}
public static void swap(int[] arr, int left , int right){
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
public static void swap(char[] arr, int left , int right){
char temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
}
public static void reverse(int[] arr){
int left = 0;
int right = arr.length-1;
while(left <= right){
swap(arr, left,right);
left++;
right--;
}
}
public static void seive(int val) {
int arr[]=new int[val+1];
for(int i=2;i<=val;i++) {
if(arr[i]==0) {
prime.add(i);
for(int j=i;j<=val;j+=i) arr[j]=1;
}
else continue;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | db8242445f99458cb9412a357d57cf63 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
/*
بسم الله الرحمن الرحيم
/$$$$$ /$$$$$$ /$$ /$$ /$$$$$$
|__ $$ /$$__ $$ |$$ |$$ /$$__ $$
| $$| $$ \ $$| $$|$$| $$ \ $$
| $$| $$$$$$$$| $$ / $$/| $$$$$$$$
/ $$ | $$| $$__ $$ \ $$ $$/ | $$__ $$
| $$ | $$| $$ | $$ \ $$$/ | $$ | $$
| $$$$$$/| $$ | $$ \ $/ | $$ | $$
\______/ |__/ |__/ \_/ |__/ |__/
/$$$$$$$ /$$$$$$$ /$$$$$$ /$$$$$$ /$$$$$$$ /$$$$$$ /$$ /$$ /$$ /$$ /$$$$$$$$ /$$$$$$$
| $$__ $$| $$__ $$ /$$__ $$ /$$__ $$| $$__ $$ /$$__ $$| $$$ /$$$| $$$ /$$$| $$_____/| $$__ $$
| $$ \ $$| $$ \ $$| $$ \ $$| $$ \__/| $$ \ $$| $$ \ $$| $$$$ /$$$$| $$$$ /$$$$| $$ | $$ \ $$
| $$$$$$$/| $$$$$$$/| $$ | $$| $$ /$$$$| $$$$$$$/| $$$$$$$$| $$ $$/$$ $$| $$ $$/$$ $$| $$$$$ | $$$$$$$/
| $$____/ | $$__ $$| $$ | $$| $$|_ $$| $$__ $$| $$__ $$| $$ $$$| $$| $$ $$$| $$| $$__/ | $$__ $$
| $$ | $$ \ $$| $$ | $$| $$ \ $$| $$ \ $$| $$ | $$| $$\ $ | $$| $$\ $ | $$| $$ | $$ \ $$
| $$ | $$ | $$| $$$$$$/| $$$$$$/| $$ | $$| $$ | $$| $$ \/ | $$| $$ \/ | $$| $$$$$$$$| $$ | $$
|__/ |__/ |__/ \______/ \______/ |__/ |__/|__/ |__/|__/ |__/|__/ |__/|________/|__/ |__/
*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Deltix281121C {
public static void main(String[] args) throws java.lang.Exception {
// your code goes here
//try {
// Scanner sc=new Scanner(System.in);
FastReader sc = new FastReader();
int t = sc.nextInt();
int N=1000000+10;
boolean prime[] = new boolean[N+1];
for(int i=0;i<=N;i++)
prime[i] = true;
prime[0]=false;
prime[1]=false;
for(int p = 2; p*p <=N; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p] == true)
{
// Update all multiples of p
for(int i = p*p; i <= N; i += p)
prime[i] = false;
}
}
while (t-- > 0) {
int n=sc.nextInt();
int e=sc.nextInt();
int[] arr=new int[n];
ArrayList<Integer> al=new ArrayList<Integer>();
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
if(arr[i]==1)al.add(i);
}
int[] pref=new int[n];
int[] suf=new int[n];
for(int i=0;i<al.size();i++){
pref[al.get(i)]=al.get(i)-e>=0?pref[al.get(i)-e]+1:1;
}
int i=0;
int j=al.size()-1;
while(i<=j){
int temp=al.get(i);
al.set(i,al.get(j));
al.set(j,temp);
i++;
j--;
}
for( i=0;i<al.size();i++){
suf[al.get(i)]=al.get(i)+e<n?suf[al.get(i)+e]+1:1;
}
long ans=0l;
for( i=0;i<n;i++){
if(prime[arr[i]]){
int pr=0;
int sf=0;
if(i-e>=0){
pr=pref[i-e];
ans+=Long.valueOf(pr);
}
if(i+e<n){
sf=suf[i+e];
ans+=Long.valueOf(sf);
}
ans+=(Long.valueOf(pr)*Long.valueOf(sf));
}
}
//for(int i)
System.out.println(ans);
/*long[] arr=new long[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextLong();
}*/
}
/*} catch (Exception e) {
return;
}*/
}
public static int lowerbound(int[] ar,int k)
{
int s=0;
int e=ar.length;
while (s !=e)
{
int mid = s+e>>1;
if (ar[mid] <k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==ar.length)
{
return -1;
}
return s;
}
public static class pair {
int ff;
int ss;
pair(int ff, int ss) {
this.ff = ff;
this.ss = ss;
}
}
static int BS(int[] arr, int l, int r, int element) {
int low = l;
int high = r;
while (high - low > 1) {
int mid = low + (high - low) / 2;
if (arr[mid] < element) {
low = mid + 1;
} else {
high = mid;
}
}
if (arr[low] == element) {
return low;
} else if (arr[high] == element) {
return high;
}
return -1;
}
static int lower_bound(int[] arr, int l, int r, int element) {
int low = l;
int high = r;
while (high - low > 1) {
int mid = low + (high - low) / 2;
if (arr[mid] < element) {
low = mid + 1;
} else {
high = mid;
}
}
if (arr[low] >= element) {
return low;
} else if (arr[high] >= element) {
return high;
}
return -1;
}
static int upper_bound(int[] arr, int l, int r, int element) {
int low = l;
int high = r;
while (high - low > 1) {
int mid = low + (high - low) / 2;
if (arr[mid] <= element) {
low = mid + 1;
} else {
high = mid;
}
}
if (arr[low] > element) {
return low;
} else if (arr[high] > element) {
return high;
}
return -1;
}
public static int upperbound(long[] arr, int k) {
int s = 0;
int e = arr.length;
while (s != e) {
int mid = s + e >> 1;
if (arr[mid] <= k) {
s = mid + 1;
} else {
e = mid;
}
}
if (s == arr.length) {
return -1;
}
return s;
}
public static long pow(long x,long y,long mod){
if(x==0)return 0l;
if(y==0)return 1l;
//(x^y)%mod
if(y%2l==1l){
return ((x%mod)*(pow(x,y-1l,mod)%mod))%mod;
}
return pow(((x%mod)*(x%mod))%mod,y/2l,mod);
}
public static long gcd_long(long a, long b) {
// a/b,a-> dividant b-> divisor
if (b == 0)
return a;
return gcd_long(b, a % b);
}
public static int gcd_int(int a, int b) {
// a/b,a-> dividant b-> divisor
if (b == 0)
return a;
return gcd_int(b, a % b);
}
public static int lcm(int a, int b) {
int gcd = gcd_int(a, b);
return (a * b) / gcd;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble(){
return Double.valueOf(Integer.parseInt(next()));
}
String nextLine() {
String s = "";
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
Long nextLong() {
return Long.parseLong(next());
}
}
public static boolean contains(String main, String Substring) {
boolean flag=false;
if(main==null && main.trim().equals("")) {
return flag;
}
if(Substring==null) {
return flag;
}
char fullstring[]=main.toCharArray();
char sub[]=Substring.toCharArray();
int counter=0;
if(sub.length==0) {
flag=true;
return flag;
}
for(int i=0;i<fullstring.length;i++) {
if(fullstring[i]==sub[counter]) {
counter++;
} else {
counter=0;
}
if(counter==sub.length) {
flag=true;
return flag;
}
}
return flag;
}
}
/*
* public static boolean lie(int n,int m,int k){ if(n==1 && m==1 && k==0){
* return true; } if(n<1 || m<1 || k<0){ return false; } boolean
* tc=lie(n-1,m,k-m); boolean lc=lie(n,m-1,k-n); if(tc || lc){ return true; }
* return false; }
*/
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 11708ad7dee9f53143a6ced741658f3e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
public class Main{
static boolean [] prime;
public static void sieveOfEratosthenes(int n)
{
prime = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
sieveOfEratosthenes((int) 1e6);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int e = sc.nextInt();
int [] arr = new int[n+1];
for (int i = 1; i <= n; i++) {
arr[i] = sc.nextInt();
}
long ans = 0;
for (int i = 1; i < n+1; i++) {
long l = 0;
long r = 0;
if(prime[arr[i]] && arr[i]!=1){
for (int j = i+e; j <= n ; j+=e) {
if(arr[j]==1)r++;
else break;
}
for (int j = i-e; j >=1 ; j-=e) {
if(arr[j]==1)l++;
else break;
}
ans += (r*l)+(r+l);
}
}
System.out.println(ans);
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 347b0ef9010172e7c78c743ca2e728a6 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
public class ComplexMarketAnalysis_C {
private static boolean isPrime(int n) {
if(n==1) {
return false;
}
for(int i = 2; i*i<=n; i++) {
if(n%i==0) {
return false;
}
}
return true;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int e = sc.nextInt();
int[] arr = new int[n];
long ans = 0;
for(int i =0 ; i<n; i++) {
arr[i] = sc.nextInt();
}
for(int i = 0; i<n; i++) {
if(isPrime(arr[i])) {
// System.out.print(i+" ");
int p = 0;
int ii = i-e;
boolean check = true;
int count = 1;
while(ii>=0) {
if(arr[ii]==1) {
count++;
}
else {
break;
}
ii=ii-e;
}
p = count;
// System.out.println(p);
int s = 0;
ii = i+e;
count = 0;
while(ii<n) {
if(arr[ii]==1) {
count++;
}
else {
break;
}
ii = ii + e;
}
s = count;
ans = ans + ((long)p-1)*((long)s+1) + ((long)s);
}
}
System.out.println(ans);
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 84d5d9f44a0a1db5ba54b69764b05a06 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.math.*;
import java.util.*;
// @author : Dinosparton
public class test {
static class Pair{
long x;
long y;
Pair(long x,long y){
this.x = x;
this.y = y;
}
}
static class Sort implements Comparator<Pair>
{
@Override
public int compare(Pair a, Pair b)
{
if(a.x!=b.x)
{
return (int)(a.x - b.x);
}
else
{
return (int)(a.y-b.y);
}
}
}
static class Compare {
void compare(Pair arr[], int n)
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
if(p1.x!=p2.x) {
return (int)(p1.x - p2.x);
}
else {
return (int)(p1.y - p2.y);
}
}
});
// for (int i = 0; i < n; i++) {
// System.out.print(arr[i].x + " " + arr[i].y + " ");
// }
// System.out.println();
}
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static Set<Integer> set = new HashSet<>();
public static void sieveOfEratosthenes()
{
int n = 1000000;
int[] a = new int[n + 1];
for (int i = 0; i <= n; i++) { // initialize all numbers as prime
a[i] = 1;
}
for (int i = 2; i <= Math.sqrt(n); i++)
{
if (a[i] == 1) // checks if `i` is prime
{
for (int j = 2; i * j <= n; j++) {
a[i * j] = 0; // multiples of `i` are not prime
}
}
}
for (int i = 2; i <= n; i++)
{
if (a[i] == 1) {
set.add(i); // prints primes
}
}
}
public static void main(String args[]) throws Exception {
Scanner sc = new Scanner();
StringBuilder res = new StringBuilder();
sieveOfEratosthenes();
int tc = sc.nextInt();
while(tc-->0) {
int n = sc.nextInt();
int e = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++) {
a[i] = sc.nextInt();
}
long cnt = 0;
for(int i=0;i<n;i++) {
if(set.contains(a[i])) {
int suf = 0;
int index = i;
while(index+e<n && a[index+e]==1) {
index+=e;
suf++;
}
cnt += suf;
suf++;
index = i;
while(index-e>=0 && a[index-e]==1) {
index -= e;
cnt += suf;
}
}
}
res.append(cnt+"\n");
}
System.out.println(res);
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 09ede864f0dbebe7a9215cd2e8cc7de6 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Set;
public class Simple{
public static Deque<Integer> possible(int a[],int n){
boolean bool = true;
Deque<Integer> deq = new LinkedList<>();
int i=0;
int j=n-1;
int max = n;
int min = 1;
while(i<=j){
if(a[i]== min){
deq.addFirst(a[i]);
i++;min++;
continue;
}
if(a[j]== min){
deq.addLast(a[j]);
j--;
min++;
continue;
}
if(a[i]==max){
deq.addFirst(a[i]);
i++;
max--;
continue;
}
if(a[j]==max){
deq.addLast(a[j]);
j--;
max--;
continue;
}
Deque<Integer> d = new LinkedList<>();
d.add(-1);
return d;
}
return deq;
}
public static class Pair implements Comparable<Pair>{
int val;
int freq = 0;
public Pair(int val,int freq){
this.val = val;
this.freq= freq;
}
public int compareTo(Pair p){
if(p.freq == this.freq)return 0;
if(this.freq > p.freq)return -1;
return 1;
}
}
public static long factorial(long n)
{
// single line to find factorial
return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
static long m = 998244353;
// Function to return the GCD of given numbers
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Recursive function to return (x ^ n) % m
static long modexp(long x, long n)
{
if (n == 0) {
return 1;
}
else if (n % 2 == 0) {
return modexp((x * x) % m, n / 2);
}
else {
return (x * modexp((x * x) % m, (n - 1) / 2) % m);
}
}
// Function to return the fraction modulo mod
static long getFractionModulo(long a, long b)
{
long c = gcd(a, b);
a = a / c;
b = b / c;
// (b ^ m-2) % m
long d = modexp(b, m - 2);
// Final answer
long ans = ((a % m) * (d % m)) % m;
return ans;
}
// public static long bs(long lo,long hi,long fact,long num){
// long help = num/fact;
// long ans = hi;
// while(lo<=hi){
// long mid = (lo+hi)/2;
// if(mid/)
// }
// }
public static boolean isPrime(int n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
public static void main(String args[]){
//System.out.println("Hello Java");
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int t1 = 1;t1<=t;t1++){
int n = s.nextInt();
int e = s.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++){
arr[i]=s.nextInt();
}
boolean bool[] = new boolean[n];
for(int i=0;i<n;i++){
bool[i] = isPrime(arr[i]);
}
long ans=0;
for(int i=0;i<n;i++){
if(!bool[i]){
continue;
}
boolean flag = false;
long right =0;
// System.out.print(i+" ");
for(int j= i+e;j<n;j=j+e){
if(arr[j]!=1){
break;
}
right++;
}
long left =0;
// if(!flag){
// }
for(int j=i-e;j>=0;j=j-e){
if(arr[j]!=1)break;
left++;
}
if(right==0 && left ==0){
ans +=0;
}
else{
ans += right + left*(right+1);
}
//System.out.println("HELLO");
}
System.out.println(ans);
}
}
}
/*abstract
1 1 2 1 1
2 1
2 1 1
2 = combn
(1 2)*2
(1 1 2)*2
*/ | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | b115536bc9d6e6df8da441346bda3224 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | // C
import java.util.*;
import java.io.*;
// THIS TEMPLATE MADE BY AKSH BANSAL.
