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PASSED | b5e5958b4ab9de244dadc4ccc1fa7c95 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.ThreadLocalRandom;
public class c731{
public static void main(String[] args) throws IOException{
BufferedWriter out = new BufferedWriter(
new OutputStreamWriter(System.out));
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
PrintWriter pt = new PrintWriter(System.out);
FastReader sc = new FastReader();
int t = sc.nextInt();
for(int o = 0; o<t;o++) {
int n = sc.nextInt();
int a = sc.nextInt();
int b = sc.nextInt();
long[] arr = new long[n+1];
for(int i = 1; i<=n;i++) {
arr[i] = sc.nextLong();
}
long[] dp = new long[n+1];
int v = 1;
for(int i = n-1 ; i>=0;i--) {
dp[i] = dp[i+1] + v *(arr[i+1] - arr[i])* b;
v++;
}
long ans = dp[0];
for(int i = 1 ; i<=n;i++ ) {
long val = arr[i]*(a+b) + dp[i];
ans = Math.min(ans, val);
}
System.out.println(ans);
}
}
//------------------------------------------------------------------------------------------------------------------------------------------------
public static boolean check(int[] arr , int n , int v , int l ) {
int x = v/2;
int y = v/2;
// System.out.println(x + " " + y);
if(v%2 == 1 ) {
x++;
}
for(int i = 0 ; i<n;i++) {
int d = l - arr[i];
int c = Math.min(d/2, y);
y -= c;
arr[i] -= c*2;
if(arr[i] > x) {
return false;
}
x -= arr[i];
}
return true;
}
public static int cnt_set(long x) {
long v = 1l;
int c =0;
int f = 0;
while(v<=x) {
if((v&x)!=0) {
c++;
}
v = v<<1;
}
return c;
}
public static int lis(int[] arr,int[] dp) {
int n = arr.length;
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(arr[0]);
dp[0]= 1;
for(int i = 1 ; i<n;i++) {
int x = al.get(al.size()-1);
if(arr[i]>x) {
al.add(arr[i]);
}else {
int v = lower_bound(al, 0, al.size(), arr[i]);
// System.out.println(v);
al.set(v, arr[i]);
}
dp[i] = al.size();
}
//return al.size();
return al.size();
}
public static int lis2(int[] arr,int[] dp) {
int n = arr.length;
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(-arr[n-1]);
dp[n-1] = 1;
// System.out.println(al);
for(int i = n-2 ; i>=0;i--) {
int x = al.get(al.size()-1);
// System.out.println(-arr[i] + " " + i + " " + x);
if((-arr[i])>x) {
al.add(-arr[i]);
}else {
int v = lower_bound(al, 0, al.size(), -arr[i]);
// System.out.println(v);
al.set(v, -arr[i]);
}
dp[i] = al.size();
}
//return al.size();
return al.size();
}
static int cntDivisors(int n){
int cnt = 0;
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i)
cnt++;
else
cnt+=2;
}
}
return cnt;
}
public 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 long ncr(long[] fac, int n , int r , long m) {
if(r>n) {
return 0;
}
return fac[n]*(modInverse(fac[r], m))%m *(modInverse(fac[n-r], m))%m;
}
public static int lower_bound(ArrayList<Integer> arr,int lo , int hi, int k)
{
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (arr.get(mid) <k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==arr.size())
{
return -1;
}
return s;
}
public static int upper_bound(ArrayList<Integer> arr,int lo , int hi, int k)
{
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (arr.get(mid) <=k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==arr.size())
{
return -1;
}
return s;
}
// -----------------------------------------------------------------------------------------------------------------------------------------------
public static long gcd(long a, long b){
if (a == 0)
return b;
return gcd(b % a, a);
}
//--------------------------------------------------------------------------------------------------------------------------------------------------------
public static long modInverse(long a, long m){
long m0 = m;
long y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1) {
// q is quotient
long q = a / m;
long t = m;
// m is remainder now, process
// same as Euclid's algo
m = a % m;
a = t;
t = y;
// Update x and y
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
//_________________________________________________________________________________________________________________________________________________________________
// private static int[] parent;
// private static int[] size;
public static int find(int[] parent, int u) {
while(u != parent[u]) {
parent[u] = parent[parent[u]];
u = parent[u];
}
return u;
}
private static void union(int[] parent,int[] size,int u, int v) {
int rootU = find(parent,u);
int rootV = find(parent,v);
if(rootU == rootV) {
return;
}
if(size[rootU] < size[rootV]) {
parent[rootU] = rootV;
size[rootV] += size[rootU];
} else {
parent[rootV] = rootU;
size[rootU] += size[rootV];
}
}
//-----------------------------------------------------------------------------------------------------------------------------------
//segment tree
//for finding minimum in range
public static void build(int [] seg,int []arr,int idx, int lo , int hi) {
if(lo == hi) {
seg[idx] = arr[lo];
return;
}
int mid = (lo + hi)/2;
build(seg,arr,2*idx+1, lo, mid);
build(seg,arr,idx*2+2, mid +1, hi);
// seg[idx] = Math.min(seg[2*idx+1],seg[2*idx+2]);
// seg[idx] = seg[idx*2+1]+ seg[idx*2+2];
// seg[idx] = Math.min(seg[idx*2+1], seg[idx*2+2]);
seg[idx] = seg[idx*2+1] * seg[idx*2+2];
}
//for finding minimum in range
public static int query(int[]seg,int idx , int lo , int hi , int l , int r) {
if(lo>=l && hi<=r) {
return seg[idx];
}
if(hi<l || lo>r) {
return 1;
}
int mid = (lo + hi)/2;
int left = query(seg,idx*2 +1, lo, mid, l, r);
int right = query(seg,idx*2 + 2, mid + 1, hi, l, r);
// return Math.min(left, right);
//return gcd(left, right);
// return Math.min(left, right);
return left * right;
}
// // for sum
//
//public static void update(int[]seg,int idx, int lo , int hi , int node , int val) {
// if(lo == hi) {
// seg[idx] += val;
// }else {
//int mid = (lo + hi )/2;
//if(node<=mid && node>=lo) {
// update(seg, idx * 2 +1, lo, mid, node, val);
//}else {
// update(seg, idx*2 + 2, mid + 1, hi, node, val);
//}
//seg[idx] = seg[idx*2 + 1] + seg[idx*2 + 2];
//
//}
//}
//---------------------------------------------------------------------------------------------------------------------------------------
//
//static void shuffleArray(long[] ar)
//{
// // If running on Java 6 or older, use `new Random()` on RHS here
// Random rnd = ThreadLocalRandom.current();
// for (int i = ar.length - 1; i > 0; i--)
// {
// int index = rnd.nextInt(i + 1);
// // Simple swap
// int a = ar[index];
// ar[index] = ar[i];
// ar[i] = a;
// }
//}
// static void shuffleArray(coup[] ar)
// {
// // If running on Java 6 or older, use `new Random()` on RHS here
// Random rnd = ThreadLocalRandom.current();
// for (int i = ar.length - 1; i > 0; i--)
// {
// int index = rnd.nextInt(i + 1);
// // Simple swap
// coup a = ar[index];
// ar[index] = ar[i];
// ar[i] = a;
// }
// }
//-----------------------------------------------------------------------------------------------------------------------------------------------------------
}
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;
}
}
//------------------------------------------------------------------------------------------------------------------------------------------------------------
//-----------------------------------------------------------------------------------------------------------------------------------------------------------
class coup{
int a;
int b;
public coup(int a , int b) {
this.a = a;
this.b = b;
}
}
class tripp{
int a;
int b;
double c;
public tripp(int a , int b, double c) {
this.a = a;
this.b = b;
this.c = c;
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 52c66b9b0aa67e91fe2a985dc8bf6556 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
public class c {
static BufferedReader bf;