public class Solution {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static void sort(int a[]){ // int -> long
ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
for(int i=0;i<a.length;i++)
arr.add(a[i]);
Collections.sort(arr);
for(int i=0;i<a.length;i++)
a[i]=arr.get(i);
}
private static long gcd(long a, long b){
if(b==0)return a;
return gcd(b,a%b);
}
private static long pow(long x,long y){
if(y==0)return 1;
long temp = pow(x, y/2);
if(y%2==1){
return x*temp*temp;
}
else{
return temp*temp;
}
}
static int log(long n){
int res = 0;
while(n>0){
res++;
n/=2;
}
return res;
}
static int mod = (int)1e9+7;
static PrintWriter out;
static FastReader sc ;
private static boolean[] isPrime;
private static void primes(){
int num = (int)1e6+1; // PRIMES FROM 1 TO NUM
isPrime = new boolean[num];
for (int i = 2; i< isPrime.length; i++) {
isPrime[i] = true;
}
for (int i = 2; i< Math.sqrt(num); i++) {
if(isPrime[i] == true) {
for(int j = (i*i); j<num; j = j+i) {
isPrime[j] = false;
}
}
}
}
public static void main(String[] args) throws IOException {
sc = new FastReader();
out = new PrintWriter(System.out);
primes();
// ________________________________
int test = sc.nextInt();
StringBuilder output = new StringBuilder();
while (test-- > 0) {
int n = sc.nextInt();
int k = sc.nextInt();
int[] arr =new int[n];
for(int i=0;i<n;i++){
arr[i] = sc.nextInt();
}
output.append(solver(arr, n, k)).append("\n");
}
out.println(output);
// _______________________________
// int n = sc.nextInt();
// out.println(solver());
// ________________________________
out.flush();
}
public static long solver(int[]arr, int n, int e) {
long res = 0;
boolean[]vis =new boolean[n];
for(int i=0;i<n-e;i++){
if(vis[i])continue;
int k=1;
vis[i] = true;
long one = arr[i]==1?1:0;
long prev = isPrime[arr[i]]?1:0;
while(true){
if(isPrime[arr[i+e*k]]){
res+=one;
prev = one+1;
one = 0;
}
else if(arr[i+e*k]==1){
one++;
res+=prev;
}
else{
prev = 0;
one = 0;
}
vis[i+e*k] = true;
// System.out.println(res+" = "+i+"__"+(i+e*k));
k++;
if(i+e*k>=n)break;
}
}
return res;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 32baab4f7e5e9055566549200de0338e | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class DIv21609C {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
int N = 1000000;
boolean[] prime = new boolean[N+1];
Arrays.fill(prime, true);
prime[0] = prime[1] = false;
for (int i = 2; i <= N; i++) {
if (prime[i] && (long)i * i <= N) {
for (int j = i * i; j <= N; j += i)
prime[j] = false;
}
}
int t = Integer.parseInt(br.readLine());
while(t --> 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int e = Integer.parseInt(st.nextToken());
int[] ar = new int[n];
boolean[] vis = new boolean[n];
st = new StringTokenizer(br.readLine());
for(int i = 0; i < n; i++)
ar[i] = Integer.parseInt(st.nextToken());
int ind = 0;
long tot = 0;
long[] left = new long[n];
while(ind < n && !vis[ind]) {
int count = 0;
for(int i = ind; i < n; i += e) {
vis[i] = true;
if(ar[i] != 1) {
if(prime[ar[i]])
left[i] += count;
count = 0;
}else
count++;
}
ind++;
}
Arrays.fill(vis, false);
ind = n-1;
while(ind >= 0 && !vis[ind]) {
long count = 0;
for(int i = ind; i >= 0; i -= e) {
vis[i] = true;
if(ar[i] != 1) {
if(prime[ar[i]])
tot += (left[i] * count) + left[i] + count;
count = 0;
}else
count++;
}
ind--;
}
pw.println(tot);
}
pw.close();
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 9fdda87d2b2010dff32b61705932e27d | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;
public class TaskC {
final static Set<Integer> primeNums = new HashSet<>();
final static Set<Integer> notPrimeNums = new HashSet<>();
public static void main(String[] arg) {
final FastScanner in = new FastScanner(System.in);
final PrintWriter out = new PrintWriter(System.out);
final int T = in.nextInt();
for (int t = 0; t < T; t++) {
final int n = in.nextInt();
final int e = in.nextInt();
final int[] a = in.readIntArr(n);
Solution sol = new Solution(n, e, a);
out.println(sol.solution());
}
out.flush();
out.close();
in.close();
}
public static class Solution {
final int n;
final int e;
final int[] a;
public Solution(final int n, final int e, final int[] a) {
this.n = n;
this.e = e;
this.a = a;
}
public long solution() {
long result = 0;
for (int i = 0; i < n; i++) {
if (a[i] == 1) {
continue;
}
if (!isPrime(a[i])) {
continue;
}
long forward = forwardLen(i);
long backward = backwardLen(i);
if (forward == 1) {
result += backward - 1;
continue;
}
if (backward == 1) {
result += forward - 1;
continue;
}
result += backward * forward - 1;
}
return result;
}
private long forwardLen(int pos) {
long len = 1;
if (!isPrime(a[pos])) {
return len;
}
int next = pos + e;
while ((next < n) && a[next] == 1) {
len++;
next += e;
}
return len;
}
private long backwardLen(int pos) {
long len = 1;
if (!isPrime(a[pos])) {
return len;
}
int next = pos - e;
while ((next >= 0) && a[next] == 1) {
len++;
next -= e;
}
return len;
}
private boolean isPrime(final int n) {
if (n == 1) {
return false;
}
if (primeNums.contains(n)) {
return true;
}
if (notPrimeNums.contains(n)) {
return false;
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
notPrimeNums.add(n);
return false;
}
}
primeNums.add(n);
return true;
}
}
private static class State {
final int cnt;
final int richCnt;
public State(int cnt, int richCnt) {
this.cnt = cnt;
this.richCnt = richCnt;
}
}
private static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner(InputStream stream) {
try {
br = new BufferedReader(new InputStreamReader(stream));
} catch (Exception e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readIntArr(int n) {
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = Integer.parseInt(next());
}
return result;
}
long[] readLongArr(int n) {
long[] result = new long[n];
for (int i = 0; i < n; i++) {
result[i] = Long.parseLong(next());
}
return result;
}
void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | f12635ba5c3aff6211eb81d10d81f916 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class C {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static boolean isP[] = new boolean[1000001];
public static void main(String[] args){
FastReader sc = new FastReader();
StringBuilder ans = new StringBuilder();
int t = sc.nextInt();
initP();
while (t-->0){
int n = sc.nextInt();
int e = sc.nextInt();
ArrayList<Integer>[] groups = new ArrayList[e];
for (int i=0;i<e;i++){
groups[i] = new ArrayList<>();
}
for (int i=0;i<n;i++){
int next = sc.nextInt();
groups[i%e].add(next);
}
long counter = 0;
for (int i=0;i<e;i++){
ArrayList<Integer> tmp = groups[i];
for (int j=0;j<tmp.size();j++){
if (isP[tmp.get(j)]){
long pos = 0;
long length = 0;
for (int k =j-1;k>=0;k--){
if (tmp.get(k)==1){
pos++;
}else {
break;
}
}
for (int k =j+1;k<tmp.size();k++){
if (tmp.get(k)==1){
length++;
}else {
break;
}
}
counter += pos*(length+1) + length;
}
}
}
ans.append(counter+"\n");
}
System.out.println(ans);
}
public static void initP(){
for (int i=2;i<isP.length;i++){
isP[i] = true;
}
for (int p=2;p*p< isP.length;p++){
if (isP[p]){
for (int i=p*p;i< isP.length;i+=p){
isP[i] = false;
}
}
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 9d26b235b7b0994e3b06388b542d549c | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.util.Scanner;
public class C {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int lines = s.nextInt();
s.nextLine();
for (int i = 0; i < lines; i += 1) {
solve(s.nextInt(), s.nextInt(), s);
}
}
public static void solve(int len, int e, Scanner s) {
int a[] = new int[len];
for (int i = 0; i < len; i++) {
a[i] = s.nextInt();
}
long pairs = 0;
long beforeO;
long thisO;
boolean lP;
for (int i = 0; i < e; i++) {
beforeO = 0;
thisO = 0;
lP = false;
for (int j = i; j < len; j += e) {
if (a[j] == 1) {
thisO++;
} else {
if (lP)
pairs += (beforeO + 1) * (thisO + 1) - 1;
lP = false;
if (prime(a[j])) {
beforeO = thisO;
lP = true;
}
thisO = 0;
}
}
if (lP)
pairs += (beforeO + 1) * (thisO + 1) - 1;
}
System.out.println(pairs);
}
public static boolean prime(int n) {
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) return false;
}
return n != 1;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | eb252edb8e0fb52bf8b7db627c0b876a | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Main {
private static FS sc = new FS();
private static class FS {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
private static class extra {
static int[] intArr(int size) {
int[] a = new int[size];
for(int i = 0; i < size; i++) a[i] = sc.nextInt();
return a;
}
static long[] longArr(int size) {
Scanner scc = new Scanner(System.in);
long[] a = new long[size];
for(int i = 0; i < size; i++) a[i] = sc.nextLong();
return a;
}
static long intSum(int[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static long longSum(long[] a) {
long sum = 0;
for(int i = 0; i < a.length; i++) {
sum += a[i];
}
return sum;
}
static LinkedList[] graphD(int vertices, int edges) {
LinkedList<Integer>[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
}
return temp;
}
static LinkedList[] graphUD(int vertices, int edges) {
LinkedList<Integer>[] temp = new LinkedList[vertices+1];
for(int i = 0; i <= vertices; i++) temp[i] = new LinkedList<>();
for(int i = 0; i < edges; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
temp[x].add(y);
temp[y].add(x);
}
return temp;
}
static void printG(LinkedList[] temp) {
for(LinkedList<Integer> aa:temp) System.out.println(aa);
}
static long cal(long val, long pow, long mod) {
if(pow == 0) return 1;
long res = cal(val, pow/2, mod);
long ret = (res*res)%mod;
if(pow%2 == 0) return ret;
return (val*ret)%mod;
}
static long gcd(long a, long b) { return b == 0 ? a:gcd(b, a%b); }
}
static int mod = 998244353;
static LinkedList<Integer>[] temp, idx;
static long inf = (long) Long.MAX_VALUE;
// static long inf = Long.MAX_VALUE;
static int max;
public static void main(String[] args) {
int t = sc.nextInt();
// int t = 1;
StringBuilder ret = new StringBuilder();
int size = (int) 1e6 + 100;
int[] prime = new int[size];
for(int i = 2; i*i < size; i++) {
if(prime[i] == 0) {
for(int j = i*i; j < size; j += i) prime[j] = 1;
}
}
// for(int i = 1; i < 100; i++) System.out.println(prime[i] + " " + i);
while(t-- > 0) {
int n = sc.nextInt(), e = sc.nextInt();
int[] a = extra.intArr(n);
// ArrayList<Integer> arr = new ArrayList<>();
// for(int i = 0; i < n; i++) if(a[i] == 1) arr.add(i);
long pairs = 0;
for(int i = 0; i < n; i++) {
if(a[i] == 1) continue;
if(prime[a[i]] == 0) {
int left = 0, right = 0;
for(int j = i+e; j < n; j += e) {
if(a[j] == 1) right++;
else break;
}
for(int j = i-e; j >= 0; j -= e) {
if(a[j] == 1) left++;
else break;
}
// System.out.println(a[i] + " " + left + " " + right);
pairs += (long) (left + right + left * 1L * right);
}
}
ret.append(pairs + "\n");
}
System.out.println(ret);
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 69c0c276cf808b050511e0b1de55be69 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
import java.io.*;
import java.util.*;
public class C {
static FastScanner sc = new FastScanner();
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
static HashSet<Integer>prime = new HashSet<>();
public static void init(){
int p = (int)1e6 + 1;
int[]arr = new int[p];
for(int i = 2;i < p;i++){
if(arr[i] == -1)
continue;
for(int j = 2;j * i < p;j++){
arr[j * i] = -1;
}
}
for(int i = 2;i < p;i++){
if(arr[i] != -1){
prime.add(i);
}
}
}
static void solve(){
int n = sc.nextInt();
int e = sc.nextInt();
int[]arr = new int[n];
for(int i = 0;i < n;i++){
int cur = sc.nextInt();
arr[i] = cur;
}
long ans = 0;
for(int i = 0;i < e;i++){
ArrayList<Integer> list = new ArrayList<>();
for(int j = i;j < n;j += e){
list.add(arr[j]);
}
for(int j = 0;j < list.size();j++){
long l = 0,r = 0;
if(prime.contains(list.get(j))){
for(int k = j - 1;k >= 0;k--){
if(list.get(k) == 1){
l++;
}else{
break;
}
}
for(int k = j + 1;k < list.size();k++){
if(list.get(k) == 1){
r++;
}else{
break;
}
}
ans += (l * r) + l + r;
}
}
}
System.out.println(ans);
}
public static void main(String[] args) {
init();
int n = sc.nextInt();
for(int i = 0;i < n;i++){
solve();
}
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | ce9cb87f4ed5f505fa02066d45e695a7 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | /*
Codeforces
Problem 1609C
*/
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.HashSet;
import java.util.Set;
public class ComplexMarketAnalysis {
public static void main(String[] args) throws IOException {
FastIO in = new FastIO(System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int e = in.nextInt();
int[] a = new int[n];
Set<Integer> l = new HashSet<>();
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
if (isPrime(a[i])) l.add(i);
}
long sum = 0;
for (int p : l) {
long i = 1;
while (p - i * e >= 0 && a[p - (int) i * e] == 1) i++;
long j = 1;
while (p + j * e < n && a[p + (int) j * e] == 1) j++;
sum += i * j - 1;
}
System.out.println(sum);
}
in.close();
}
public static boolean isPrime(int n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i += 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static class FastIO {
private InputStream dis;
private byte[] buffer = new byte[1 << 17];
private int pointer = 0;
public FastIO(String fileName) throws IOException {
dis = new FileInputStream(fileName);
}
public FastIO(InputStream is) throws IOException {
dis = is;
}
public int nextInt() throws IOException {
int ret = 0;
byte b;
do {
b = nextByte();
} while (b <= ' ');
boolean negative = false;
if (b == '-') {
negative = true;
b = nextByte();
}
while (b >= '0' && b <= '9') {
ret = 10 * ret + b - '0';
b = nextByte();
}
return (negative) ? -ret : ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte b;
do {
b = nextByte();
} while (b <= ' ');
boolean negative = false;
if (b == '-') {
negative = true;
b = nextByte();
}
while (b >= '0' && b <= '9') {
ret = 10 * ret + b - '0';
b = nextByte();
}
return (negative) ? -ret : ret;
}
public byte nextByte() throws IOException {
if (pointer == buffer.length) {
dis.read(buffer, 0, buffer.length);
pointer = 0;
}
return buffer[pointer++];
}
public String next() throws IOException {
StringBuffer ret = new StringBuffer();
byte b;
do {
b = nextByte();
} while (b <= ' ');
while (b > ' ') {
ret.appendCodePoint(b);
b = nextByte();
}
return ret.toString();
}
public void close() throws IOException {
dis.close();
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | c9ed3695c37491628632d55ca0bcadfc | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.text.DecimalFormat;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class C {
static long mod = (long) 1e9 + 7;
static long mod1 = 998244353;
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
public static void main(String[] args) throws IOException {
int t = in.nextInt();
boolean[] prime = new boolean[1000000+1];
for(int i=0;i<=1000000;i++)
prime[i] = true;
for(int p = 2; p*p <=1000000; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p])
{
// Update all multiples of p
for(int i = p*p; i <= 1000000; i += p)
prime[i] = false;
}
}
while(t-->0) {
int n = in.nextInt();
int e = in.nextInt();
int[] arr = in.readArray(n);
long[] one = new long[n];
for(int i = 0;i<n;i++){
if(arr[i] == 1){
if(i-e>=0)
one[i] = one[i-e]+1;
else
one[i] = 1;
}
}
long[] oneSuffix = new long[n];
for(int i = n-1;i>=0;i--){
if(arr[i] == 1){
if(i+e<n)
oneSuffix[i] = oneSuffix[i+e]+1;
else
oneSuffix[i] = 1;
}
}
// out.println(Arrays.toString(one));
// out.println(Arrays.toString(oneSuffix));
long ans = 0;
for(int i = 0;i<n;i++){
if(prime[arr[i]] && arr[i]!=1){
long left = 0;
long right = 0;
if(i-e>=0) {
ans += one[i - e];
left = one[i-e];
}
if(i+e<n) {
ans += oneSuffix[i + e];
right = oneSuffix[i + e];
}
ans += left*right;
// out.println(arr[i]+" "+ans);
}
}
out.println(ans);
}
out.close();
}
static final Random random = new Random();
static void ruffleSort(int[] a) {
int n = a.length;//shuffle, then sort
for (int i = 0; i < n; i++) {
int oi = random.nextInt(n), temp = a[oi];
a[oi] = a[i];
a[i] = temp;
}
Arrays.sort(a);
}
static long gcd(long x, long y) {
if (x == 0)
return y;
if (y == 0)
return x;
long r = 0, a, b;
a = Math.max(x, y);
b = Math.min(x, y);
r = b;
while (a % b != 0) {
r = a % b;
a = b;
b = r;
}
return r;
}
static long modulo(long a, long b, long c) {
long x = 1, y = a % c;
while (b > 0) {
if (b % 2 == 1)
x = (x * y) % c;
y = (y * y) % c;
b = b >> 1;
}
return x % c;
}
public static void debug(Object... o) {
System.err.println(Arrays.deepToString(o));
}
static int upper_bound(int[] arr, int n, int x) {
int mid;
int low = 0;
int high = n;
while (low < high) {
mid = low + (high - low) / 2;
if (x >= arr[mid])
low = mid + 1;
else
high = mid;
}
return low;
}
static int lower_bound(int[] arr, int n, int x) {
int mid;
int low = 0;
int high = n;
while (low < high) {
mid = low + (high - low) / 2;
if (x <= arr[mid])
high = mid;
else
low = mid + 1;
}
return low;
}
static String printPrecision(double d) {
DecimalFormat ft = new DecimalFormat("0.00000000000");
return String.valueOf(ft.format(d));
}
static int countBit(long mask) {
int ans = 0;
while (mask != 0) {
mask &= (mask - 1);
ans++;
}
return ans;
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = nextInt();
return arr;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 8880c02b10a74db42c683eaf5dafa8eb | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Khater
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
CComplexMarketAnalysis solver = new CComplexMarketAnalysis();
solver.solve(1, in, out);
out.close();
}
static class CComplexMarketAnalysis {
int[] spf;
int MAXN;
void sieve() {
spf[1] = 1;
for (int i = 2; i < MAXN; i++)
// marking smallest prime factor for every
// number to be itself.
spf[i] = i;
// separately marking spf for every even
// number as 2
for (int i = 4; i < MAXN; i += 2)
spf[i] = 2;
for (int i = 3; i * i < MAXN; i++) {
// checking if i is prime
if (spf[i] == i) {
// marking SPF for all numbers divisible by i
for (int j = i * i; j < MAXN; j += i)
// marking spf[j] if it is not
// previously marked
if (spf[j] == j)
spf[j] = i;
}
}
}
boolean prime(int x) {
return x != 1 && spf[x] == x;
}
public void solve(int testNumber, Scanner sc, PrintWriter pw) {
int t = 1;
t = sc.nextInt();
MAXN = (int) (1e6 + 5);
spf = new int[MAXN];
sieve();
while (t-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
int[] arr = sc.nextIntArr(n);
long ans = 0;
ArrayList<Integer>[] list = new ArrayList[e];
for (int i = 0; i < e; i++) {
list[i] = new ArrayList<>();
}
for (int i = 0; i < n; i++) {
list[i % e].add(arr[i]);
}
for (int i = 0; i < e; i++) {
ans += solve(list[i]);
}
pw.println(ans);
}
}
long solve(ArrayList<Integer> arr) {
int[] pre = new int[arr.size()];
int[] suf = new int[arr.size()];
int c = 0;
for (int i = 0; i < arr.size(); i++) {
c += arr.get(i).intValue() == 1 ? 1 : 0;
if (arr.get(i).intValue() != 1) c = 0;
pre[i] = c;
}
c = 0;
for (int i = arr.size() - 1; i >= 0; i--) {
c += arr.get(i).intValue() == 1 ? 1 : 0;
if (arr.get(i).intValue() != 1) c = 0;
suf[i] = c;
}
long ret = 0;
for (int i = 0; i < arr.size(); i++) {
if (prime(arr.get(i))) {
ret += ((1l + (i == 0 ? 0 : pre[i - 1])) * (1l + (i == arr.size() - 1 ? 0 : suf[i + 1])) - 1);
}
}
return ret;
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public int[] nextIntArr(int n) {
try {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
} catch (Exception e) {
throw new RuntimeException(e);
}
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 2278b286afd16cbc2abe6b0db72d4506 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
import java.io.*;
public class Deltix {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
linearSieve((int) ((1e6) + 1));
while (t-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
long [] a = new long[n];
for (int i = 0; i < n; i++) a[i] = sc.nextLong();
ArrayList<Long> [] b = new ArrayList[e];
for (int i = 0; i < e; i++) {
b[i] = new ArrayList<>();
for (int j = i; j < n; j += e) {
b[i].add(a[j]);
}
}
long res = 0;
for (int i = 0; i < e; i++) {
res += solve(b[i]);
}
out.println(res);
}
out.close();
}
static boolean isPrime(long x) {
if (x == 1) return false;
int y = (int) x;
return lp[y] == y;
}
static long solve(ArrayList<Long> a) {
int n = a.size();
int run = 0;
ArrayList<Integer> runs = new ArrayList<>();
ArrayList<Integer> starts = new ArrayList<>();
for (int i = 0; i < n; i++) {
if (a.get(i) == 1) {
if (run == 0) starts.add(i);
++run;
} else {
if (run > 0) runs.add(run);
run = 0;
}
}
if (run > 0) runs.add(run);
long res = 0;
int sz = runs.size();
if (runs.size() == 1) {
if (starts.get(0) != 0 && isPrime(a.get(starts.get(0) - 1))) {
res += runs.get(0);
}
if (starts.get(0) + runs.get(0) < n && isPrime(a.get(starts.get(0) + runs.get(0)))) {
res += runs.get(0);
}
} else {
for (int i = 0; i < sz; i++) {
if (i == 0) {
if (isPrime(a.get(starts.get(i) + runs.get(i)))) res += runs.get(i);
if (starts.get(0) != 0 && isPrime(a.get(starts.get(0) - 1))) res += runs.get(i);
} else if (i == sz - 1) {
if (isPrime(a.get(starts.get(i) - 1))) res += runs.get(i);
if (starts.get(sz - 1) + runs.get(sz - 1) < n && isPrime(a.get(starts.get(sz - 1) + runs.get(sz - 1)))) res += runs.get(i);
} else {
if (isPrime(a.get(starts.get(i) + runs.get(i)))) res += runs.get(i);
if (isPrime(a.get(starts.get(i) - 1))) res += runs.get(i);
}
}
for (int i = 0; i < sz - 1; i++) {
if (starts.get(i) + runs.get(i) + 1 == starts.get(i + 1) && isPrime(a.get(starts.get(i) + runs.get(i)))) {
res += (long) runs.get(i) * runs.get(i + 1);
}
}
}
return res;
}
static int [] lp;
static ArrayList<Integer> primes;
static void linearSieve(int sz) {
lp = new int[sz + 1]; primes = new ArrayList<>();
for (int i = 2; i <= sz; i++) {
if (lp[i] == 0) {
primes.add(i);
lp[i] = i;
}
for (int j = 0; j < primes.size() && j < lp[i] && i * primes.get(j) <= sz; j++) {
lp[primes.get(j) * i] = primes.get(j);
}
}
}
static void sort(int[] a) {
ArrayList<Integer> q = new ArrayList<>();
for (int i : a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
static void sort(long[] a) {
ArrayList<Long> q = new ArrayList<>();
for (long i : a) q.add(i);
Collections.sort(q);
for (int i = 0; i < a.length; i++) a[i] = q.get(i);
}
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | f01e5afd33b0eff995588ed3fd2c0ca9 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.util.*;
public class C {
public static final int MOD998 = 998244353;
public static final int MOD100 = 1000000007;
public static void main(String[] args) throws Exception {
ContestScanner sc = new ContestScanner();
ContestPrinter cp = new ContestPrinter();
int T = sc.nextInt();
boolean[] prime = new boolean[1000001];
Arrays.fill(prime, true);
prime[0] = false;
prime[1] = false;
for (int n = 2; n <= 1000000; n++) {
if (prime[n]) {
for (int m = 2; m * n <= 1000000; m++) {
prime[n * m] = false;
}
}
}
for (int t = 0; t < T; t++) {
int N = sc.nextInt();
int e = sc.nextInt();
int[] A = sc.nextIntArray(N);
long[] dpone = new long[N + 1];
long[] dppri = new long[N + 1];
int pri = 0;
for (int n = 0; n < N; n++) {
int bef = Math.max(n + 1 - e, 0);
if (A[n] == 1) {
dpone[n + 1] = dpone[bef] + 1;
dppri[n + 1] = dppri[bef];
} else if (prime[A[n]]) {
dppri[n + 1] = dpone[bef] + 1;
dpone[n + 1] = 0;
pri++;
} else {
dpone[n + 1] = 0;
dppri[n + 1] = 0;
}
}
long ans = 0;
for (long p : dppri) {
ans += p;
}
cp.println(ans - pri);
}
cp.close();
}
//////////////////
// My Library //
//////////////////
public static int zeroOneBFS(int[][][] weighted_graph, int start, int goal) {
int[] dist = new int[weighted_graph.length];
Arrays.fill(dist, Integer.MAX_VALUE);
dist[start] = 0;
LinkedList<Integer> queue = new LinkedList<>();
queue.add(start);
while (!queue.isEmpty()) {
int now = queue.poll();
if (now == goal) {
return dist[goal];
}
for (int[] info : weighted_graph[now]) {
if (dist[info[0]] > dist[now] + info[1]) {
dist[info[0]] = dist[now] + info[1];
if (info[1] == 0) {
queue.addFirst(info[0]);
} else {
queue.addLast(info[0]);
}
}
}
}
return -1;
}
public static int[] zeroOneBFSAll(int[][][] weighted_graph, int start) {
int[] dist = new int[weighted_graph.length];
Arrays.fill(dist, Integer.MAX_VALUE);
dist[start] = 0;
LinkedList<Integer> queue = new LinkedList<>();
queue.add(start);
while (!queue.isEmpty()) {
int now = queue.poll();
for (int[] info : weighted_graph[now]) {
if (dist[info[0]] > dist[now] + info[1]) {
dist[info[0]] = dist[now] + info[1];
if (info[1] == 0) {
queue.addFirst(info[0]);
} else {
queue.addLast(info[0]);
}
}
}
}
return dist;
}