static PrintWriter out;
public static void main (String[] args)throws IOException {
bf = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
int t= nextInt();
while(t-- > 0){
solve();
}
}
public static void solve()throws IOException{
String[]temp = bf.readLine().split(" ");
long n = toLong(temp[0]);
long a = toLong(temp[1]);
long b = toLong(temp[2]);
long []arr = nextLongArray((int)n);
if(n == 1){
println(arr[0]*b);
return;
}
long count = 0;
long cur = 0;
for(int i =0;i<n;i++){
count += (Math.abs(arr[i] - cur)*b);
if(Math.abs(arr[i] - cur)*a <= ((n-1)-i) * b * (Math.abs(arr[i] - cur))){
count += Math.abs(arr[i] - cur)*a;
cur = arr[i];
}
}
println(count);
}
// code for input
public static void print(String s ){
System.out.print(s);
}
public static void print(int num ){
System.out.print(num);
}
public static void print(long num ){
System.out.print(num);
}
public static void println(String s){
System.out.println(s);
}
public static void println(int num){
System.out.println(num);
}
public static void println(long num){
System.out.println(num);
}
public static void println(){
System.out.println();
}
public static int toInt(String s){
return Integer.parseInt(s);
}
public static long toLong(String s){
return Long.parseLong(s);
}
public static String[] nextStringArray()throws IOException{
return bf.readLine().split(" ");
}
public static int nextInt()throws IOException{
return Integer.parseInt(bf.readLine());
}
public static long nextLong()throws IOException{
return Long.parseLong(bf.readLine());
}
public static String nextString()throws IOException{
return bf.readLine();
}
public static int[] nextIntArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
int[]arr = new int[n];
for(int i =0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
return arr;
}
public static long[] nextLongArray(int n)throws IOException{
String[]str = bf.readLine().split(" ");
long[]arr = new long[n];
for(int i =0;i<n;i++){
arr[i] = Long.parseLong(str[i]);
}
return arr;
}
public static int[][] newIntMatrix(int r,int c)throws IOException{
int[][]arr = new int[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Integer.parseInt(str[j]);
}
}
return arr;
}
public static long[][] newLongMatrix(int r,int c)throws IOException{
long[][]arr = new long[r][c];
for(int i =0;i<r;i++){
String[]str = bf.readLine().split(" ");
for(int j =0;j<c;j++){
arr[i][j] = Long.parseLong(str[j]);
}
}
return arr;
}
//someuseFull functions to use
public static long gcd(long a,long b){
if(b == 0)return a;
return gcd(b,a%b);
}
public static int gcd(int a,int b){
if(b == 0)return a;
return gcd(b,a%b);
}
public static long lcm(long a,long b){
return (a*b)/(gcd(a,b));
}
}
// if a problem is related to binary string it could also be related to parenthesis
// try to use binary search in the question it might work
// try sorting
// try to think in opposite direction of question it might work in your way
// if a problem is related to maths try to relate some of the continuous subarray with variables like - > a+b+c+ or a,b,c,d in general
// gcd(1.p1,2.p2,3.p3,4.p4....n.pn) cannot be greater than 2 it has been proved | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | d266fda914ee5f4b26d96a471559b09d | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class Contest_yandexA{
//static final int MAXN = (int)1e6;
public static void main(String[] args) throws IOException{
Scanner input = new Scanner(System.in);
/*int n = input.nextInt();
int k = input.nextInt();
k = k%4;
int[] a = new int[n];
for(int i = 0;i<n;i++){
a[i] = input.nextInt();
}
int[] count = new int[n];
int sum = 0;
for(int tt = 0;tt<k;tt++){
for(int i = 0;i<n;i++){
count[a[i]-1]++;
}
for(int i = 0;i<n;i++){
sum+= count[i];
}
for(int i = 0;i<n;i++){
a[i] = sum;
sum-= count[i];
}
}
for(int i = 0;i<n;i++){
System.out.print(a[i] + " ");
}*/
int t = input.nextInt();
for(int tt = 0;tt<t;tt++){
int n = input.nextInt();
long a = input.nextLong();
long b = input.nextLong();
long[] x = new long[n];
long[] dp = new long[n+1];
for(int i = 0;i<n;i++){
x[i] = input.nextLong();
dp[i+1]+= dp[i] + x[i];
}
long min = Long.MAX_VALUE;
for(int i = 0;i<n-1;i++){
min = Math.min(min,x[i]*(a+b)+(dp[n]-dp[i+1]-(x[i]*(n-i-1)))*b);
}
min = Math.min(min,dp[n]*b);
System.out.println(min);
}
}
public static long gcd(long a,long b){
if(b == 0){
return a;
}
return gcd(b,a%b);
}
/*public static int lcm(int a,int b){
return (a / gcd(a, b)) * b;
}*/
}
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 p){
return this.x-p.x;
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | ab57aed5a170572cb9da151e1d33d86d | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static FastReader sc=new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=1000000007;
// static long mod=998244353;
static int MAX=Integer.MAX_VALUE;
static int MIN=Integer.MIN_VALUE;
static long MAXL=Long.MAX_VALUE;
static long MINL=Long.MIN_VALUE;
static ArrayList<Integer> graph[];
// static ArrayList<ArrayList<Integer>> graph;
static long fact[];
static int seg[];
static int dp[][];
// static long dp[][];
public static void main (String[] args) throws java.lang.Exception
{
// code goes here
int t=I();
outer:while(t-->0)
{
int n=I();
long a=L(),b=L();
long arr[]=new long[n+1];
for(int i=1;i<=n;i++){
arr[i]=L();
}
long temp1[]=new long[n+1];
long temp2[]=new long[n+1];
for(int i=1;i<=n;i++){
temp1[i]+=arr[i]+temp1[i-1];
}
for(int i=1;i<=n;i++){
temp2[i]+=(b*(arr[i]-arr[i-1]))+temp2[i-1];
}
long ans=temp1[n]*b;
for(int i=1;i<=n;i++){
long temp=b*(temp1[n]-temp1[i]-(n-i)*arr[i]);
temp+=temp2[i];
temp+=a*arr[i];
ans=min(ans,temp);
}
out.println(ans);
}
out.close();
}
public static class pair
{
long a;
long b;
public pair(long aa,long bb)
{
a=aa;
b=bb;
}
}
public static class myComp implements Comparator<pair>
{
//sort in ascending order.
public int compare(pair p1,pair p2)
{
if(p1.a==p2.a)
return 0;
else if(p1.a<p2.a)
return -1;
else
return 1;
}
// sort in descending order.
// public int compare(pair p1,pair p2)
// {
// if(p1.a==p2.a)
// return 0;
// else if(p1.a<p2.a)
// return 1;
// else
// return -1;
// }
}
public static void DFS(int s,boolean visited[])
{
visited[s]=true;
for(int i:graph[s]){
if(!visited[i]){
DFS(i,visited);
}
}
}
public static void setGraph(int n,int m)
{
graph=new ArrayList[n+1];
for(int i=0;i<=n;i++){
graph[i]=new ArrayList<>();
}
for(int i=0;i<m;i++){
int u=I(),v=I();
graph[u].add(v);
graph[v].add(u);
}
}
//LOWER_BOUND and UPPER_BOUND functions
//It returns answer according to zero based indexing.
public static int lower_bound(long arr[],long X,int start, int end) //start=0,end=n-1
{
if(start>end)return -1;
if(arr[arr.length-1]<X)return end;
if(arr[0]>X)return -1;
int left=start,right=end;
while(left<right){
int mid=(left+right)/2;
// if(arr[mid]==X){ //Returns last index of lower bound value.
// if(mid<end && arr[mid+1]==X){
// left=mid+1;
// }else{
// return mid;
// }
// }
if(arr[mid]==X){ //Returns first index of lower bound value.
if(mid>start && arr[mid-1]==X){
right=mid-1;
}else{
return mid;
}
}
else if(arr[mid]>X){
if(mid>start && arr[mid-1]<X){
return mid-1;
}else{
right=mid-1;
}
}else{
if(mid<end && arr[mid+1]>X){
return mid;
}else{
left=mid+1;
}
}
}
return left;
}
//It returns answer according to zero based indexing.
public static int upper_bound(long arr[],long X,int start,int end) //start=0,end=n-1
{
if(arr[0]>=X)return start;
if(arr[arr.length-1]<X)return -1;
int left=start,right=end;
while(left<right){
int mid=(left+right)/2;
if(arr[mid]==X){ //returns first index of upper bound value.
if(mid>start && arr[mid-1]==X){
right=mid-1;
}else{
return mid;
}
}
// if(arr[mid]==X){ //returns last index of upper bound value.
// if(mid<end && arr[mid+1]==X){
// left=mid+1;
// }else{
// return mid;
// }
// }
else if(arr[mid]>X){
if(mid>start && arr[mid-1]<X){
return mid;
}else{
right=mid-1;
}
}else{
if(mid<end && arr[mid+1]>X){
return mid+1;
}else{
left=mid+1;
}
}
}
return left;
}
//END
//Segment Tree Code
public static void buildTree(int a[],int si,int ss,int se)
{
if(ss==se){
// seg[si]=ss;
if(a[ss]%2==0){
seg[si]=1;
}else{
seg[si]=0;
}
return;
}
int mid=(ss+se)/2;
buildTree(a,2*si+1,ss,mid);
buildTree(a,2*si+2,mid+1,se);
// seg[si]=max(seg[2*si+1],seg[2*si+2]);
seg[si]=(seg[2*si+1]&seg[2*si+2]);
}
public static void merge(ArrayList<Integer> f,ArrayList<Integer> a,ArrayList<Integer> b)
{
int i=0,j=0;
while(i<a.size() && j<b.size()){
if(a.get(i)<=b.get(j)){
f.add(a.get(i));
i++;
}else{
f.add(b.get(j));
j++;
}
}
while(i<a.size()){
f.add(a.get(i));
i++;
}
while(j<b.size()){
f.add(b.get(j));
j++;
}
}
public static void update(int si,int ss,int se,int pos,int val)
{
if(ss==se){
seg[si]=(int)add(seg[si],val);
return;
}
int mid=(ss+se)/2;
if(pos<=mid){
update(2*si+1,ss,mid,pos,val);
}else{
update(2*si+2,mid+1,se,pos,val);
}
seg[si]=(int)add(seg[2*si+1],seg[2*si+2]);
}
public static int query(int a[],int si,int ss,int se,int qs,int qe)
{
if(qs>se || qe<ss)return -1;
if(ss>=qs && se<=qe)return seg[si];
int mid=(ss+se)/2;
// return max(query(2*si+1,ss,mid,qs,qe),query(2*si+2,mid+1,se,qs,qe));
int p1=query(a,2*si+1,ss,mid,qs,qe);
int p2=query(a,2*si+2,mid+1,se,qs,qe);
if(p1==-1 && p2==-1)return -1;
else if(p1==-1)return p2;
else if(p2==-1)return p1;
else return (p1&p2);
}
//Segment Tree Code end
//Prefix Function of KMP Algorithm
public static int[] KMP(char c[],int n)
{
int pi[]=new int[n];
for(int i=1;i<n;i++){
int j=pi[i-1];
while(j>0 && c[i]!=c[j]){
j=pi[j-1];
}
if(c[i]==c[j])j++;
pi[i]=j;
}
return pi;
}
public static long kadane(long a[],int n) //largest sum subarray
{
long max_sum=Long.MIN_VALUE,max_end=0;
for(int i=0;i<n;i++){
max_end+=a[i];
if(max_sum<max_end){max_sum=max_end;}
if(max_end<0){max_end=0;}
}
return max_sum;
}
public static long nPr(int n,int r)
{
long ans=divide(fact(n),fact(n-r),mod);
return ans;
}
public static long nCr(int n,int r)
{
long ans=divide(fact[n],mul(fact[n-r],fact[r]),mod);
return ans;
}
public static boolean isSorted(int a[])
{
int n=a.length;
for(int i=0;i<n-1;i++){
if(a[i]>a[i+1])return false;
}
return true;
}
public static boolean isSorted(long a[])
{
int n=a.length;
for(int i=0;i<n-1;i++){
if(a[i]>a[i+1])return false;
}
return true;
}
public static int computeXOR(int n) //compute XOR of all numbers between 1 to n.
{
if (n % 4 == 0)
return n;
if (n % 4 == 1)
return 1;
if (n % 4 == 2)
return n + 1;
return 0;
}
public static int np2(int x)
{
x--;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x++;
return x;
}
public static int hp2(int x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
public static long hp2(long x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
public static ArrayList<Integer> primeSieve(int n)
{
ArrayList<Integer> arr=new ArrayList<>();
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;
}
}
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
// Fenwick / BinaryIndexed Tree USE IT - FenwickTree ft1=new FenwickTree(n);
public static class FenwickTree
{
int farr[];
int n;
public FenwickTree(int c)
{
n=c+1;
farr=new int[n];
}
// public void update_range(int l,int r,long p)
// {
// update(l,p);
// update(r+1,(-1)*p);
// }
public void update(int x,int p)
{
for(;x<n;x+=x&(-x))
{
farr[x]+=p;
}
}
public int get(int x)
{
int ans=0;
for(;x>0;x-=x&(-x))
{
ans=ans+farr[x];
}
return ans;
}
}
//Disjoint Set Union
public static class DSU
{
int par[],rank[];
public DSU(int c)
{
par=new int[c+1];
rank=new int[c+1];
for(int i=0;i<=c;i++)
{
par[i]=i;
rank[i]=0;
}
}
public int find(int a)
{
if(a==par[a])
return a;
return par[a]=find(par[a]);
}
public void union(int a,int b)
{
int a_rep=find(a),b_rep=find(b);
if(a_rep==b_rep)
return;
if(rank[a_rep]<rank[b_rep])
par[a_rep]=b_rep;
else if(rank[a_rep]>rank[b_rep])
par[b_rep]=a_rep;
else
{
par[b_rep]=a_rep;
rank[a_rep]++;
}
}
}
public static HashMap<Integer,Integer> primeFact(int a)
{
// HashSet<Long> arr=new HashSet<>();
HashMap<Integer,Integer> hm=new HashMap<>();
int p=0;
while(a%2==0){
// arr.add(2L);
p++;
a=a/2;
}
hm.put(2,hm.getOrDefault(2,0)+p);
for(int i=3;i*i<=a;i+=2){
p=0;
while(a%i==0){
// arr.add(i);
p++;
a=a/i;
}
hm.put(i,hm.getOrDefault(i,0)+p);
}
if(a>2){
// arr.add(a);
hm.put(a,hm.getOrDefault(a,0)+1);
}
// return arr;
return hm;
}
public static boolean isInteger(double N)
{
int X = (int)N;
double temp2 = N - X;
if (temp2 > 0)
{
return false;
}
return true;
}
public static boolean isPalindrome(String s)
{
int n=s.length();
for(int i=0;i<=n/2;i++){
if(s.charAt(i)!=s.charAt(n-i-1)){
return false;
}
}
return true;
}
public static int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long gcd(long a,long b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long fact(long n)
{
long fact=1;
for(long i=2;i<=n;i++){
fact=((fact%mod)*(i%mod))%mod;
}
return fact;
}
public static long fact(int n)
{
long fact=1;
for(int i=2;i<=n;i++){
fact=((fact%mod)*(i%mod))%mod;
}
return fact;
}
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;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static boolean isPrime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static void printArray(long a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(int a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(char a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]);
}
out.println();
}
public static void printArray(String a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(boolean a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(pair a[])
{
for(pair p:a){
out.println(p.a+"->"+p.b);
}
}
public static void printArray(int a[][])
{
for(int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.print(a[i][j]+" ");
}out.println();
}
}
public static void printArray(long a[][])
{
for(int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.print(a[i][j]+" ");
}out.println();
}
}
public static void printArray(char a[][])
{
for(int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.print(a[i][j]+" ");
}out.println();
}
}
public static void printArray(ArrayList<?> arr)
{
for(int i=0;i<arr.size();i++){
out.print(arr.get(i)+" ");
}
out.println();
}
public static void printMapI(HashMap<?,?> hm){
for(Map.Entry<?,?> e:hm.entrySet()){
out.println(e.getKey()+"->"+e.getValue());
}out.println();
}
public static void printMap(HashMap<Long,ArrayList<Integer>> hm){
for(Map.Entry<Long,ArrayList<Integer>> e:hm.entrySet()){
out.print(e.getKey()+"->");
ArrayList<Integer> arr=e.getValue();
for(int i=0;i<arr.size();i++){
out.print(arr.get(i)+" ");
}out.println();
}
}
//Modular Arithmetic
public static long add(long a,long b)
{
a+=b;
if(a>=mod)a-=mod;
return a;
}
public static long sub(long a,long b)
{
a-=b;
if(a<0)a+=mod;
return a;
}
public static long mul(long a,long b)
{
return ((a%mod)*(b%mod))%mod;
}
public static long divide(long a,long b,long m)
{
a=mul(a,modInverse(b,m));
return a;
}
public static long modInverse(long a,long m)
{
int x=0,y=0;
own p=new own(x,y);
long g=gcdExt(a,m,p);
if(g!=1){
out.println("inverse does not exists");
return -1;
}else{
long res=((p.a%m)+m)%m;
return res;
}
}
public static long gcdExt(long a,long b,own p)
{
if(b==0){
p.a=1;
p.b=0;
return a;
}
int x1=0,y1=0;
own p1=new own(x1,y1);
long gcd=gcdExt(b,a%b,p1);
p.b=p1.a - (a/b) * p1.b;
p.a=p1.b;
return gcd;
}
public static long pwr(long m,long n)
{
long res=1;
if(m==0)
return 0;
while(n>0)
{
if((n&1)!=0)
{
res=(res*m);
}
n=n>>1;
m=(m*m);
}
return res;
}
public static long modpwr(long m,long n)
{
long res=1;
m=m%mod;
if(m==0)
return 0;
while(n>0)
{
if((n&1)!=0)
{
res=(res*m)%mod;
}
n=n>>1;
m=(m*m)%mod;
}
return res;
}
public static class own
{
long a;
long b;
public own(long val,long index)
{
a=val;
b=index;
}
}
//Modular Airthmetic
public static void sort(int[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(char[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
char tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(long[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
//max & min
public static int max(int a,int b){return Math.max(a,b);}
public static int min(int a,int b){return Math.min(a,b);}
public static int max(int a,int b,int c){return Math.max(a,Math.max(b,c));}
public static int min(int a,int b,int c){return Math.min(a,Math.min(b,c));}
public static long max(long a,long b){return Math.max(a,b);}
public static long min(long a,long b){return Math.min(a,b);}
public static long max(long a,long b,long c){return Math.max(a,Math.max(b,c));}