public static long dijkstra(int[][][] weighted_graph, int start, int goal) {
long[] dist = new long[weighted_graph.length];
Arrays.fill(dist, 0, dist.length, Long.MAX_VALUE);
dist[start] = 0;
PriorityQueue<Pair<Integer, Long>> unsettled = new PriorityQueue<>((u, v) -> (int) (u.cdr - v.cdr));
unsettled.offer(new Pair<Integer, Long>(start, 0L));
while (!unsettled.isEmpty()) {
Pair<Integer, Long> pair = unsettled.poll();
int now = pair.car;
if (now == goal) {
return dist[goal];
}
if (dist[now] < pair.cdr) {
continue;
}
for (int[] info : weighted_graph[now]) {
if (dist[info[0]] > dist[now] + info[1]) {
dist[info[0]] = dist[now] + info[1];
unsettled.offer(new Pair<Integer, Long>(info[0], dist[info[0]]));
}
}
}
return -1;
}
public static long[] dijkstraAll(int[][][] weighted_graph, int start) {
long[] dist = new long[weighted_graph.length];
Arrays.fill(dist, 0, dist.length, Long.MAX_VALUE);
dist[start] = 0;
PriorityQueue<Pair<Integer, Long>> unsettled = new PriorityQueue<>((u, v) -> (int) (u.cdr - v.cdr));
unsettled.offer(new Pair<Integer, Long>(start, 0L));
while (!unsettled.isEmpty()) {
Pair<Integer, Long> pair = unsettled.poll();
int now = pair.car;
if (dist[now] < pair.cdr) {
continue;
}
for (int[] info : weighted_graph[now]) {
if (dist[info[0]] > dist[now] + info[1]) {
dist[info[0]] = dist[now] + info[1];
unsettled.offer(new Pair<Integer, Long>(info[0], dist[info[0]]));
}
}
}
return dist;
}
public static class Pair<A, B> {
public final A car;
public final B cdr;
public Pair(A car_, B cdr_) {
car = car_;
cdr = cdr_;
}
private static boolean eq(Object o1, Object o2) {
return o1 == null ? o2 == null : o1.equals(o2);
}
private static int hc(Object o) {
return o == null ? 0 : o.hashCode();
}
@Override
public boolean equals(Object o) {
if (!(o instanceof Pair))
return false;
Pair<?, ?> rhs = (Pair<?, ?>) o;
return eq(car, rhs.car) && eq(cdr, rhs.cdr);
}
@Override
public int hashCode() {
return hc(car) ^ hc(cdr);
}
}
public static class Tuple1<A> extends Pair<A, Object> {
public Tuple1(A a) {
super(a, null);
}
}
public static class Tuple2<A, B> extends Pair<A, Tuple1<B>> {
public Tuple2(A a, B b) {
super(a, new Tuple1<>(b));
}
}
public static class Tuple3<A, B, C> extends Pair<A, Tuple2<B, C>> {
public Tuple3(A a, B b, C c) {
super(a, new Tuple2<>(b, c));
}
}
public static class Tuple4<A, B, C, D> extends Pair<A, Tuple3<B, C, D>> {
public Tuple4(A a, B b, C c, D d) {
super(a, new Tuple3<>(b, c, d));
}
}
public static class Tuple5<A, B, C, D, E> extends Pair<A, Tuple4<B, C, D, E>> {
public Tuple5(A a, B b, C c, D d, E e) {
super(a, new Tuple4<>(b, c, d, e));
}
}
public static class PriorityQueueLogTime<T> {
private PriorityQueue<T> queue;
private Multiset<T> total;
private int size = 0;
public PriorityQueueLogTime() {
queue = new PriorityQueue<>();
total = new Multiset<>();
}
public PriorityQueueLogTime(Comparator<T> c) {
queue = new PriorityQueue<>(c);
total = new Multiset<>();
}
public void clear() {
queue.clear();
total.clear();
size = 0;
}
public boolean contains(T e) {
return total.count(e) > 0;
}
public boolean isEmpty() {
return size == 0;
}
public boolean offer(T e) {
total.addOne(e);
size++;
return queue.offer(e);
}
public T peek() {
if (total.isEmpty()) {
return null;
}
simplify();
return queue.peek();
}
public T poll() {
if (total.isEmpty()) {
return null;
}
simplify();
size--;
T res = queue.poll();
total.removeOne(res);
return res;
}
public void remove(T e) {
total.removeOne(e);
size--;
}
public int size() {
return size;
}
private void simplify() {
while (total.count(queue.peek()) == 0) {
queue.poll();
}
}
}
static int[][] scanGraphOneIndexed(ContestScanner sc, int node, int edge, boolean undirected) {
int[][] arr = sc.nextIntArrayMulti(edge, 2);
for (int n = 0; n < edge; n++) {
arr[0][n]--;
arr[1][n]--;
}
return GraphBuilder.makeGraph(node, edge, arr[0], arr[1], undirected);
}
static int[][][] scanWeightedGraphOneIndexed(ContestScanner sc, int node, int edge, boolean undirected) {
int[][] arr = sc.nextIntArrayMulti(edge, 3);
for (int n = 0; n < edge; n++) {
arr[0][n]--;
arr[1][n]--;
}
return GraphBuilder.makeGraphWithWeight(node, edge, arr[0], arr[1], arr[2], undirected);
}
static class EdgeData {
private int capacity;
private int[] from, to, weight;
private int p = 0;
private boolean weighted;
public EdgeData(boolean weighted) {
this(weighted, 500000);
}
public EdgeData(boolean weighted, int initial_capacity) {
capacity = initial_capacity;
from = new int[capacity];
to = new int[capacity];
weight = new int[capacity];
this.weighted = weighted;
}
public void addEdge(int u, int v) {
if (weighted) {
System.err.println("The graph is weighted!");
return;
}
if (p == capacity) {
int[] newfrom = new int[capacity * 2];
int[] newto = new int[capacity * 2];
System.arraycopy(from, 0, newfrom, 0, capacity);
System.arraycopy(to, 0, newto, 0, capacity);
capacity *= 2;
}
from[p] = u;
to[p] = v;
p++;
}
public void addEdge(int u, int v, int w) {
if (!weighted) {
System.err.println("The graph is NOT weighted!");
return;
}
if (p == capacity) {
int[] newfrom = new int[capacity * 2];
int[] newto = new int[capacity * 2];
int[] newweight = new int[capacity * 2];
System.arraycopy(from, 0, newfrom, 0, capacity);
System.arraycopy(to, 0, newto, 0, capacity);
System.arraycopy(weight, 0, newweight, 0, capacity);
capacity *= 2;
}
from[p] = u;
to[p] = v;
weight[p] = w;
p++;
}
public int[] getFrom() {
int[] result = new int[p];
System.arraycopy(from, 0, result, 0, p);
return result;
}
public int[] getTo() {
int[] result = new int[p];
System.arraycopy(to, 0, result, 0, p);
return result;
}
public int[] getWeight() {
int[] result = new int[p];
System.arraycopy(weight, 0, result, 0, p);
return result;
}
public int size() {
return p;
}
}
////////////////////////////////
// Atcoder Library for Java //
////////////////////////////////
static class MathLib {
private static long safe_mod(long x, long m) {
x %= m;
if (x < 0)
x += m;
return x;
}
private static long[] inv_gcd(long a, long b) {
a = safe_mod(a, b);
if (a == 0)
return new long[] { b, 0 };
long s = b, t = a;
long m0 = 0, m1 = 1;
while (t > 0) {
long u = s / t;
s -= t * u;
m0 -= m1 * u;
long tmp = s;
s = t;
t = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
if (m0 < 0)
m0 += b / s;
return new long[] { s, m0 };
}
public static long gcd(long a, long b) {
a = java.lang.Math.abs(a);
b = java.lang.Math.abs(b);
return inv_gcd(a, b)[0];
}
public static long lcm(long a, long b) {
a = java.lang.Math.abs(a);
b = java.lang.Math.abs(b);
return a / gcd(a, b) * b;
}
public static long pow_mod(long x, long n, int m) {
assert n >= 0;
assert m >= 1;
if (m == 1)
return 0L;
x = safe_mod(x, m);
long ans = 1L;
while (n > 0) {
if ((n & 1) == 1)
ans = (ans * x) % m;
x = (x * x) % m;
n >>>= 1;
}
return ans;
}
public static long[] crt(long[] r, long[] m) {
assert (r.length == m.length);
int n = r.length;
long r0 = 0, m0 = 1;
for (int i = 0; i < n; i++) {
assert (1 <= m[i]);
long r1 = safe_mod(r[i], m[i]), m1 = m[i];
if (m0 < m1) {
long tmp = r0;
r0 = r1;
r1 = tmp;
tmp = m0;
m0 = m1;
m1 = tmp;
}
if (m0 % m1 == 0) {
if (r0 % m1 != r1)
return new long[] { 0, 0 };
continue;
}
long[] ig = inv_gcd(m0, m1);
long g = ig[0], im = ig[1];
long u1 = m1 / g;
if ((r1 - r0) % g != 0)
return new long[] { 0, 0 };
long x = (r1 - r0) / g % u1 * im % u1;
r0 += x * m0;
m0 *= u1;
if (r0 < 0)
r0 += m0;
// System.err.printf("%d %d\n", r0, m0);
}
return new long[] { r0, m0 };
}
public static long floor_sum(long n, long m, long a, long b) {
long ans = 0;
if (a >= m) {
ans += (n - 1) * n * (a / m) / 2;
a %= m;
}
if (b >= m) {
ans += n * (b / m);
b %= m;
}
long y_max = (a * n + b) / m;
long x_max = y_max * m - b;
if (y_max == 0)
return ans;
ans += (n - (x_max + a - 1) / a) * y_max;
ans += floor_sum(y_max, a, m, (a - x_max % a) % a);
return ans;
}
public static java.util.ArrayList<Long> divisors(long n) {
java.util.ArrayList<Long> divisors = new ArrayList<>();
java.util.ArrayList<Long> large = new ArrayList<>();
for (long i = 1; i * i <= n; i++)
if (n % i == 0) {
divisors.add(i);
if (i * i < n)
large.add(n / i);
}
for (int p = large.size() - 1; p >= 0; p--) {
divisors.add(large.get(p));
}
return divisors;
}
}
static class Multiset<T> extends java.util.TreeMap<T, Long> {
public Multiset() {
super();
}
public Multiset(java.util.List<T> list) {
super();
for (T e : list)
this.addOne(e);
}
public long count(Object elm) {
return getOrDefault(elm, 0L);
}
public void add(T elm, long amount) {
if (!this.containsKey(elm))
put(elm, amount);
else
replace(elm, get(elm) + amount);
if (this.count(elm) == 0)
this.remove(elm);
}
public void addOne(T elm) {
this.add(elm, 1);
}
public void removeOne(T elm) {
this.add(elm, -1);
}
public void removeAll(T elm) {
this.add(elm, -this.count(elm));
}
public static <T> Multiset<T> merge(Multiset<T> a, Multiset<T> b) {
Multiset<T> c = new Multiset<>();
for (T x : a.keySet())
c.add(x, a.count(x));
for (T y : b.keySet())
c.add(y, b.count(y));
return c;
}
}
static class GraphBuilder {
public static int[][] makeGraph(int NumberOfNodes, int NumberOfEdges, int[] from, int[] to,
boolean undirected) {
int[][] graph = new int[NumberOfNodes][];
int[] outdegree = new int[NumberOfNodes];
for (int i = 0; i < NumberOfEdges; i++) {
outdegree[from[i]]++;
if (undirected)
outdegree[to[i]]++;
}
for (int i = 0; i < NumberOfNodes; i++)
graph[i] = new int[outdegree[i]];
for (int i = 0; i < NumberOfEdges; i++) {
graph[from[i]][--outdegree[from[i]]] = to[i];
if (undirected)
graph[to[i]][--outdegree[to[i]]] = from[i];
}
return graph;
}
public static int[][][] makeGraphWithWeight(int NumberOfNodes, int NumberOfEdges, int[] from, int[] to,
int[] weight, boolean undirected) {
int[][][] graph = new int[NumberOfNodes][][];
int[] outdegree = new int[NumberOfNodes];
for (int i = 0; i < NumberOfEdges; i++) {
outdegree[from[i]]++;
if (undirected)
outdegree[to[i]]++;
}
for (int i = 0; i < NumberOfNodes; i++)
graph[i] = new int[outdegree[i]][];
for (int i = 0; i < NumberOfEdges; i++) {
graph[from[i]][--outdegree[from[i]]] = new int[] { to[i], weight[i] };
if (undirected)
graph[to[i]][--outdegree[to[i]]] = new int[] { from[i], weight[i] };
}
return graph;
}
public static int[][][] makeGraphWithEdgeInfo(int NumberOfNodes, int NumberOfEdges, int[] from, int[] to,
boolean undirected) {
int[][][] graph = new int[NumberOfNodes][][];
int[] outdegree = new int[NumberOfNodes];
for (int i = 0; i < NumberOfEdges; i++) {
outdegree[from[i]]++;
if (undirected)
outdegree[to[i]]++;
}
for (int i = 0; i < NumberOfNodes; i++)
graph[i] = new int[outdegree[i]][];
for (int i = 0; i < NumberOfEdges; i++) {
graph[from[i]][--outdegree[from[i]]] = new int[] { to[i], i, 0 };
if (undirected)
graph[to[i]][--outdegree[to[i]]] = new int[] { from[i], i, 1 };
}
return graph;
}
public static int[][][] makeGraphWithWeightAndEdgeInfo(int NumberOfNodes, int NumberOfEdges, int[] from,
int[] to, int[] weight, boolean undirected) {
int[][][] graph = new int[NumberOfNodes][][];
int[] outdegree = new int[NumberOfNodes];
for (int i = 0; i < NumberOfEdges; i++) {
outdegree[from[i]]++;
if (undirected)
outdegree[to[i]]++;
}
for (int i = 0; i < NumberOfNodes; i++)
graph[i] = new int[outdegree[i]][];
for (int i = 0; i < NumberOfEdges; i++) {
graph[from[i]][--outdegree[from[i]]] = new int[] { to[i], weight[i], i, 0 };
if (undirected)
graph[to[i]][--outdegree[to[i]]] = new int[] { from[i], weight[i], i, 1 };
}
return graph;
}
}
static class DSU {
private int n;
private int[] parentOrSize;
public DSU(int n) {
this.n = n;
this.parentOrSize = new int[n];
java.util.Arrays.fill(parentOrSize, -1);
}
int merge(int a, int b) {
if (!(0 <= a && a < n))
throw new IndexOutOfBoundsException("a=" + a);
if (!(0 <= b && b < n))
throw new IndexOutOfBoundsException("b=" + b);
int x = leader(a);
int y = leader(b);
if (x == y)
return x;
if (-parentOrSize[x] < -parentOrSize[y]) {
int tmp = x;
x = y;
y = tmp;
}
parentOrSize[x] += parentOrSize[y];
parentOrSize[y] = x;
return x;
}
boolean same(int a, int b) {
if (!(0 <= a && a < n))
throw new IndexOutOfBoundsException("a=" + a);
if (!(0 <= b && b < n))
throw new IndexOutOfBoundsException("b=" + b);
return leader(a) == leader(b);
}
int leader(int a) {
if (parentOrSize[a] < 0) {
return a;