public static long min(long a,long b,long c){return Math.min(a,Math.min(b,c));}
public static int maxinA(int a[]){int n=a.length;int mx=a[0];for(int i=1;i<n;i++){mx=max(mx,a[i]);}return mx;}
public static long maxinA(long a[]){int n=a.length;long mx=a[0];for(int i=1;i<n;i++){mx=max(mx,a[i]);}return mx;}
public static int mininA(int a[]){int n=a.length;int mn=a[0];for(int i=1;i<n;i++){mn=min(mn,a[i]);}return mn;}
public static long mininA(long a[]){int n=a.length;long mn=a[0];for(int i=1;i<n;i++){mn=min(mn,a[i]);}return mn;}
public static long suminA(int a[]){int n=a.length;long sum=0;for(int i=0;i<n;i++){sum+=a[i];}return sum;}
public static long suminA(long a[]){int n=a.length;long sum=0;for(int i=0;i<n;i++){sum+=a[i];}return sum;}
//end
public static int[] I(int n){int a[]=new int[n];for(int i=0;i<n;i++){a[i]=I();}return a;}
public static long[] IL(int n){long a[]=new long[n];for(int i=0;i<n;i++){a[i]=L();}return a;}
public static long[] prefix(int a[]){int n=a.length;long pre[]=new long[n];pre[0]=a[0];for(int i=1;i<n;i++){pre[i]=pre[i-1]+a[i];}return pre;}
public static long[] prefix(long a[]){int n=a.length;long pre[]=new long[n];pre[0]=a[0];for(int i=1;i<n;i++){pre[i]=pre[i-1]+a[i];}return pre;}
public static int I(){return sc.I();}
public static long L(){return sc.L();}
public static String S(){return sc.S();}
public static double D(){return sc.D();}
}
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 I(){ return Integer.parseInt(next());}
long L(){ return Long.parseLong(next());}
double D(){return Double.parseDouble(next());}
String S(){
String str = "";
try {
str = br.readLine();
}
catch (IOException e){
e.printStackTrace();
}
return str;
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | e43e17b7a9f3eca05e8247ae43c7f485 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int T=sc.nextInt();
while(T-->0) {
int n=sc.nextInt();
int a=sc.nextInt();
int b=sc.nextInt();
long[] x=new long[n+1];
long[] pre=new long[n+1];
for(int i=1;i<=n;i++) {
x[i]=sc.nextLong();
pre[i]=pre[i-1]+x[i];
}
long ans=b*pre[n];
for(int i=1;i<=n;i++) {
long t1=(pre[n]-pre[i]-(n-i)*x[i])*b;
long t2=x[i]*a;
long t3=x[i]*b;
ans=Math.min(ans,t1+t2+t3);
}
System.out.println(ans);
}
sc.close();
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 0118d6345a0f7835a8214bffcea9e4f7 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String arggs[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt(), a = sc.nextInt(), b = sc.nextInt();
long[] arr = new long[n+1], sum = new long[n+1];
for(int i=1; i<=n; i++)
sum[i] = sum[i-1] + (arr[i] = sc.nextInt());
long cost = sum[n]*b, ans = cost;
for(int i=1; i<=n; i++) {
cost -= (n-i)*(arr[i]-arr[i-1])*b;
cost += (arr[i]-arr[i-1])*a;
ans = Math.min(ans, cost);
}
System.out.println(ans);
}
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 578b2bd7c3e58040cb349be6737251af | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | /*
author:Karan
created:18.04.2022 11:26:13
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.*;
public class C_Line_Empire {
public static void main(String[] args) throws IOException {
try {
FastScanner fs = new FastScanner();
long T = fs.nextInt();
for (long tt = 0; tt < T; tt++) {
long a,b;
int n;
n=fs.nextInt();
a=fs.nextLong();
b=fs.nextLong();
long arr[] = fs.readArray(n);
long pref[] = new long[(int)n+1];
pref[n-1]=arr[n-1];
for(int i=n-2;i>=0;i--) {
pref[i]=pref[i+1]+arr[i];
}
pref[n]=pref[n-1];
long cur=0;
long ans = b*pref[0];
for(int i=0;i<n;i++) {
long temp=arr[i];
if(i!=0) temp-=arr[i-1];
cur+=temp*a;
cur+=temp*b;
//System.out.println(cur+" "+b*(pref[i+1]-((n-i-1)*arr[i])));
ans=Math.min(ans,cur+b*(pref[i+1]-((n-i-1)*arr[i])));
}
System.out.println(ans);
}
}
catch (Exception e) {
return;
}
}
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[] readArray(long n) {
long[] a = new long[(int)n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | d734273bc22cca0e1efaf71fc5b44d58 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class C_Line_Empire {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int n, testCases;
static long a, b;
static long arr[];
static long max = Long.MAX_VALUE, mod = (long) (1e9);
static long solve(int index, long max, long summation) {
if (index >= n) {
return max;
}
long current = (a + b) * arr[index];
summation -= arr[index];
long X = summation - (n - index - 1) * arr[index];
current += X * b;
max = Math.min(max, current);
return solve(index + 1, max, summation);
}
static void solve(int t) {
long sum = detect_sum(0, arr, 0);
max = Long.MAX_VALUE;
ans.append(solve(0, max, sum));
if (t != testCases) {
ans.append("\n");
}
}
public static void main(String[] priya) throws IOException {
testCases = in.nextInt();
for (int t = 0; t < testCases; ++t) {
n = in.nextInt() + 1;
a = in.nextLong();
b = in.nextLong();
arr = new long[n];
for (int i = 1; i < n; ++i) {
arr[i] = in.nextLong();
}
solve(t + 1);
}
out.print(ans.toString());
out.flush();
in.close();
}
static boolean isSmaller(String str1, String str2) {
// Calculate lengths of both string
int n1 = str1.length(), n2 = str2.length();
if (n1 < n2) {
return true;
}
if (n2 < n1) {
return false;
}
for (int i = 0; i < n1; i++) {
if (str1.charAt(i) < str2.charAt(i)) {
return true;
} else if (str1.charAt(i) > str2.charAt(i)) {
return false;
}
}
return false;
}
static String sub(String str1, String str2) {
if (isSmaller(str1, str2)) {
String t = str1;
str1 = str2;
str2 = t;
}
String str = "";
int n1 = str1.length(), n2 = str2.length();
int diff = n1 - n2;
int carry = 0;
for (int i = n2 - 1; i >= 0; i--) {
int sub
= (((int) str1.charAt(i + diff) - (int) '0')
- ((int) str2.charAt(i) - (int) '0')
- carry);
if (sub < 0) {
sub = sub + 10;
carry = 1;
} else {
carry = 0;
}
str += String.valueOf(sub);
}
for (int i = n1 - n2 - 1; i >= 0; i--) {
if (str1.charAt(i) == '0' && carry > 0) {
str += "9";
continue;
}
int sub = (((int) str1.charAt(i) - (int) '0')
- carry);
if (i > 0 || sub > 0) {
str += String.valueOf(sub);
}
carry = 0;
}
return new StringBuilder(str).reverse().toString();
}
static String sum(String str1, String str2) {
if (str1.length() > str2.length()) {
String t = str1;
str1 = str2;
str2 = t;
}
String str = "";
int n1 = str1.length(), n2 = str2.length();
int diff = n2 - n1;
int carry = 0;
for (int i = n1 - 1; i >= 0; i--) {
int sum = ((int) (str1.charAt(i) - '0')
+ (int) (str2.charAt(i + diff) - '0') + carry);
str += (char) (sum % 10 + '0');
carry = sum / 10;
}
for (int i = n2 - n1 - 1; i >= 0; i--) {
int sum = ((int) (str2.charAt(i) - '0') + carry);
str += (char) (sum % 10 + '0');
carry = sum / 10;
}
if (carry > 0) {
str += (char) (carry + '0');
}
return new StringBuilder(str).reverse().toString();
}
static long detect_sum(int i, long a[], long sum) {
if (i >= a.length) {
return sum;
}
return detect_sum(i + 1, a, sum + a[i]);
}
static String mul(String num1, String num2) {
int len1 = num1.length();
int len2 = num2.length();
if (len1 == 0 || len2 == 0) {
return "0";
}
int result[] = new int[len1 + len2];
int i_n1 = 0;
int i_n2 = 0;
for (int i = len1 - 1; i >= 0; i--) {
int carry = 0;
int n1 = num1.charAt(i) - '0';
i_n2 = 0;
for (int j = len2 - 1; j >= 0; j--) {
int n2 = num2.charAt(j) - '0';
int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
carry = sum / 10;
result[i_n1 + i_n2] = sum % 10;
i_n2++;
}
if (carry > 0) {
result[i_n1 + i_n2] += carry;
}
i_n1++;
}
int i = result.length - 1;
while (i >= 0 && result[i] == 0) {
i--;
}
if (i == -1) {
return "0";
}
String s = "";
while (i >= 0) {
s += (result[i--]);
}
return s;
}
static class Node<T> {
T data;
Node<T> next;
public Node() {
this.next = null;
}
public Node(T data) {
this.data = data;
this.next = null;
}
public T getData() {
return data;
}
public void setData(T data) {
this.data = data;
}
public Node<T> getNext() {
return next;
}
public void setNext(Node<T> next) {
this.next = next;
}
@Override
public String toString() {
return this.getData().toString() + " ";
}
}
static class ArrayList<T> {
Node<T> head, tail;
int len;
public ArrayList() {
this.head = null;
this.tail = null;
this.len = 0;
}
int size() {
return len;
}
boolean isEmpty() {
return len == 0;
}
int indexOf(T data) {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
int index = -1, i = 0;
while (temp != null) {
if (temp.getData() == data) {
index = i;
}
i++;
temp = temp.getNext();
}
return index;
}
void add(T data) {
Node<T> newNode = new Node<>(data);
if (isEmpty()) {
head = newNode;
tail = newNode;
len++;
} else {
tail.setNext(newNode);
tail = newNode;
len++;
}
}
void see() {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
while (temp != null) {
out.print(temp.getData().toString() + " ");