} else {
parentOrSize[a] = leader(parentOrSize[a]);
return parentOrSize[a];
}
}
int size(int a) {
if (!(0 <= a && a < n))
throw new IndexOutOfBoundsException("" + a);
return -parentOrSize[leader(a)];
}
java.util.ArrayList<java.util.ArrayList<Integer>> groups() {
int[] leaderBuf = new int[n];
int[] groupSize = new int[n];
for (int i = 0; i < n; i++) {
leaderBuf[i] = leader(i);
groupSize[leaderBuf[i]]++;
}
java.util.ArrayList<java.util.ArrayList<Integer>> result = new java.util.ArrayList<>(n);
for (int i = 0; i < n; i++) {
result.add(new java.util.ArrayList<>(groupSize[i]));
}
for (int i = 0; i < n; i++) {
result.get(leaderBuf[i]).add(i);
}
result.removeIf(java.util.ArrayList::isEmpty);
return result;
}
}
static class ModIntFactory {
private final ModArithmetic ma;
private final int mod;
private final boolean usesMontgomery;
private final ModArithmetic.ModArithmeticMontgomery maMontgomery;
private ArrayList<Integer> factorial;
private ArrayList<Integer> factorial_inversion;
public ModIntFactory(int mod) {
this.ma = ModArithmetic.of(mod);
this.usesMontgomery = ma instanceof ModArithmetic.ModArithmeticMontgomery;
this.maMontgomery = usesMontgomery ? (ModArithmetic.ModArithmeticMontgomery) ma : null;
this.mod = mod;
this.factorial = new ArrayList<>();
this.factorial_inversion = new ArrayList<>();
}
public ModInt create(long value) {
if ((value %= mod) < 0)
value += mod;
if (usesMontgomery) {
return new ModInt(maMontgomery.generate(value));
} else {
return new ModInt((int) value);
}
}
private void prepareFactorial(int max) {
factorial.ensureCapacity(max + 1);
if (factorial.size() == 0)
factorial.add(1);
for (int i = factorial.size(); i <= max; i++) {
factorial.add(ma.mul(factorial.get(i - 1), i));
}
}
public ModInt factorial(int i) {
prepareFactorial(i);
return create(factorial.get(i));
}
public ModInt permutation(int n, int r) {
if (n < 0 || r < 0 || n < r)
return create(0);
prepareFactorial(n);
if (factorial_inversion.size() > n) {
return create(ma.mul(factorial.get(n), factorial_inversion.get(n - r)));
}
return create(ma.div(factorial.get(n), factorial.get(n - r)));
}
public ModInt combination(int n, int r) {
if (n < 0 || r < 0 || n < r)
return create(0);
prepareFactorial(n);
if (factorial_inversion.size() > n) {
return create(
ma.mul(factorial.get(n), ma.mul(factorial_inversion.get(n - r), factorial_inversion.get(r))));
}
return create(ma.div(factorial.get(n), ma.mul(factorial.get(r), factorial.get(n - r))));
}
public void prepareFactorialInv(int max) {
prepareFactorial(max);
factorial_inversion.ensureCapacity(max + 1);
for (int i = factorial_inversion.size(); i <= max; i++) {
factorial_inversion.add(ma.inv(factorial.get(i)));
}
}
public int getMod() {
return mod;
}
public class ModInt {
private int value;
private ModInt(int value) {
this.value = value;
}
public int mod() {
return mod;
}
public int value() {
if (ma instanceof ModArithmetic.ModArithmeticMontgomery) {
return ((ModArithmetic.ModArithmeticMontgomery) ma).reduce(value);
}
return value;
}
public ModInt add(ModInt mi) {
return new ModInt(ma.add(value, mi.value));
}
public ModInt add(ModInt mi1, ModInt mi2) {
return new ModInt(ma.add(value, mi1.value)).addAsg(mi2);
}
public ModInt add(ModInt mi1, ModInt mi2, ModInt mi3) {
return new ModInt(ma.add(value, mi1.value)).addAsg(mi2).addAsg(mi3);
}
public ModInt add(ModInt mi1, ModInt mi2, ModInt mi3, ModInt mi4) {
return new ModInt(ma.add(value, mi1.value)).addAsg(mi2).addAsg(mi3).addAsg(mi4);
}
public ModInt add(ModInt mi1, ModInt... mis) {
ModInt mi = add(mi1);
for (ModInt m : mis)
mi.addAsg(m);
return mi;
}
public ModInt add(long mi) {
return new ModInt(ma.add(value, ma.remainder(mi)));
}
public ModInt sub(ModInt mi) {
return new ModInt(ma.sub(value, mi.value));
}
public ModInt sub(long mi) {
return new ModInt(ma.sub(value, ma.remainder(mi)));
}
public ModInt mul(ModInt mi) {
return new ModInt(ma.mul(value, mi.value));
}
public ModInt mul(ModInt mi1, ModInt mi2) {
return new ModInt(ma.mul(value, mi1.value)).mulAsg(mi2);
}
public ModInt mul(ModInt mi1, ModInt mi2, ModInt mi3) {
return new ModInt(ma.mul(value, mi1.value)).mulAsg(mi2).mulAsg(mi3);
}
public ModInt mul(ModInt mi1, ModInt mi2, ModInt mi3, ModInt mi4) {
return new ModInt(ma.mul(value, mi1.value)).mulAsg(mi2).mulAsg(mi3).mulAsg(mi4);
}
public ModInt mul(ModInt mi1, ModInt... mis) {
ModInt mi = mul(mi1);
for (ModInt m : mis)
mi.mulAsg(m);
return mi;
}
public ModInt mul(long mi) {
return new ModInt(ma.mul(value, ma.remainder(mi)));
}
public ModInt div(ModInt mi) {
return new ModInt(ma.div(value, mi.value));
}
public ModInt div(long mi) {
return new ModInt(ma.div(value, ma.remainder(mi)));
}
public ModInt inv() {
return new ModInt(ma.inv(value));
}
public ModInt pow(long b) {
return new ModInt(ma.pow(value, b));
}
public ModInt addAsg(ModInt mi) {
this.value = ma.add(value, mi.value);
return this;
}
public ModInt addAsg(ModInt mi1, ModInt mi2) {
return addAsg(mi1).addAsg(mi2);
}
public ModInt addAsg(ModInt mi1, ModInt mi2, ModInt mi3) {
return addAsg(mi1).addAsg(mi2).addAsg(mi3);
}
public ModInt addAsg(ModInt mi1, ModInt mi2, ModInt mi3, ModInt mi4) {
return addAsg(mi1).addAsg(mi2).addAsg(mi3).addAsg(mi4);
}
public ModInt addAsg(ModInt... mis) {
for (ModInt m : mis)
addAsg(m);
return this;
}
public ModInt addAsg(long mi) {
this.value = ma.add(value, ma.remainder(mi));
return this;
}
public ModInt subAsg(ModInt mi) {
this.value = ma.sub(value, mi.value);
return this;
}
public ModInt subAsg(long mi) {
this.value = ma.sub(value, ma.remainder(mi));
return this;
}
public ModInt mulAsg(ModInt mi) {
this.value = ma.mul(value, mi.value);
return this;
}
public ModInt mulAsg(ModInt mi1, ModInt mi2) {
return mulAsg(mi1).mulAsg(mi2);
}
public ModInt mulAsg(ModInt mi1, ModInt mi2, ModInt mi3) {
return mulAsg(mi1).mulAsg(mi2).mulAsg(mi3);
}
public ModInt mulAsg(ModInt mi1, ModInt mi2, ModInt mi3, ModInt mi4) {
return mulAsg(mi1).mulAsg(mi2).mulAsg(mi3).mulAsg(mi4);
}
public ModInt mulAsg(ModInt... mis) {
for (ModInt m : mis)
mulAsg(m);
return this;
}
public ModInt mulAsg(long mi) {
this.value = ma.mul(value, ma.remainder(mi));
return this;
}
public ModInt divAsg(ModInt mi) {
this.value = ma.div(value, mi.value);
return this;
}
public ModInt divAsg(long mi) {
this.value = ma.div(value, ma.remainder(mi));
return this;
}
@Override
public String toString() {
return String.valueOf(value());
}
@Override
public boolean equals(Object o) {
if (o instanceof ModInt) {
ModInt mi = (ModInt) o;
return mod() == mi.mod() && value() == mi.value();
}
return false;
}
@Override
public int hashCode() {
return (1 * 37 + mod()) * 37 + value();
}
}
private static abstract class ModArithmetic {
abstract int mod();
abstract int remainder(long value);
abstract int add(int a, int b);
abstract int sub(int a, int b);
abstract int mul(int a, int b);
int div(int a, int b) {
return mul(a, inv(b));
}
int inv(int a) {
int b = mod();
if (b == 1)
return 0;
long u = 1, v = 0;
while (b >= 1) {
int t = a / b;
a -= t * b;
int tmp1 = a;
a = b;
b = tmp1;
u -= t * v;
long tmp2 = u;
u = v;
v = tmp2;
}
if (a != 1) {
throw new ArithmeticException("divide by zero");
}
return remainder(u);
}
int pow(int a, long b) {
if (b < 0)
throw new ArithmeticException("negative power");
int r = 1;
int x = a;
while (b > 0) {
if ((b & 1) == 1)
r = mul(r, x);
x = mul(x, x);
b >>= 1;
}
return r;
}
static ModArithmetic of(int mod) {
if (mod <= 0) {
throw new IllegalArgumentException();
} else if (mod == 1) {
return new ModArithmetic1();
} else if (mod == 2) {
return new ModArithmetic2();
} else if (mod == 998244353) {
return new ModArithmetic998244353();
} else if (mod == 1000000007) {
return new ModArithmetic1000000007();
} else if ((mod & 1) == 1) {
return new ModArithmeticMontgomery(mod);
} else {
return new ModArithmeticBarrett(mod);
}
}
private static final class ModArithmetic1 extends ModArithmetic {
int mod() {
return 1;
}
int remainder(long value) {
return 0;
}
int add(int a, int b) {
return 0;
}
int sub(int a, int b) {
return 0;
}
int mul(int a, int b) {
return 0;
}
int pow(int a, long b) {
return 0;
}
}
private static final class ModArithmetic2 extends ModArithmetic {
int mod() {
return 2;
}
int remainder(long value) {
return (int) (value & 1);
}
int add(int a, int b) {
return a ^ b;
}
int sub(int a, int b) {
return a ^ b;
}
int mul(int a, int b) {
return a & b;
}
}
private static final class ModArithmetic998244353 extends ModArithmetic {
private final int mod = 998244353;
int mod() {
return mod;
}
int remainder(long value) {
return (int) ((value %= mod) < 0 ? value + mod : value);
}
int add(int a, int b) {
int res = a + b;
return res >= mod ? res - mod : res;
}
int sub(int a, int b) {
int res = a - b;
return res < 0 ? res + mod : res;
}
int mul(int a, int b) {
return (int) (((long) a * b) % mod);
}
}
private static final class ModArithmetic1000000007 extends ModArithmetic {
private final int mod = 1000000007;
int mod() {
return mod;
}
int remainder(long value) {
return (int) ((value %= mod) < 0 ? value + mod : value);
}
int add(int a, int b) {
int res = a + b;
return res >= mod ? res - mod : res;
}
int sub(int a, int b) {
int res = a - b;
return res < 0 ? res + mod : res;
}
int mul(int a, int b) {
return (int) (((long) a * b) % mod);
}
}
private static final class ModArithmeticMontgomery extends ModArithmeticDynamic {
private final long negInv;
private final long r2;
private ModArithmeticMontgomery(int mod) {
super(mod);
long inv = 0;
long s = 1, t = 0;
for (int i = 0; i < 32; i++) {
if ((t & 1) == 0) {
t += mod;
inv += s;
}
t >>= 1;
s <<= 1;
}
long r = (1l << 32) % mod;
this.negInv = inv;
this.r2 = (r * r) % mod;
}
private int generate(long x) {
return reduce(x * r2);
}
private int reduce(long x) {
x = (x + ((x * negInv) & 0xffff_ffffl) * mod) >>> 32;
return (int) (x < mod ? x : x - mod);
}
@Override
int remainder(long value) {
return generate((value %= mod) < 0 ? value + mod : value);
}
@Override
int mul(int a, int b) {
return reduce((long) a * b);
}
@Override
int inv(int a) {
return super.inv(reduce(a));
}
@Override
int pow(int a, long b) {
return generate(super.pow(a, b));
}
}
private static final class ModArithmeticBarrett extends ModArithmeticDynamic {
private static final long mask = 0xffff_ffffl;
private final long mh;
private final long ml;
private ModArithmeticBarrett(int mod) {
super(mod);
/**
* m = floor(2^64/mod) 2^64 = p*mod + q, 2^32 = a*mod + b => (a*mod + b)^2 =
* p*mod + q => p = mod*a^2 + 2ab + floor(b^2/mod)
*/
long a = (1l << 32) / mod;
long b = (1l << 32) % mod;
long m = a * a * mod + 2 * a * b + (b * b) / mod;
mh = m >>> 32;
ml = m & mask;
}
private int reduce(long x) {
long z = (x & mask) * ml;
z = (x & mask) * mh + (x >>> 32) * ml + (z >>> 32);
z = (x >>> 32) * mh + (z >>> 32);
x -= z * mod;
return (int) (x < mod ? x : x - mod);
}
@Override
int remainder(long value) {
return (int) ((value %= mod) < 0 ? value + mod : value);
}
@Override
int mul(int a, int b) {
return reduce((long) a * b);
}
}
private static class ModArithmeticDynamic extends ModArithmetic {
final int mod;
ModArithmeticDynamic(int mod) {
this.mod = mod;
}
int mod() {
return mod;
}
int remainder(long value) {
return (int) ((value %= mod) < 0 ? value + mod : value);
}
int add(int a, int b) {
int sum = a + b;
return sum >= mod ? sum - mod : sum;
}
int sub(int a, int b) {
int sum = a - b;
return sum < 0 ? sum + mod : sum;
}
int mul(int a, int b) {
return (int) (((long) a * b) % mod);
}
}
}
}
static class Convolution {
/**
* Find a primitive root.
*
* @param m A prime number.
* @return Primitive root.
*/
private static int primitiveRoot(int m) {
if (m == 2)
return 1;
if (m == 167772161)
return 3;
if (m == 469762049)
return 3;
if (m == 754974721)
return 11;
if (m == 998244353)
return 3;
int[] divs = new int[20];
divs[0] = 2;
int cnt = 1;
int x = (m - 1) / 2;
while (x % 2 == 0)
x /= 2;
for (int i = 3; (long) (i) * i <= x; i += 2) {
if (x % i == 0) {
divs[cnt++] = i;
while (x % i == 0) {
x /= i;
}
}
}
if (x > 1) {
divs[cnt++] = x;
}
for (int g = 2;; g++) {
boolean ok = true;
for (int i = 0; i < cnt; i++) {
if (pow(g, (m - 1) / divs[i], m) == 1) {
ok = false;
break;
}
}
if (ok)
return g;
}
}
/**
* Power.
*
* @param x Parameter x.
* @param n Parameter n.