out.flush();
temp = temp.getNext();
}
out.println();
out.flush();
}
void inserFirst(T data) {
Node<T> newNode = new Node<>(data);
Node<T> temp = head;
if (isEmpty()) {
head = newNode;
tail = newNode;
len++;
} else {
newNode.setNext(temp);
head = newNode;
len++;
}
}
T get(int index) {
if (isEmpty() || index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> temp = head;
int i = 0;
T data = null;
while (temp != null) {
if (i == index) {
data = temp.getData();
}
i++;
temp = temp.getNext();
}
return data;
}
void addAt(T data, int index) {
if (index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
Node<T> newNode = new Node<>(data);
int i = 0;
Node<T> temp = head;
while (temp.next != null) {
if (i == index) {
newNode.setNext(temp.next);
temp.next = newNode;
}
i++;
temp = temp.getNext();
}
// temp.setNext(temp);
len++;
}
void popFront() {
if (isEmpty()) {
throw new ArrayIndexOutOfBoundsException();
}
if (head == tail) {
head = null;
tail = null;
} else {
head = head.getNext();
}
len--;
}
void removeAt(int index) {
if (index >= len) {
throw new ArrayIndexOutOfBoundsException();
}
if (index == 0) {
this.popFront();
return;
}
Node<T> temp = head;
int i = 0;
Node<T> n = new Node<>();
while (temp != null) {
if (i == index) {
n.next = temp.next;
temp.next = n;
break;
}
i++;
n = temp;
temp = temp.getNext();
}
tail = n;
--len;
}
void clearAll() {
this.head = null;
this.tail = null;
}
}
static void merge(long a[], int left, int right, int mid) {
int n1 = mid - left + 1, n2 = right - mid;
long L[] = new long[n1];
long R[] = new long[n2];
for (int i = 0; i < n1; i++) {
L[i] = a[left + i];
}
for (int i = 0; i < n2; i++) {
R[i] = a[mid + 1 + i];
}
int i = 0, j = 0, k1 = left;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
a[k1] = L[i];
i++;
} else {
a[k1] = R[j];
j++;
}
k1++;
}
while (i < n1) {
a[k1] = L[i];
i++;
k1++;
}
while (j < n2) {
a[k1] = R[j];
j++;
k1++;
}
}
static void sort(long a[], int left, int right) {
if (left >= right) {
return;
}
int mid = (left + right) / 2;
sort(a, left, mid);
sort(a, mid + 1, right);
merge(a, left, right, mid);
}
static class Scanner {
BufferedReader in;
StringTokenizer st;
public Scanner() {
in = new BufferedReader(new InputStreamReader(System.in));
}
String next() throws IOException {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
String nextLine() throws IOException {
return in.readLine();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
double nextDouble() throws IOException {
return Double.parseDouble(next());
}
long nextLong() throws IOException {
return Long.parseLong(next());
}
void close() throws IOException {
in.close();
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | af50a05191e9153cd5cd5652b63a15bf | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
public class a {
public static void main(String[] args){
FastScanner sc = new FastScanner();
int t = sc.nextInt();
while(t-- > 0){
int n = sc.nextInt();
long a = sc.nextInt();
long b = sc.nextInt();
int arr[] = new int[n];
for(int i=0; i<n; i++){
arr[i] = sc.nextInt();
}
int selected[] = new int[n];
for(int i=0; i<n-1; i++){
if(a * (arr[i]-0) < (n-1-i)*b*(arr[i]-0)){
selected[i] = 1;
}
}
long ans = 0;
long cap = 0;
for(int i=0; i<n; i++){
if(selected[i] == 1){
ans += b*(arr[i]-cap);
ans += a*(arr[i]-cap);
cap = arr[i];
}
else{
ans += b*(arr[i]-cap);
}
}
System.out.println(ans);
}
}
}
class FastScanner
{
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | d6fe3d1334137792a9632ceff2e5b56e | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
//import java.math.BigInteger;
public class code{
public static class Pair{
int a;
int b;
Pair(int i,int j){
a=i;
b=j;
}
}
public static int GCD(int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
public static void shuffle(int a[], int n)
{
for (int i = 0; i < n; i++) {
// getting the random index
int t = (int)Math.random() * a.length;
// and swapping values a random index
// with the current index
int x = a[t];
a[t] = a[i];
a[i] = x;
}
}
public static PrintWriter out = new PrintWriter(System.out);
public static long[][] dp;
//@SuppressWarnings("unchecked")
public static void main(String[] arg) throws IOException{
//Reader in=new Reader();
//PrintWriter out = new PrintWriter(System.out);
//Scanner in = new Scanner(System.in);
FastScanner in=new FastScanner();
int t=in.nextInt();
while(t-- > 0){
int n=in.nextInt();
long a =in.nextLong();
long b=in.nextLong();
long[] arr=new long[n+1];
for(int i=1;i<=n;i++) arr[i]=in.nextLong();
for(int i=1;i<=n;i++) arr[i]+=arr[i-1];
long res=arr[n]*b;
for(int i=1;i<n;i++){
long d=arr[i]-arr[i-1];
long rem=n-i;
long val=b*((arr[n]-arr[i])-(rem*d))+d*(a+b);
//out.println(i+" "+val+" "+res);
res=Math.min(res,val);
}
out.println(res);
}
out.flush();
}
}
class Fenwick{
int[] bit;
public Fenwick(int n){
bit=new int[n];
//int sum=0;
}
public void update(int index,int val){
index++;
for(;index<bit.length;index += index&(-index)) bit[index]+=val;
}
public int presum(int index){
int sum=0;
for(;index>0;index-=index&(-index)) sum+=bit[index];
return sum;
}
public int sum(int l,int r){
return presum(r+1)-presum(l);
}
}
class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1) return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] nextInts(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC) c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-') neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32) c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32) return true;
while (true) {
c = getChar();
if (c == NC) return false;
else if (c > 32) return true;
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 71234fb1bb113d03f5d69025c5c09294 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 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.*;
/**
*
* @Har_Har_Mahadev
*/
/**
* Main , Solution , Remove Public
*/
public class Solution {
public static void process() throws IOException {
int n=sc.nextInt();
long c1=sc.nextInt(), c2=sc.nextInt();
int[] a=sc.readArray(n);
long ans=0;
for (int i=0; i<n; i++) {
long nToRight=n-i;
int last=i==0?0:a[i-1];
long dist=a[i]-last;
ans+=dist*min(nToRight*c2, c1+c2);
}
System.out.println(ans);
}
//=============================================================================
//--------------------------The End---------------------------------
//=============================================================================
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
// google(TTT++);
process();
}
out.flush();
// tr(System.currentTimeMillis()-s+"ms");
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
/*
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof Pair)) return false;
Pair key = (Pair) o;
return x == key.x && y == key.y;
}
@Override
public int hashCode() {
int result = x;
result = 31 * result + y;
return result;
}
*/
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
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];
}
//custom multiset (replace with HashMap if needed)
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
//map[k] += v;
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
//assumes map[k] >= v
//map[k] -= v
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
// compress Big value to Time Limit
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1; //min value
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
// Fast Writer
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static 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;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static 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;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
// if(x < 0){ x = 0; }
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
// Fast Inputs
static class FastScanner {
//I don't understand how this works lmao
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | c96ff85da2ac2c500b45f35311bf783b | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.lang.*;
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int a=sc.nextInt();
int b=sc.nextInt();
long arr[]=new long[n];
long sum=0;
for(int i=0;i<n;i++){
arr[i]=sc.nextLong();
sum+=arr[i];
}
long ans=sum*b;
long curr=0;
for(int i=1;i<=n;i++){
if(i==1){
curr=arr[i-1];
}
else curr=curr+arr[i-1]-arr[i-2];
sum=sum-arr[i-1];
ans=Math.min(ans,curr*a+curr*b+(sum-(n-i)*arr[i-1])*b);
}
System.out.println(ans);
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 21f0f716ce5b12e88a5dee9566632d2a | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | /*
I am dead inside
Do you like NCT, sKz, BTS?
5 4 3 2 1 Moonwalk
Imma knock it down like domino
Is this what you want? Is this what you want?