* @param m Mod.
* @return n-th power of x mod m.
*/
private static long pow(long x, long n, int m) {
if (m == 1)
return 0;
long r = 1;
long y = x % m;
while (n > 0) {
if ((n & 1) != 0)
r = (r * y) % m;
y = (y * y) % m;
n >>= 1;
}
return r;
}
/**
* Ceil of power 2.
*
* @param n Value.
* @return Ceil of power 2.
*/
private static int ceilPow2(int n) {
int x = 0;
while ((1L << x) < n)
x++;
return x;
}
/**
* Garner's algorithm.
*
* @param c Mod convolution results.
* @param mods Mods.
* @return Result.
*/
private static long garner(long[] c, int[] mods) {
int n = c.length + 1;
long[] cnst = new long[n];
long[] coef = new long[n];
java.util.Arrays.fill(coef, 1);
for (int i = 0; i < n - 1; i++) {
int m1 = mods[i];
long v = (c[i] - cnst[i] + m1) % m1;
v = v * pow(coef[i], m1 - 2, m1) % m1;
for (int j = i + 1; j < n; j++) {
long m2 = mods[j];
cnst[j] = (cnst[j] + coef[j] * v) % m2;
coef[j] = (coef[j] * m1) % m2;
}
}
return cnst[n - 1];
}
/**
* Pre-calculation for NTT.
*
* @param mod NTT Prime.
* @param g Primitive root of mod.
* @return Pre-calculation table.
*/
private static long[] sumE(int mod, int g) {
long[] sum_e = new long[30];
long[] es = new long[30];
long[] ies = new long[30];
int cnt2 = Integer.numberOfTrailingZeros(mod - 1);
long e = pow(g, (mod - 1) >> cnt2, mod);
long ie = pow(e, mod - 2, mod);
for (int i = cnt2; i >= 2; i--) {
es[i - 2] = e;
ies[i - 2] = ie;
e = e * e % mod;
ie = ie * ie % mod;
}
long now = 1;
for (int i = 0; i < cnt2 - 2; i++) {
sum_e[i] = es[i] * now % mod;
now = now * ies[i] % mod;
}
return sum_e;
}
/**
* Pre-calculation for inverse NTT.
*
* @param mod Mod.
* @param g Primitive root of mod.
* @return Pre-calculation table.
*/
private static long[] sumIE(int mod, int g) {
long[] sum_ie = new long[30];
long[] es = new long[30];
long[] ies = new long[30];
int cnt2 = Integer.numberOfTrailingZeros(mod - 1);
long e = pow(g, (mod - 1) >> cnt2, mod);
long ie = pow(e, mod - 2, mod);
for (int i = cnt2; i >= 2; i--) {
es[i - 2] = e;
ies[i - 2] = ie;
e = e * e % mod;
ie = ie * ie % mod;
}
long now = 1;
for (int i = 0; i < cnt2 - 2; i++) {
sum_ie[i] = ies[i] * now % mod;
now = now * es[i] % mod;
}
return sum_ie;
}
/**
* Inverse NTT.
*
* @param a Target array.
* @param sumIE Pre-calculation table.
* @param mod NTT Prime.
*/
private static void butterflyInv(long[] a, long[] sumIE, int mod) {
int n = a.length;
int h = ceilPow2(n);
for (int ph = h; ph >= 1; ph--) {
int w = 1 << (ph - 1), p = 1 << (h - ph);
long inow = 1;
for (int s = 0; s < w; s++) {
int offset = s << (h - ph + 1);
for (int i = 0; i < p; i++) {
long l = a[i + offset];
long r = a[i + offset + p];
a[i + offset] = (l + r) % mod;
a[i + offset + p] = (mod + l - r) * inow % mod;
}
int x = Integer.numberOfTrailingZeros(~s);
inow = inow * sumIE[x] % mod;
}
}
}
/**
* Inverse NTT.
*
* @param a Target array.
* @param sumE Pre-calculation table.
* @param mod NTT Prime.
*/
private static void butterfly(long[] a, long[] sumE, int mod) {
int n = a.length;
int h = ceilPow2(n);
for (int ph = 1; ph <= h; ph++) {
int w = 1 << (ph - 1), p = 1 << (h - ph);
long now = 1;
for (int s = 0; s < w; s++) {
int offset = s << (h - ph + 1);
for (int i = 0; i < p; i++) {
long l = a[i + offset];
long r = a[i + offset + p] * now % mod;
a[i + offset] = (l + r) % mod;
a[i + offset + p] = (l - r + mod) % mod;
}
int x = Integer.numberOfTrailingZeros(~s);
now = now * sumE[x] % mod;
}
}
}
/**
* Convolution.
*
* @param a Target array 1.
* @param b Target array 2.
* @param mod NTT Prime.
* @return Answer.
*/
public static long[] convolution(long[] a, long[] b, int mod) {
int n = a.length;
int m = b.length;
if (n == 0 || m == 0)
return new long[0];
int z = 1 << ceilPow2(n + m - 1);
{
long[] na = new long[z];
long[] nb = new long[z];
System.arraycopy(a, 0, na, 0, n);
System.arraycopy(b, 0, nb, 0, m);
a = na;
b = nb;
}
int g = primitiveRoot(mod);
long[] sume = sumE(mod, g);
long[] sumie = sumIE(mod, g);
butterfly(a, sume, mod);
butterfly(b, sume, mod);
for (int i = 0; i < z; i++) {
a[i] = a[i] * b[i] % mod;
}
butterflyInv(a, sumie, mod);
a = java.util.Arrays.copyOf(a, n + m - 1);
long iz = pow(z, mod - 2, mod);
for (int i = 0; i < n + m - 1; i++)
a[i] = a[i] * iz % mod;
return a;
}
/**
* Convolution.
*
* @param a Target array 1.
* @param b Target array 2.
* @param mod Any mod.
* @return Answer.
*/
public static long[] convolutionLL(long[] a, long[] b, int mod) {
int n = a.length;
int m = b.length;
if (n == 0 || m == 0)
return new long[0];
int mod1 = 754974721;
int mod2 = 167772161;
int mod3 = 469762049;
long[] c1 = convolution(a, b, mod1);
long[] c2 = convolution(a, b, mod2);
long[] c3 = convolution(a, b, mod3);
int retSize = c1.length;
long[] ret = new long[retSize];
int[] mods = { mod1, mod2, mod3, mod };
for (int i = 0; i < retSize; ++i) {
ret[i] = garner(new long[] { c1[i], c2[i], c3[i] }, mods);
}
return ret;
}
/**
* Convolution by ModInt.
*
* @param a Target array 1.
* @param b Target array 2.
* @return Answer.
*/
public static java.util.List<ModIntFactory.ModInt> convolution(java.util.List<ModIntFactory.ModInt> a,
java.util.List<ModIntFactory.ModInt> b) {
int mod = a.get(0).mod();
long[] va = a.stream().mapToLong(ModIntFactory.ModInt::value).toArray();
long[] vb = b.stream().mapToLong(ModIntFactory.ModInt::value).toArray();
long[] c = convolutionLL(va, vb, mod);
ModIntFactory factory = new ModIntFactory(mod);
return java.util.Arrays.stream(c).mapToObj(factory::create).collect(java.util.stream.Collectors.toList());
}
/**
* Naive convolution. (Complexity is O(N^2)!!)
*
* @param a Target array 1.
* @param b Target array 2.
* @param mod Mod.
* @return Answer.
*/
public static long[] convolutionNaive(long[] a, long[] b, int mod) {
int n = a.length;
int m = b.length;
int k = n + m - 1;
long[] ret = new long[k];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
ret[i + j] += a[i] * b[j] % mod;
ret[i + j] %= mod;
}
}
return ret;
}
}
static class ContestScanner {
private final java.io.InputStream in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private static final long LONG_MAX_TENTHS = 922337203685477580L;
private static final int LONG_MAX_LAST_DIGIT = 7;
private static final int LONG_MIN_LAST_DIGIT = 8;
public ContestScanner(java.io.InputStream in) {
this.in = in;
}
public ContestScanner() {
this(System.in);
}
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
} else {
ptr = 0;
try {
buflen = in.read(buffer);
} catch (java.io.IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte())
return buffer[ptr++];
else
return -1;
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
public boolean hasNext() {
while (hasNextByte() && !isPrintableChar(buffer[ptr]))
ptr++;
return hasNextByte();
}
public String next() {
if (!hasNext())
throw new java.util.NoSuchElementException();
StringBuilder sb = new StringBuilder();
int b = readByte();
while (isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext())
throw new java.util.NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
int digit = b - '0';
if (n >= LONG_MAX_TENTHS) {
if (n == LONG_MAX_TENTHS) {
if (minus) {
if (digit <= LONG_MIN_LAST_DIGIT) {
n = -n * 10 - digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("%d%s... is not number", n, Character.toString(b)));
}
}
} else {
if (digit <= LONG_MAX_LAST_DIGIT) {
n = n * 10 + digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("%d%s... is not number", n, Character.toString(b)));
}
}
}
}
throw new ArithmeticException(
String.format("%s%d%d... overflows long.", minus ? "-" : "", n, digit));
}
n = n * 10 + digit;
} else if (b == -1 || !isPrintableChar(b)) {
return minus ? -n : n;
} else {
throw new NumberFormatException();
}
b = readByte();
}
}
public int nextInt() {
long nl = nextLong();
if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
throw new NumberFormatException();
return (int) nl;
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long[] nextLongArray(int length) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = this.nextLong();
return array;
}
public long[] nextLongArray(int length, java.util.function.LongUnaryOperator map) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsLong(this.nextLong());
return array;
}
public int[] nextIntArray(int length) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = this.nextInt();
return array;
}
public int[][] nextIntArrayMulti(int length, int width) {
int[][] arrays = new int[width][length];
for (int i = 0; i < length; i++) {
for (int j = 0; j < width; j++)
arrays[j][i] = this.nextInt();
}
return arrays;
}
public int[] nextIntArray(int length, java.util.function.IntUnaryOperator map) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsInt(this.nextInt());
return array;
}
public double[] nextDoubleArray(int length) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = this.nextDouble();
return array;
}
public double[] nextDoubleArray(int length, java.util.function.DoubleUnaryOperator map) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsDouble(this.nextDouble());
return array;
}
public long[][] nextLongMatrix(int height, int width) {
long[][] mat = new long[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextLong();
}
return mat;
}
public int[][] nextIntMatrix(int height, int width) {
int[][] mat = new int[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextInt();
}
return mat;
}
public double[][] nextDoubleMatrix(int height, int width) {
double[][] mat = new double[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextDouble();
}
return mat;
}
public char[][] nextCharMatrix(int height, int width) {
char[][] mat = new char[height][width];
for (int h = 0; h < height; h++) {
String s = this.next();
for (int w = 0; w < width; w++) {
mat[h][w] = s.charAt(w);
}
}
return mat;
}
}
static class ContestPrinter extends java.io.PrintWriter {
public ContestPrinter(java.io.PrintStream stream) {
super(stream);
}
public ContestPrinter() {
super(System.out);
}
private static String dtos(double x, int n) {
StringBuilder sb = new StringBuilder();
if (x < 0) {
sb.append('-');
x = -x;
}
x += Math.pow(10, -n) / 2;
sb.append((long) x);
sb.append(".");
x -= (long) x;
for (int i = 0; i < n; i++) {
x *= 10;
sb.append((int) x);
x -= (int) x;
}
return sb.toString();
}
@Override
public void print(float f) {
super.print(dtos(f, 20));
}
@Override
public void println(float f) {
super.println(dtos(f, 20));
}
@Override
public void print(double d) {
super.print(dtos(d, 20));
}
@Override
public void println(double d) {
super.println(dtos(d, 20));
}
public void printArray(int[] array, String separator) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(array[i]);
super.print(separator);
}
super.println(array[n - 1]);
}
public void printArray(int[] array) {
this.printArray(array, " ");
}
public void printArray(int[] array, String separator, java.util.function.IntUnaryOperator map) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(map.applyAsInt(array[i]));
super.print(separator);
}
super.println(map.applyAsInt(array[n - 1]));
}
public void printArray(int[] array, java.util.function.IntUnaryOperator map) {
this.printArray(array, " ", map);
}
public void printArray(long[] array, String separator) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(array[i]);
super.print(separator);
}
super.println(array[n - 1]);
}
public void printArray(long[] array) {
this.printArray(array, " ");
}
public void printArray(long[] array, String separator, java.util.function.LongUnaryOperator map) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(map.applyAsLong(array[i]));
super.print(separator);
}
super.println(map.applyAsLong(array[n - 1]));
}
public void printArray(long[] array, java.util.function.LongUnaryOperator map) {
this.printArray(array, " ", map);
}
}
public static void safeSort(int[] array) {
Integer[] temp = new Integer[array.length];
for (int n = 0; n < array.length; n++) {
temp[n] = array[n];
}
Arrays.sort(temp);
for (int n = 0; n < array.length; n++) {
array[n] = temp[n];
}
}
public static void safeSort(long[] array) {
Long[] temp = new Long[array.length];
for (int n = 0; n < array.length; n++) {
temp[n] = array[n];
}
Arrays.sort(temp);
for (int n = 0; n < array.length; n++) {
array[n] = temp[n];
}
}
public static void safeSort(double[] array) {
Double[] temp = new Double[array.length];
for (int n = 0; n < array.length; n++) {
temp[n] = array[n];
}
Arrays.sort(temp);
for (int n = 0; n < array.length; n++) {
array[n] = temp[n];
}
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 4e92f8586772238550dc748bd4ba4de2 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes |
public class Main {
public static void main(String[] args) {
new Main();
}
public Main() {
FastScanner fs = new FastScanner();
java.io.PrintWriter out = new java.io.PrintWriter(System.out);
solve(fs, out);
out.flush();
}
public void solve(FastScanner fs, java.io.PrintWriter out) {
int t = fs.nextInt();
boolean[] isPrime = new boolean[3000000];
java.util.Arrays.fill(isPrime, true);
isPrime[1] = false;
for (int i = 2;i < isPrime.length;++ i) {
if (!isPrime[i]) continue;
for (int j = 2 * i;j < isPrime.length;j += i) isPrime[j] = false;
}
while(t --> 0) {
int n = fs.nextInt(), e = fs.nextInt();
int[] a = new int[n];
for (int i = 0;i < n;++ i) a[i] = fs.nextInt();
long ans = 0;
for (int i = 0;i < e;++ i) {
int first = 0, second = 0, k = 0;
for (;i + e * k < n;++ k) {
++ second;
if (isPrime[a[i + e * k]]) {
ans += Math.max(0, (long)first * second - 1);
first = second;
second = 0;
} else if (a[i + e * k] == 1) {
} else {
ans += Math.max(0, (long)first * second - 1);
first = second = 0;
}
}
++ second;
ans += Math.max(0, (long)first * second - 1);
}
out.println(ans);
}
}
boolean isABC(char[] s, int i) {
if (i < 0 || i >= s.length - 2) return false;
if (s[i] == 'a' && s[i + 1] == 'b' && s[i + 2] == 'c') return true;
return false;
}
final int MOD = 998_244_353;
int plus(int n, int m) {
int sum = n + m;
if (sum >= MOD) sum -= MOD;
return sum;
}
int minus(int n, int m) {
int sum = n - m;
if (sum < 0) sum += MOD;
return sum;
}
int times(int n, int m) {
return (int)((long)n * m % MOD);
}
int divide(int n, int m) {
return times(n, IntMath.pow(m, MOD - 2, MOD));
}
int[] fact, invf;
void calc(int len) {
len += 2;
fact = new int[len];
invf = new int[len];
fact[0] = fact[1] = invf[0] = invf[1] = 1;
for (int i = 2;i < fact.length;++ i) fact[i] = times(fact[i - 1], i);
invf[len - 1] = divide(1, fact[len - 1]);
for (int i = len - 1;i > 1;-- i) invf[i - 1] = times(i, invf[i]);
}
int comb(int n, int m) {
if (n < m) return 0;
return times(fact[n], times(invf[n - m], invf[m]));
}
}
class FastScanner {
private final java.io.InputStream in = System.in;
private final byte[] buffer = new byte[8192];
private int ptr = 0;
private int buflen = 0;
private boolean hasNextByte() {
if (ptr < buflen) return true;
ptr = 0;
try {
buflen = in.read(buffer);
} catch (java.io.IOException e) {
e.printStackTrace();
}
return buflen > 0;
}
private byte readByte() {
return hasNextByte() ? buffer[ptr++ ] : -1;
}
private static boolean isPrintableChar(byte c) {
return 32 < c || c < 0;
}
private static boolean isNumber(int c) {
return '0' <= c && c <= '9';
}
public boolean hasNext() {
while (hasNextByte() && !isPrintableChar(buffer[ptr]))
ptr++ ;
return hasNextByte();
}
public String next() {
if (!hasNext()) throw new java.util.NoSuchElementException();
StringBuilder sb = new StringBuilder();
byte b;
while (isPrintableChar(b = readByte()))
sb.appendCodePoint(b);
return sb.toString();
}
public final char nextChar() {
if (!hasNext()) throw new java.util.NoSuchElementException();
return (char)readByte();
}
public final long nextLong() {
if (!hasNext()) throw new java.util.NoSuchElementException();
long n = 0;
try {
byte b = readByte();
if (b == '-') {
while (isNumber(b = readByte()))
n = n * 10 + '0' - b;
return n;
} else if (!isNumber(b)) throw new NumberFormatException();
do
n = n * 10 + b - '0';
while (isNumber(b = readByte()));
} catch (java.util.NoSuchElementException e) {}
return n;
}
public final int nextInt() {
if (!hasNext()) throw new java.util.NoSuchElementException();
int n = 0;
try {
byte b = readByte();
if (b == '-') {
while (isNumber(b = readByte()))
n = n * 10 + '0' - b;
return n;
} else if (!isNumber(b)) throw new NumberFormatException();
do
n = n * 10 + b - '0';
while (isNumber(b = readByte()));
} catch (java.util.NoSuchElementException e) {}
return n;
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
class Arrays {
public static void sort(final int[] array) {
sort(array, 0, array.length);
}
public static void sort(final int[] array, int fromIndex, int toIndex) {
if (toIndex - fromIndex <= 512) {
java.util.Arrays.sort(array, fromIndex, toIndex);
return;
}
sort(array, fromIndex, toIndex, 0, new int[array.length]);
}
private static final void sort(int[] a, final int from, final int to, final int l, final int[] bucket) {
if (to - from <= 512) {
java.util.Arrays.sort(a, from, to);
return;
}
final int BUCKET_SIZE = 256;
final int INT_RECURSION = 4;
final int MASK = 0xff;
final int shift = l << 3;
final int[] cnt = new int[BUCKET_SIZE + 1];
final int[] put = new int[BUCKET_SIZE];
for (int i = from; i < to; i++) ++ cnt[(a[i] >>> shift & MASK) + 1];
for (int i = 0; i < BUCKET_SIZE; i++) cnt[i + 1] += cnt[i];
for (int i = from; i < to; i++) {
int bi = a[i] >>> shift & MASK;
bucket[cnt[bi] + put[bi]++] = a[i];
}
for (int i = BUCKET_SIZE - 1, idx = from; i >= 0; i--) {
int begin = cnt[i];
int len = cnt[i + 1] - begin;
System.arraycopy(bucket, begin, a, idx, len);
idx += len;
}
final int nxtL = l + 1;
if (nxtL < INT_RECURSION) {
sort(a, from, to, nxtL, bucket);
if (l == 0) {
int lft, rgt;
lft = from - 1; rgt = to;
while (rgt - lft > 1) {
int mid = lft + rgt >> 1;
if (a[mid] < 0) lft = mid;
else rgt = mid;
}
reverse(a, from, rgt);
reverse(a, rgt, to);
}
}
}
public static void sort(final long[] array) {
sort(array, 0, array.length);
}
public static void sort(final long[] array, int fromIndex, int toIndex) {
if (toIndex - fromIndex <= 512) {
java.util.Arrays.sort(array, fromIndex, toIndex);
return;
}
sort(array, fromIndex, toIndex, 0, new long[array.length]);
}
private static final void sort(long[] a, final int from, final int to, final int l, final long[] bucket) {
final int BUCKET_SIZE = 256;
final int LONG_RECURSION = 8;
final int MASK = 0xff;
final int shift = l << 3;
final int[] cnt = new int[BUCKET_SIZE + 1];
final int[] put = new int[BUCKET_SIZE];
for (int i = from; i < to; i++) ++ cnt[(int) ((a[i] >>> shift & MASK) + 1)];
for (int i = 0; i < BUCKET_SIZE; i++) cnt[i + 1] += cnt[i];
for (int i = from; i < to; i++) {
int bi = (int) (a[i] >>> shift & MASK);
bucket[cnt[bi] + put[bi]++] = a[i];
}
for (int i = BUCKET_SIZE - 1, idx = from; i >= 0; i--) {
int begin = cnt[i];
int len = cnt[i + 1] - begin;
System.arraycopy(bucket, begin, a, idx, len);
idx += len;
}
final int nxtL = l + 1;
if (nxtL < LONG_RECURSION) {
sort(a, from, to, nxtL, bucket);
if (l == 0) {
int lft, rgt;
lft = from - 1; rgt = to;
while (rgt - lft > 1) {
int mid = lft + rgt >> 1;
if (a[mid] < 0) lft = mid;
else rgt = mid;
}
reverse(a, from, rgt);
reverse(a, rgt, to);
}
}
}
public static void reverse(int[] array) {
reverse(array, 0, array.length);
}
public static void reverse(int[] array, int fromIndex, int toIndex) {
for (-- toIndex;fromIndex < toIndex;++ fromIndex, -- toIndex) {
int swap = array[fromIndex];
array[fromIndex] = array[toIndex];
array[toIndex] = swap;
}
}
public static void reverse(long[] array) {
reverse(array, 0, array.length);
}
public static void reverse(long[] array, int fromIndex, int toIndex) {
for (-- toIndex;fromIndex < toIndex;++ fromIndex, -- toIndex) {
long swap = array[fromIndex];
array[fromIndex] = array[toIndex];
array[toIndex] = swap;
}
}
public static void shuffle(int[] array) {
java.util.Random rnd = new java.util.Random();
for (int i = 0;i < array.length;++ i) {
int j = rnd.nextInt(array.length - i) + i;
int swap = array[i];
array[i] = array[j];
array[j] = swap;
}
}
public static void shuffle(long[] array) {
java.util.Random rnd = new java.util.Random();
for (int i = 0;i < array.length;++ i) {
int j = rnd.nextInt(array.length - i) + i;
long swap = array[i];
array[i] = array[j];
array[j] = swap;
}
}
}
class IntMath {
public static int gcd(int a, int b) {
while (a != 0)
if ((b %= a) != 0) a %= b;
else return a;
return b;
}
public static int gcd(int... array) {
int ret = array[0];
for (int i = 1; i < array.length; ++i)
ret = gcd(ret, array[i]);
return ret;
}
public static long gcd(long a, long b) {
while (a != 0)
if ((b %= a) != 0) a %= b;
else return a;
return b;
}
public static long gcd(long... array) {
long ret = array[0];
for (int i = 1; i < array.length; ++i)
ret = gcd(ret, array[i]);
return ret;
}
public static long lcm(long a, long b) {
return a / gcd(a, b) * b;
}
public static int pow(int a, int b) {
int ans = 1;
for (int mul = a; b > 0; b >>= 1, mul *= mul)
if ((b & 1) != 0) ans *= mul;
return ans;
}
public static long pow(long a, long b) {
long ans = 1;
for (long mul = a; b > 0; b >>= 1, mul *= mul)
if ((b & 1) != 0) ans *= mul;
return ans;
}
public static int pow(int a, long b, int mod) {
if (b < 0) b = b % (mod - 1) + mod - 1;
long ans = 1;
for (long mul = a; b > 0; b >>= 1, mul = mul * mul % mod)
if ((b & 1) != 0) ans = ans * mul % mod;
return (int)ans;
}
public static int pow(long a, long b, int mod) {
return pow((int)(a % mod), b, mod);
}
public static int floorsqrt(long n) {
return (int)Math.sqrt(n + 0.1);
}
public static int ceilsqrt(long n) {
return n <= 1 ? (int)n : (int)Math.sqrt(n - 0.1) + 1;
}
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 3887d79aff0682de1676bd420e134210 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
public static void main(String[] args){
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
solve(in, out);
out.close();
}
static String reverse(String s) {
return (new StringBuilder(s)).reverse().toString();
}
static void sieveOfEratosthenes(int n, int factors[], ArrayList<Integer> ar)
{
factors[1]=1;
int p;
for(p = 2; p*p <=n; p++)
{
if(factors[p] == 0)
{
ar.add(p);
factors[p]=p;
for(int i = p*p; i <= n; i += p)
if(factors[i]==0)
factors[i] = p;
}
}
for(;p<=n;p++){
if(factors[p] == 0)
{
factors[p] = p;
ar.add(p);
}
}
}
static void sort(int ar[]) {
int n = ar.length;
ArrayList<Integer> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static void sort1(long ar[]) {
int n = ar.length;
ArrayList<Long> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static void sort2(char ar[]) {
int n = ar.length;
ArrayList<Character> a = new ArrayList<>();
for (int i = 0; i < n; i++)
a.add(ar[i]);
Collections.sort(a);
for (int i = 0; i < n; i++)
ar[i] = a.get(i);
}
static long ncr(long n, long r, long mod) {
if (r == 0)
return 1;
long val = ncr(n - 1, r - 1, mod);
val = (n * val) % mod;
val = (val * modInverse(r, mod)) % mod;
return val;
}
static long max[];
static long lz[];
// static void build(long a[], int v, int tl, int tr) {
// if (tl == tr) {
// max[v] = a[tl];
// } else {
// int tm = (tl + tr) / 2;
// build(a, v*2, tl, tm);
// build(a, v*2+1, tm+1, tr);
// max[v] = max[v*2] + max[v*2+1];
// }
// }
// static void push(int v, int tl, int tr){
// if(tl==tr){
// lz[v] = 0;
// return;
// }
// max[2*v] += lz[v];
// // max[2*v] %= m;
// max[2*v+1] += lz[v];
// // max[2*v+1] %= m;
// lz[2*v] += lz[v];
// lz[2*v] %= m;
// lz[2*v+1] += lz[v];
// lz[2*v+1] %= m;
// lz[v] = 0;
// }
static void update(int v, int tl, int tr, int l, int r, long val) {
if(tl>r||tr<l)
return;
if(tl>=l&&tr<=r){
max[v] = Math.max(val, max[v]);
return;
}
int tm = ((tl + tr) >> 1);
update(v*2, tl, tm, l, r, val);
update(v*2+1, tm+1, tr, l, r, val);
max[v] = Math.max(max[2*v],max[2*v+1]);
}
static long get(int v, int tl, int tr, int l, int r) {
if(l>tr||r<tl)
return 0;
if (tl >= l && tr <= r) {
return max[v];
}
int tm = ((tl + tr) >> 1);
return Math.max(get(2*v, tl, tm, l, r), get(2*v+1, tm+1, tr, l, r));
}
// static int y = 0;
static int parent[];
static int rank[];
static long m = 0;
static int find_set(int v) {
if (v == parent[v])
return v;
return find_set(parent[v]);
}
static void make_set(int v) {
parent[v] = v;
rank[v] = 1;
}
static void union_sets(int a, int b) {
a = find_set(a);
b = find_set(b);
if (a != b) {
if (rank[a] < rank[b]){
int tmp = a;
a = b;
b = tmp;
}
parent[b] = a;
// if (rank[a] == rank[b])
// rank[a]++;
rank[a] += rank[b];
}
}
static int parent1[];
static int rank1[];
static int find_set1(int v) {
if (v == parent1[v])
return v;
return find_set1(parent1[v]);
}
static void make_set1(int v) {
parent1[v] = v;
rank1[v] = 1;
}
static void union_sets1(int a, int b) {
a = find_set1(a);
b = find_set1(b);
if (a != b) {
if (rank1[a] < rank1[b]){
int tmp = a;
a = b;
b = tmp;
}
parent1[b] = a;
// if (rank1[a] == rank1[b])
// rank1[a]++;
rank1[a] += rank1[b];
}
}
static int count = 0;
static int count1 = 0;
public static void solve(InputReader sc, PrintWriter pw){
int i, j = 0;
// int t = 1;
long mod = 1000000007;
int factors[] = new int[1000001];
ArrayList<Integer> ar = new ArrayList<>();
sieveOfEratosthenes(1000000, factors, ar);
HashSet<Integer> set = new HashSet<>();
for(int x:ar){
set.add(x);
}
int t = sc.nextInt();
u: while (t-- > 0) {
int n = sc.nextInt();
int e = sc.nextInt();
int a[] = new int[n];
for(i=0;i<n;i++){
a[i] = sc.nextInt();
}
long ans = 0;
for(i=0;i<e;i++){
int prime = 0, one = 0, prev = 0;
for(j=i;j<n;j+=e){
if(a[j]==1){
one++;
if(prime==1){
ans += 1 + prev;
}
}
else if(set.contains(a[j])){