Let's ttalkbocky about that
*/
import static java.lang.Math.*;
import java.util.*;
import java.io.*;
import java.math.*;
public class NewTimeC
{
public static void main(String hi[]) throws Exception
{
BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(infile.readLine());
int T = Integer.parseInt(st.nextToken());
StringBuilder sb = new StringBuilder();
while(T-->0)
{
st = new StringTokenizer(infile.readLine());
int N = Integer.parseInt(st.nextToken());
long A = Integer.parseInt(st.nextToken());
long B = Integer.parseInt(st.nextToken());
int[] arr = readArr(N, infile, st);
long[] suffix = new long[N];
suffix[N-1] = arr[N-1];
for(int i=N-2; i >= 0; i--)
suffix[i] = suffix[i+1]+arr[i];
long res = B*suffix[0];
long temp = 0L;
long curr = 0L;
for(int i=0; i < N; i++)
{
temp += (A+B)*(arr[i]-curr);
curr = arr[i];
long val = 0L;
if(i+1 < N)
val = suffix[i+1]-(long)(N-i-1)*arr[i];
res = min(res, temp+B*val);
}
sb.append(res+"\n");
}
System.out.print(sb);
}
public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
{
int[] arr = new int[N];
st = new StringTokenizer(infile.readLine());
for(int i=0; i < N; i++)
arr[i] = Integer.parseInt(st.nextToken());
return arr;
}
}
/*
dp[y] = min time needed to capture suffix [y..n] if current capital is city y
dp[x] = min(b*[abs(c_b-c_{b+1})+abs(c_b-c_{b+2})+...+abs(c_b-c_y)]+a*(c_b-c_y)+dp[y])
for any y > x
is it ever optimal to go to y > x+1?
*/ | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | e285a2679f98896fea7de46da8d79dda | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main {
static int t;
static int n;
static int[] a;
static String s;
static FastReader fr = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
public static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
public static void main(String[] args) {
int t = fr.nextInt();
while ((t --) > 0) {
int n = fr.nextInt();
int a = fr.nextInt();
int b = fr.nextInt();
int[] x = fr.nextIntArray(n);
long[] sum = new long[n + 1];
for (int i = 0; i < n; i ++) sum[i + 1] = sum[i] + x[i];
long ans = Long.MAX_VALUE;
ans = 1l * sum[n] * b;
for (int i = 0; i < n; i ++) {
long costa = 1l * a * (x[i]);
long costb1 = 1l * b * (x[i]);
long costb2 = 1l * b * (sum[n] - sum[i + 1] - 1l * (n - i - 1) * x[i]);
ans = Math.min(ans, costb1 + costb2 + costa);
}
System.out.println(ans);
}
return;
}
static long ask(String op, int a, int b) {
System.out.println(op + " " + a + " " + b);
System.out.flush();
long gcd = fr.nextLong();
return gcd;
// return gcd(12354 + a, 12354 + b);
}
static class FastReader {
private BufferedReader bfr;
private StringTokenizer st;
public FastReader() {
bfr = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
if (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(bfr.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());
}
char nextChar() {
return next().toCharArray()[0];
}
String nextString() {
return next();
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
double[] nextDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextDouble();
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
int[][] nextIntGrid(int nL, int m) {
int[][] grid = new int[n][m];
for (int i = 0; i < n; i++) {
char[] line = fr.next().toCharArray();
for (int j = 0; j < m; j++)
grid[i][j] = line[j] - 48;
}
return grid;
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 0b6363d64ef1e7e05f9f0b7a63b22998 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Map.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
import static java.lang.System.*;
public class Main
{
public void tq() throws Exception
{
st=new StringTokenizer(bq.readLine());
int tq=i();
sb=new StringBuilder(2000000);
o:
while(tq-->0)
{
int n=i();
long a=l();
long b=l();
int ar[]=new int[n+1];
for(int x=1;x<=n;x++)ar[x]=i();
long c=0l;
for(int x:ar)c+=b*x;
long m=c;
for(int x=1;x<=n;x++)
{
c-=1l*(n-x)*b*(ar[x]-ar[x-1]);
c+=a*(ar[x]-ar[x-1]);
m=min(m,c);
}
pl(m);
}
p(sb);
}
long mod=1000000007l;
int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
BufferedReader bq = new BufferedReader(new InputStreamReader(in));StringTokenizer st;StringBuilder sb;
public static void main(String[] a)throws Exception{new Main().tq();}
int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
void f1(int ar[],int v){fill(ar,v);}
void f2(int ar[][],int v){for(int a[]:ar)fill(a,v);}
void f3(int ar[][][],int v){for(int a[][]:ar)for(int b[]:a)fill(b,v);}
void f1(long ar[],long v){fill(ar,v);}
void f2(long ar[][],int v){for(long a[]:ar)fill(a,v);}
void f3(long ar[][][],int v){for(long a[][]:ar)for(long b[]:a)fill(b,v);}
void f(){out.flush();}
int p(int i,int p[]){return p[i]<0?i:(p[i]=p(p[i],p));}
boolean c(int x,int y,int n,int m){return x>=0&&x<n&&y>=0&&y<m;}
int[] so(int ar[])
{
Integer r[] = new Integer[ar.length];
for (int x = 0; x < ar.length; x++) r[x] = ar[x];
sort(r);
for (int x = 0; x < ar.length; x++) ar[x] = r[x];
return ar;
}
long[] so(long ar[])
{
Long r[] = new Long[ar.length];
for (int x = 0; x < ar.length; x++) r[x] = ar[x];
sort(r);
for (int x = 0; x < ar.length; x++) ar[x] = r[x];
return ar;
}
char[] so(char ar[])
{
Character r[] = new Character[ar.length];
for (int x = 0; x < ar.length; x++) r[x] = ar[x];
sort(r);
for (int x = 0; x < ar.length; x++) ar[x] = r[x];
return ar;
}
void p(Object p) {out.print(p);}void pl(Object p) {out.println(p);}void pl() {out.println();}
void s(String s) {sb.append(s);}
void s(int s) {sb.append(s);}
void s(long s) {sb.append(s);}
void s(char s) {sb.append(s);}
void s(double s) {sb.append(s);}
void ss() {sb.append(' ');}
void sl(String s) {s(s);sb.append("\n");}
void sl(int s) {s(s);sb.append("\n");}
void sl(long s) {s(s);sb.append("\n");}
void sl(char s) {s(s);sb.append("\n");}
void sl(double s) {s(s);sb.append("\n");}
void sl() {sb.append("\n");}
int l(int v) {return 31 - Integer.numberOfLeadingZeros(v);}
long l(long v) {return 63 - Long.numberOfLeadingZeros(v);}
int sq(int a) {return (int) sqrt(a);}
long sq(long a) {return (long) sqrt(a);}
int gcd(int a, int b)
{
while (b > 0)
{
int c = a % b;
a = b;
b = c;
}
return a;
}
long gcd(long a, long b)
{
while (b > 0l)
{
long c = a % b;
a = b;
b = c;
}
return a;
}
boolean p(String s, int i, int j)
{
while (i < j) if (s.charAt(i++) != s.charAt(j--)) return false;
return true;
}
boolean[] si(int n)
{
boolean bo[] = new boolean[n + 1];
bo[0] = bo[1] = true;
for (int x = 4; x <= n; x += 2) bo[x] = true;
for (int x = 3; x * x <= n; x += 2)
{
if (!bo[x])
{
int vv = (x << 1);
for (int y = x * x; y <= n; y += vv) bo[y] = true;
}
}
return bo;
}
long mul(long a, long b, long m)
{
long r = 1l;
a %= m;
while (b > 0)
{
if ((b & 1) == 1) r = (r * a) % m;
b >>= 1;
a = (a * a) % m;
}
return r;
}
int i() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return Integer.parseInt(st.nextToken());
}
long l() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return Long.parseLong(st.nextToken());
}
String s() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return st.nextToken();
}
double d() throws IOException
{
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
return Double.parseDouble(st.nextToken());
}
void s(int a[])
{
for (int e : a){sb.append(e);sb.append(' ');}
sb.append("\n");
}
void s(long a[])
{
for (long e : a){sb.append(e);sb.append(' ');}
sb.append("\n");
}
void s(char a[])
{
for (char e : a){sb.append(e);sb.append(' ');}
sb.append("\n");
}
void s(int ar[][]) {for (int a[] : ar) s(a);}
void s(long ar[][]) {for (long a[] : ar) s(a);}
void s(char ar[][]) {for (char a[] : ar) s(a);}
int[] ari(int n) throws IOException
{
int ar[] = new int[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = Integer.parseInt(st.nextToken());
return ar;
}
long[] arl(int n) throws IOException
{
long ar[] = new long[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = Long.parseLong(st.nextToken());
return ar;
}
char[] arc(int n) throws IOException
{
char ar[] = new char[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = st.nextToken().charAt(0);
return ar;
}
double[] ard(int n) throws IOException
{
double ar[] = new double[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = Double.parseDouble(st.nextToken());
return ar;
}
String[] ars(int n) throws IOException
{
String ar[] = new String[n];
if (!st.hasMoreTokens()) st = new StringTokenizer(bq.readLine());
for (int x = 0; x < n; x++) ar[x] = st.nextToken();
return ar;
}
int[][] ari(int n, int m) throws IOException
{
int ar[][] = new int[n][m];
for (int x = 0; x < n; x++)ar[x]=ari(m);
return ar;
}
long[][] arl(int n, int m) throws IOException
{
long ar[][] = new long[n][m];
for (int x = 0; x < n; x++)ar[x]=arl(m);
return ar;
}
char[][] arc(int n, int m) throws IOException
{
char ar[][] = new char[n][m];
for (int x = 0; x < n; x++)ar[x]=arc(m);
return ar;
}
double[][] ard(int n, int m) throws IOException
{
double ar[][] = new double[n][m];
for (int x = 0; x < n; x++)ar[x]=ard(m);
return ar;
}
void p(int ar[])
{
sb = new StringBuilder(11 * ar.length);
for (int a : ar) {sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(long ar[])
{
StringBuilder sb = new StringBuilder(20 * ar.length);
for (long a : ar){sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(double ar[])
{
StringBuilder sb = new StringBuilder(22 * ar.length);
for (double a : ar){sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(char ar[])
{
StringBuilder sb = new StringBuilder(2 * ar.length);
for (char aa : ar){sb.append(aa);sb.append(' ');}
out.println(sb);
}
void p(String ar[])
{
int c = 0;
for (String s : ar) c += s.length() + 1;
StringBuilder sb = new StringBuilder(c);
for (String a : ar){sb.append(a);sb.append(' ');}
out.println(sb);
}
void p(int ar[][])
{
StringBuilder sb = new StringBuilder(11 * ar.length * ar[0].length);
for (int a[] : ar)
{
for (int aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void p(long ar[][])
{
StringBuilder sb = new StringBuilder(20 * ar.length * ar[0].length);
for (long a[] : ar)
{
for (long aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void p(double ar[][])
{
StringBuilder sb = new StringBuilder(22 * ar.length * ar[0].length);
for (double a[] : ar)
{
for (double aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void p(char ar[][])
{
StringBuilder sb = new StringBuilder(2 * ar.length * ar[0].length);
for (char a[] : ar)
{
for (char aa : a){sb.append(aa);sb.append(' ');}
sb.append("\n");
}
p(sb);
}
void pl(Object... ar) {for (Object e : ar) p(e + " ");pl();}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 427781624c3ce87514f1e8291466878c | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
//import javafx.util.*;
public class Main
{
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
static int INF = Integer.MAX_VALUE;
static int NINF = Integer.MIN_VALUE;
public static void main (String[] args) throws java.lang.Exception
{
//check if you have to take product or the constraints are big
int t = i();
while(t-- > 0){
int n = i();
long a = in.nextLong();
long b = in.nextLong();
long[] arr = new long[n + 1];
for(int i = 1;i < n + 1;i++){
arr[i] = in.nextLong();
}
long sum = 0l;
for(int i = 1;i < n + 1;i++){
sum += arr[i];
}
long ans = Long.MAX_VALUE;
for(int i = 0;i < n;i++){
long curr = a*arr[i] + b*arr[i];
sum = sum - arr[i];
long remaining = sum - arr[i]*(n - i);
curr += remaining * b;
ans = Math.min(ans,curr);
}
out.println(ans);
}
out.close();
}
public static int solve(String s){
int ans = -1;
int count = 0;
int n = s.length();
for(int i = 0;i < n;i++){
if(s.charAt(i) == 'R'){
count++;
}else{
ans = Math.max(ans,count);
count = 0;
}
}
ans = Math.max(ans,count);
return ans;
}
public static void sort(int[] arr){
ArrayList<Integer> ls = new ArrayList<>();
for(int x : arr){
ls.add(x);
}
Collections.sort(ls);
for(int i = 0;i < arr.length;i++){
arr[i] = ls.get(i);
}
}
static int i()
{
return in.nextInt();
}
static long l()
{
return in.nextLong();
}
static int[] input(int N){
int A[]=new int[N];
for(int i=0; i<N; i++)
{
A[i]=in.nextInt();
}
return A;
}
static long[] inputLong(int N) {
long A[]=new long[N];
for(int i=0; i<A.length; i++)A[i]=in.nextLong();
return A;
}
}
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 obj)
{
// we sort objects on the basis of Student Id
return (this.x - obj.x);
}
}
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 | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 22641c1b0e548a5436d6c6a8226dfbef | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class C782 {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
solver.solve(in, out);
out.close();
}
private static class Solver {
void solve(InputReader in, PrintWriter out) {
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int a = in.nextInt();
int b = in.nextInt();
long[] x = new long[n];
for (int i = 0; i < n; i++) {
x[i] = in.nextLong();
}
// A 移动首都
// B 攻占城市
// 当前首都的绝对位置
long current = 0;
// 已攻占城市坐标
int vis = -1;
long ans = 0;
for (int i = 0; i < n; i++) {
// 直接攻占的代价
long directCost = (x[i] - current) * b;
// 先移动再攻占的代价
long sumCost = Long.MAX_VALUE;
long sumCost2 = Long.MAX_VALUE;
if (vis != -1) {
// 移动的代价
long costA = (x[vis] - current) * a;
// 移动为未来节省的代价
long saveB = (x[vis] - current) * (n - 1 - i) * b;
// 攻占i的代价
long costB = (x[i] - x[vis]) * b;
sumCost = costA + costB;
sumCost2 = sumCost - saveB;
}
if (sumCost2 <= directCost) {
current = x[vis];
ans += sumCost;
} else {
ans += directCost;
}
vis = i;
}
out.println(ans);
}
}
}
private static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | a18e0aa592da6c96d5ed2f8a25e0207f | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 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.StringTokenizer;
/**
*
* @author eslam
*/
public class IceCave {
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 FastReader input = new FastReader();
static BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
public static void main(String[] args) throws IOException {
int t = input.nextInt();
while (t-- > 0) {
int n = input.nextInt();
long a = input.nextInt();
long b = input.nextInt();
long ar[] = new long[n + 1];
long br[] = new long[n + 1];
long sum = 0;
long ans = Long.MAX_VALUE;
long s = 0;
for (int i = 1; i < ar.length; i++) {
ar[i] = input.nextInt();
sum += ar[i];
}
for (int i = 1; i < ar.length; i++) {
br[i] = (ar[i] - ar[i - 1]) * b;
}
for (int i = 1; i < br.length; i++) {
br[i] = br[i] + br[i - 1];
}
for (int i = 0; i < n; i++) {
long c = ((sum - ar[i] * (n - i)));
long d = br[i] + c * b;
ans = Math.min(ans, d + s);
sum -= ar[i + 1];
s += (ar[i + 1] - ar[i]) * a;
}
log.write(ans + "\n");
}
log.flush();
}
public static long binaryToDecimal(String w) {
long r = 0;
long l = 0;
for (int i = w.length() - 1; i > -1; i--) {
long x = (w.charAt(i) - '0') * (long) Math.pow(2, l);
r = r + x;
l++;
}
return r;
}
public static String decimalToBinary(long n) {
String w = "";
while (n > 0) {
w = n % 2 + w;
n /= 2;
}
return w;
}
public static boolean isSorted(int[] a) {
for (int i = 0; i < a.length - 1; i++) {
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
public static void print(int a[]) throws IOException {
for (int i = 0; i < a.length; i++) {
log.write(a[i] + " ");
}
log.write("\n");
}
public static void read(long[] a) {
for (int i = 1; i < a.length; i++) {
a[i] = input.nextInt();
}
}
static class pair {
int x;
int y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
}
public static long LCM(long x, long y) {
return (x * y) / GCD(x, y);
}
public static long GCD(long x, long y) {
if (y == 0) {
return x;
}
return GCD(y, x % y);
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | e3a9e12dab18f39faf3ce13369083c9a | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes |
import static java.lang.Math.min;
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;
// snippet
public class CF782C {
// Question Link :
static FastScanner sc=new FastScanner();
static PrintWriter out=new PrintWriter(System.out);
public static void main(String[] args) {
int T=sc.nextInt();
for (int tt=0; tt<T; tt++) {
solve();
}
}
private static void solve() {
// TODO Auto-generated method stub
int n = sc.nextInt()+1;
int a = sc.nextInt();
int b = sc.nextInt();
int arr[] = new int[n];
long sum =0;
for(int i=1;i<arr.length;i++) {
arr[i] = sc.nextInt();
sum+=arr[i];
}
//System.out.println(sum);
long res = Long.MAX_VALUE;
for(int i=0;i<arr.length;i++) {
long curr = (a+b)*1L*(long)arr[i];
//System.out.println(curr+" .");
sum-=arr[i];
// System.out.println(sum+" ..");
// System.out.println(sum-(n-i-1)+" ...");
curr+=(sum-(n-i-1)*(long)arr[i])*b;
// System.out.println(curr);
res = Math.min(res,curr);
// System.out.println(curr+" ....");
}
System.out.println(res);
}
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\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 2e2f94c457625037eb12088b597314e3 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static long factorial(int number) {
long f = 1;
int j = 1;
while (j <= number) {
f = (f * j) % 998244353;
j++;
}
return f;
}
public static long combination(int n, int r) {