prime = 1;
ans += one;
prev = one;
one = 0;
}
else{
prev = one = prime = 0;
}
// pw.println(a[i]+" "+prime+" "+one);
}
}
pw.println(ans);
}
}
static void KMPSearch(char pat[], char txt[], int pres[]){
int M = pat.length;
int N = txt.length;
int lps[] = new int[M];
int j = 0;
computeLPSArray(pat, M, lps);
int i = 0;
while (i < N) {
if (pat[j] == txt[i]) {
j++;
i++;
}
if (j == M) {
pres[i-1] = 1;
j = lps[j - 1];
}
else if (i < N && pat[j] != txt[i]) {
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
}
static void computeLPSArray(char pat[], int M, int lps[])
{
int len = 0;
int i = 1;
lps[0] = 0;
while (i < M) {
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
else
{
if (len != 0) {
len = lps[len - 1];
}
else
{
lps[i] = len;
i++;
}
}
}
}
static long[][] matrixMult(long a[][], long b[][], long mod){
int n = a.length;
int m = a[0].length;
int p = b[0].length;
long c[][] = new long[n][p];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
for(int k=0;k<p;k++){
c[i][k] += a[i][j]*b[j][k];
c[i][k] %= mod;
}
}
}
return c;
}
static long[][] exp(long mat[][], long b, long mod){
if(b==0){
int n = mat.length;
long res[][] = new long[n][n];
for(int i=0;i<n;i++){
res[i][i] = 1;
}
return res;
}
long half[][] = exp(mat, b/2, mod);
long res[][] = matrixMult(half, half, mod);
if(b%2==1){
res = matrixMult(res, mat, mod);
}
return res;
}
static void countPrimeFactors(int num, int a[], HashMap<Integer,Integer> pos){
for(int i=2;i*i<num;i++){
if(num%i==0){
int y = pos.get(i);
while(num%i==0){
a[y]++;
num/=i;
}
}
}
if(num>1){
int y = pos.get(num);
a[y]++;
}
}
static void assignAnc(ArrayList<Integer> ar[], int depth[], int sz[], int par[][] ,int curr, int parent, int d){
depth[curr] = d;
sz[curr] = 1;
par[curr][0] = parent;
for(int v:ar[curr]){
if(parent==v)
continue;
assignAnc(ar, depth, sz, par, v, curr, d+1);
sz[curr] += sz[v];
}
}
static int findLCA(int a, int b, int par[][], int depth[]){
if(depth[a]>depth[b]){
a = a^b;
b = a^b;
a = a^b;
}
int diff = depth[b] - depth[a];
for(int i=19;i>=0;i--){
if((diff&(1<<i))>0){
b = par[b][i];
}
}
if(a==b)
return a;
for(int i=19;i>=0;i--){
if(par[b][i]!=par[a][i]){
b = par[b][i];
a = par[a][i];
}
}
return par[a][0];
}
static void formArrayForBinaryLifting(int n, int par[][]){
for(int j=1;j<20;j++){
for(int i=0;i<n;i++){
if(par[i][j-1]==-1)
continue;
par[i][j] = par[par[i][j-1]][j-1];
}
}
}
static long lcm(int a, int b){
return a*b/gcd(a,b);
}
static class Pair1 {
long a;
long b;
Pair1(long a, long b) {
this.a = a;
this.b = b;
}
}
static class Pair implements Comparable<Pair> {
int a;
int b;
// int c;
Pair(int a, int b) {
this.a = a;
this.b = b;
// this.c = c;
}
public int compareTo(Pair p) {
if(a!=p.a)
return p.a-a;
return b-p.b;
}
}
static class Pair2 implements Comparable<Pair2> {
int a;
int b;
int c;
Pair2(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}
public int compareTo(Pair2 p) {
return a-p.a;
}
}
static boolean isPrime(int n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static long fast_pow(long base, long n, long M) {
if (n == 0)
return 1;
if (n == 1)
return base % M;
long halfn = fast_pow(base, n / 2, M);
if (n % 2 == 0)
return (halfn * halfn) % M;
else
return (((halfn * halfn) % M) * base) % M;
}
static long modInverse(long n, long M) {
return fast_pow(n, M - 2, M);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 9992768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output | |
PASSED | 2900f42d4f452fc685b77c6a6faac3b3 | train_110.jsonl | 1638110100 | While performing complex market analysis William encountered the following problem:For a given array $$$a$$$ of size $$$n$$$ and a natural number $$$e$$$, calculate the number of pairs of natural numbers $$$(i, k)$$$ which satisfy the following conditions: $$$1 \le i, k$$$ $$$i + e \cdot k \le n$$$. Product $$$a_i \cdot a_{i + e} \cdot a_{i + 2 \cdot e} \cdot \ldots \cdot a_{i + k \cdot e} $$$ is a prime number. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. | 256 megabytes | import java.io.*;
import java.util.*;
public class zia
{
static boolean prime[] = new boolean[25001];
static void ruffleSort(int[] a) {
int n=a.length;
Random random = new Random();
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
public static int Lcm(int a,int b)
{ int max=Math.max(a,b);
for(int i=1;;i++)
{
if((max*i)%a==0&&(max*i)%b==0)
return (max*i);
}
}
static void sieve(int n,boolean prime[])
{
// boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i =i+ p)
prime[i] = false;
}
}
}
// public static String run(int ar[],int n)
// {
// }
public static long lcm(long a,long b)
{
long max=Math.max(a, b);
long min=Math.min(a, b);
for(int i=1;;i++)
{
if((max*i)%min==0)
return max;
}
}
public static long calculate(long a,long b,long x,long y,long n)
{
if(a-x>=n)
{
a-=n;
}
else
{
b=b-(n-(a-x));
a=x;
if(b<y)
b=y;
}
return a*b;
}
public static int upperbound(int s,int e, long ar[],long x)
{
int res=-1;
while(s<=e)
{ int mid=((s-e)/2)+e;
if(ar[mid]>x)
{e=mid-1;res=mid;}
else if(ar[mid]<x)
{s=mid+1;}
else
{e=mid-1;res=mid;
if(mid>0&&ar[mid]==ar[mid-1])
e=mid-1;
else
break;
}
}
return res;
}
public static int lowerbound(int s,int e, long ar[],long x)
{
int res=-1;
while(s<=e)
{ int mid=((s-e)/2)+e;
if(ar[mid]>x)
{e=mid-1;}
else if(ar[mid]<x)
{s=mid+1;res=mid;}
else
{res=mid;
if(mid+1<ar.length&&ar[mid]==ar[mid+1])
s=mid+1;
else
break;}
}
return res;
}
static long modulo=1000000007;
public static long power(long a, long b)
{
if(b==0) return 1;
long temp=power(a,b/2)%modulo;
if((b&1)==0)
return (temp*temp)%modulo;
else
return (((temp*temp)%modulo)*a)%modulo;
}
public static long powerwithoutmod(long a, long b)
{
if(b==0) return 1;
long temp=power(a,b/2);
if((b&1)==0)
return (temp*temp);
else
return ((temp*temp)*a);
}
public static double log2(long a)
{ double d=Math.log(a)/Math.log(2);
return d;
}
public static int log10(long a)
{ double d=Math.log(a)/Math.log(10);
return (int)d;
}
public static int find(int ar[],int x)
{
int count=0;
for(int i=0;i<ar.length;i+=2)
{
if((ar[i]&1)!=x)
count++;
}
return count;
}
public static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long frogK(int index,long height[],long output[],int k)
{
if(index==1)
return 0;
if(output[index]==0)
{
long MinEnergy=Long.MAX_VALUE;
for(int j=1;j<=k&&index-j>=1;j++)
MinEnergy=Math.min(MinEnergy,frogK(index-j, height, output, k)+Math.abs(height[index]-height[index-j]));
output[index]=MinEnergy;
}
return output[index];
}
public static void tree(int s,int e,int ar[],int c)
{
if(s<=e)
{
int max=s;
for(int i=s;i<=e;i++)
if(ar[i]>ar[max])
max=i;
ar[max]=c++;
tree(s,max-1,ar,c);
tree(max+1,e,ar,c);
}
}
public static void solve(int s,int m,ArrayList<Integer> digits)
{
for(int i=0;i<m;i++)
{
if(s>=9)
{
digits.add(9);
s-=9;
}
else
{
if(s!=0)
{digits.add(s);s=0;}
else
digits.add(0);
}
}
}
public static boolean isValid(int hour,int minute,int h,int m)
{
int ar[]={0,1,5,-1,-1,2,-1,-1,8,-1};
if(ar[minute%10]==-1||ar[minute/10]==-1 ||ar[hour%10]==-1||ar[hour/10]==-1)
return false;
int mirrorHour=ar[minute%10]*10+ar[minute/10];
int mirrorMinute=ar[hour%10]*10+ar[hour/10];
if(mirrorMinute>=m||mirrorHour>=h)
return false;
return true;
}
public static void main(String[] args) throws Exception
{
FastIO sc = new FastIO();
//sc.println();
//xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx CODE xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
int test=sc.nextInt();
// // double c=Math.log(10);
boolean prime[]=new boolean[1000001];
sieve(1000000, prime);
while(test-->0)
{
int n=sc.nextInt();
int e=sc.nextInt();
int ar[]=new int[n];
int left[]=new int[n];
int right[]=new int[n];
for(int i=0;i<n;i++)
{
ar[i]=sc.nextInt();
if(ar[i]==1)
{
if(i-e>=0&&ar[i-e]==1)
left[i]=left[i-e]+1;
}
}
for(int i=n-1;i>=0;i--)
{
if(ar[i]==1)
{
if(i+e<n&&ar[i+e]==1)
right[i]=right[i+e]+1;
}
}
// sc.println("left="+Arrays.toString(left)+"\nright="+Arrays.toString(right));
long res=0;
for(int i=0;i<n;i++)
{
long leftcount=0,rightcount=0;
if(ar[i]!=1&&prime[ar[i]])
{
if(i-e>=0&&ar[i-e]==1)
leftcount+=left[i-e]+1;
if(i+e<n&&ar[i+e]==1)
rightcount+=right[i+e]+1;
// if(count!=0)
res=res+leftcount+rightcount+leftcount*rightcount;
}
}
sc.println(res);
}
// xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx CODE xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
sc.close();
}
}
class pair implements Comparable<pair>{
long a;
long b;
pair(long a,long b)
{this.a=a;
this.b=b;
}
public int compareTo(pair p)
{return (int)(-this.a+p.a);}
}
class triplet implements Comparable<triplet>{
int first,second,third;
triplet(int first,int second,int third)
{this.first=first;
this.second=second;
this.third=third;
}
public int compareTo(triplet p)
{return this.third-p.third;}
}
// class triplet
// {
// int x1,x2,i;
// triplet(int a,int b,int c)
// {x1=a;x2=b;i=c;}
// }
class FastIO extends PrintWriter {
private InputStream stream;
private byte[] buf = new byte[1<<16];
private int curChar, numChars;
// standard input
public FastIO() { this(System.in,System.out); }
public FastIO(InputStream i, OutputStream o) {
super(o);
stream = i;
}
// file input
public FastIO(String i, String o) throws IOException {
super(new FileWriter(o));
stream = new FileInputStream(i);
}
// throws InputMismatchException() if previously detected end of file
private int nextByte() {
if (numChars == -1) throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars == -1) return -1; // end of file
}
return buf[curChar++];
}
public String nextLine() {
int c; do { c = nextByte(); } while (c <= '\n');
StringBuilder res = new StringBuilder();
do { res.appendCodePoint(c); c = nextByte(); } while (c > '\n');
return res.toString();
}
public String next() {
int c; do { c = nextByte(); } while (c <= ' ');
StringBuilder res = new StringBuilder();
do { res.appendCodePoint(c); c = nextByte(); } while (c > ' ');
return res.toString();
}
public int nextInt() {
int c; do { c = nextByte(); } while (c <= ' ');
int sgn = 1; if (c == '-') { sgn = -1; c = nextByte(); }
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10*res+c-'0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public long nextLong() {
int c; do { c = nextByte(); } while (c <= ' ');
long sgn = 1; if (c == '-') { sgn = -1; c = nextByte(); }
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res = 10*res+c-'0';
c = nextByte();
} while (c > ' ');
return res * sgn;
}
public double nextDouble() { return Double.parseDouble(next()); }
}
| Java | ["6\n7 3\n10 2 1 3 1 19 3\n3 2\n1 13 1\n9 3\n2 4 2 1 1 1 1 4 2\n3 1\n1 1 1\n4 1\n1 2 1 1\n2 2\n1 2"] | 2 seconds | ["2\n0\n4\n0\n5\n0"] | NoteIn the first example test case two pairs satisfy the conditions: $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{5} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 19$$$ which is a prime number. In the second example test case there are no pairs that satisfy the conditions.In the third example test case four pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{4} \cdot a_{7} = 2$$$ which is a prime number. $$$i = 3, k = 1$$$, for which the product is: $$$a_{3} \cdot a_{6} = 2$$$ which is a prime number. $$$i = 6, k = 1$$$, for which the product is: $$$a_{6} \cdot a_{9} = 2$$$ which is a prime number. In the fourth example test case there are no pairs that satisfy the conditions.In the fifth example test case five pairs satisfy the conditions: $$$i = 1, k = 1$$$, for which the product is: $$$a_{1} \cdot a_{2} = 2$$$ which is a prime number. $$$i = 1, k = 2$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 1, k = 3$$$, for which the product is: $$$a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. $$$i = 2, k = 1$$$, for which the product is: $$$a_{2} \cdot a_{3} = 2$$$ which is a prime number. $$$i = 2, k = 2$$$, for which the product is: $$$a_{2} \cdot a_{3} \cdot a_{4} = 2$$$ which is a prime number. In the sixth example test case there are no pairs that satisfy the conditions. | Java 11 | standard input | [
"binary search",
"dp",
"implementation",
"number theory",
"schedules",
"two pointers"
] | 32130f939336bb6f2deb4dfa5402867d | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10\,000$$$). Description of the test cases follows. The first line of each test case contains two integers $$$n$$$ and $$$e$$$ $$$(1 \le e \le n \le 2 \cdot 10^5)$$$, the number of items in the array and number $$$e$$$, respectively. The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ $$$(1 \le a_i \le 10^6)$$$, the contents of the array. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,400 | For each test case output the answer in the following format: Output one line containing the number of pairs of numbers $$$(i, k)$$$ which satisfy the conditions. | standard output |
Subsets and Splits