return factorial(n) / (factorial(n - r) * factorial(r));
}
public static void main(String[] args) throws Exception {
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int a = sc.nextInt();
int b = sc.nextInt();
n++;
int[]x = new int[n];
long sum=0;
for(int i=1;i<n;i++) {
x[i]=sc.nextInt();
sum+=x[i];
}
long ans = (long)1e18;
for(int i=0;i<n;i++) {
long tmp = (a+b)*1l*x[i];
sum-=x[i];
tmp+=(sum-(n-i-1)*(1l)*x[i])*b;
ans = Math.min(ans, tmp);
}
pw.println(ans);
}
pw.flush();
}
static class SegmentTree {
int[] arr, sTree, lazy;
int N;
public SegmentTree(int[] in) {
arr = in;
N = in.length - 1;
sTree = new int[2 * N];
lazy = new int[2 * N];
build(1, 1, N);
}
public void build(int Node, int left, int right) { // O(n)
if (left == right)
sTree[Node] = arr[left];
else {
int r = 2 * Node + 1;
int l = 2 * Node;
int mid = (left + right) / 2;
build(l, left, mid);
build(r, mid + 1, right);
sTree[Node] = sTree[l] + sTree[r];
}
}
public int query(int i, int j) {
return query(1, 1, N, i, j);
}
public int query(int Node, int left, int right, int i, int j) {
if (right < i || left > j)
return 0;
if (right <= j && i <= left) {
return sTree[Node];
}
int mid = (left + right) / 2;
int l = 2 * Node;
int r = 2 * Node + 1;
return query(l, left, mid, i, j) + query(r, mid + 1, right, i, j);
}
public void updatePoint(int idx, int value) {
int node = idx + N - 1;
arr[idx] = value;
sTree[node] = value;
while (node > 1) {
node /= 2;
int l = 2 * node;
int r = 2 * node + 1;
sTree[node] = sTree[l] + sTree[r];
}
}
public void updateRange(int i, int j, int val) {
updateRange(1, 1, N, i, j, val);
}
public void updateRange(int Node, int left, int right, int i, int j, int val) {
if (i > right || left > j)
return;
if (right <= j && i <= left) {
sTree[Node] += val * (right - left + 1);
lazy[Node] += val;
} else {
int l = 2 * Node;
int r = 2 * Node + 1;
int mid = (left + right) / 2;
propagate(Node, left, right);
updateRange(l, left, mid, i, j, val);
updateRange(r, mid + 1, right, mid, j, val);
sTree[Node] = sTree[l] + sTree[r];
}
}
public void propagate(int node, int left, int right) {
int l = 2 * node;
int r = 2 * node + 1;
int mid = (left + right) / 2;
lazy[l] = lazy[node];
lazy[r] = lazy[node];
sTree[l] += lazy[node] * (mid - left + 1);
sTree[r] += lazy[node] * (right - mid);
lazy[node] = 0;
}
}
public static void sort(int[] in) {
shuffle(in);
Arrays.sort(in);
}
public static void shuffle(int[] in) {
for (int i = 0; i < in.length; i++) {
int idx = (int) (Math.random() * in.length);
int tmp = in[i];
in[i] = in[idx];
in[idx] = tmp;
}
}
static class Pair implements Comparable<Pair> {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o) {
return x - o.x;
}
public String toString() {
return x + " " + y;
}
}
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
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();
}
}
public static void display(char[][] a) {
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++) {
System.out.print(a[i][j]);
}
System.out.println();
}
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 802f6b8ddf0133e0270f97a7bf21eb31 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import com.sun.security.jgss.GSSUtil;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class CF1{
public static void main(String[] args) {
FastScanner sc=new FastScanner();
int T=sc.nextInt();
// int T=1;
for (int tt=1; tt<=T; tt++){
long n = sc.nextLong();
long a = sc.nextLong();
long b= sc.nextLong();
long x[]= new long[(int) n+1];
long arr[]= new long[(int)n+1];
long sum=0;
for (int i=1; i<=n; i++){
arr[i]= sc.nextLong();
sum+=arr[i];
}
long ans=0;
int pos=0;
for (int i=1; i<=n; i++){
x[i]=arr[i-1]+x[i-1];
}
for (int i=1; i<=n; i++){
ans+=b*(arr[i]-arr[pos]);
long y=((x[(int)n]-x[i])-arr[pos]*(n-i))*b;
long z = a*(arr[i]-arr[pos])+(x[(int)n]-x[i]-arr[i]*(n-i))*b;
if (y>z){
ans+=a*(arr[i]-arr[pos]);
pos=i;
}
}
System.out.println(ans);
}
}
static class LPair{
long x,y;
LPair(long x , long y){
this.x=x;
this.y=y;
}
}
static long prime(long n){
for (long i=3; i*i<=n; i+=2){
if (n%i==0) return i;
}
return -1;
}
static long factorial (int x){
if (x==0) return 1;
long ans =x;
for (int i=x-1; i>=1; i--){
ans*=i;
ans%=mod;
}
return ans;
}
static long mod =1000000007L;
static long power2 (long a, long b){
long res=1;
while (b>0){
if ((b&1)== 1){
res= (res * a % mod)%mod;
}
a=(a%mod * a%mod)%mod;
b=b>>1;
}
return res;
}
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;
}
}
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 void sortLong(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 long gcd (long n, long m){
if (m==0) return n;
else return gcd(m, n%m);
}
static class Pair implements Comparable<Pair>{
int x,y;
private static final int hashMultiplier = BigInteger.valueOf(new Random().nextInt(1000) + 100).nextProbablePrime().intValue();
public Pair(int x, int y){
this.x = x;
this.y = y;
}
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair pii = (Pair) o;
if (x != pii.x) return false;
return y == pii.y;
}
public int hashCode() {
return hashMultiplier * x + y;
}
public int compareTo(Pair o){
if (this.x==o.x) return Integer.compare(this.y,o.y);
else return Integer.compare(this.x,o.x);
}
// this.x-o.x is ascending
}
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\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | bebc5674a182d09a73f2801c4f633cdf | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
// BEFORE 31ST MARCH 2022 !!
//MAX RATING EVER ACHIEVED-1622(LETS SEE WHEN WILL I GET TO CHANGE THIS)
////***************************************************************************
/* public class E_Gardener_and_Tree implements Runnable{
public static void main(String[] args) throws Exception {
new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
}
public void run(){
WRITE YOUR CODE HERE!!!!
JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
}
}
*/
/////**************************************************************************
public class C_Line_Empire{
public static void main(String[] args) {
FastScanner s= new FastScanner();
//PrintWriter out=new PrintWriter(System.out);
//end of program
//out.println(answer);
//out.close();
StringBuilder res = new StringBuilder();
int t=s.nextInt();
int p=0;
while(p<t){
int n=s.nextInt();
long a=s.nextLong();
long b=s.nextLong();
n++;
long array[]= new long[n];
for(int i=1;i<n;i++){
array[i]=s.nextLong();
}
long nice[]= new long[n];
nice[n-1]=0;
for(int i=n-2;i>=0;i--){
long hh=(array[i+1]-array[i]);
long yo=n-i-1;
hh=(hh*yo);
long yoyo=hh+nice[i+1];
nice[i]=yoyo;
// System.out.print("nn "+nice[i]);
}
long ans=Long.MAX_VALUE;
long till=0;
for(int i=0;i<n;i++){
long yoyo=b*nice[i];
long yoyo2=till+yoyo;
ans=Math.min(ans,yoyo2);
if(i+1<n){
long hh=array[i+1]-array[i];
hh=(hh*(a+b));
till+=hh;
}
}
res.append(ans+" \n");
p++;
}
System.out.println(res);
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(String s) {
try {
br = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public FastScanner() {
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();
}
int nextInt() {
return Integer.parseInt(nextToken());
}
long nextLong() {
return Long.parseLong(nextToken());
}
double nextDouble() {
return Double.parseDouble(nextToken());
}
}
static long modpower(long x, long y, long p)
{
long res = 1; // Initialize result
x = x % p; // 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) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// SIMPLE POWER FUNCTION=>
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;
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 97130245cb87b2ec154b321818df76a3 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
/*
*/
public class C {
public static void main(String[] args) {
FastReader sc = new FastReader();
int t = sc.nextInt();
while (t-- > 0) {
int n, a, b;
n = sc.nextInt() + 1;
a = sc.nextInt();
b = sc.nextInt();
long aa[] = new long[n];
for (int i = 1; i < n; i++)
aa[i] = sc.nextLong();
long sum = 0;
for (int i = 1; i < n; ++i) {
sum += aa[i];
}
long ans = help(aa,n,a,b,sum);
System.out.println(ans);
}
}
public static long help(long[]aa,int n,int a,int b,long sum)
{
long ans = Long.MAX_VALUE;
for (int i = 0; i < n; ++i) {
sum -= aa[i];
long now = (a + b) * aa[i];
now = now + (sum - (n - i - 1) * aa[i]) * b;
ans = min(ans, now);
}
return ans;
}
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] = 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();
}
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 reverse(int[] a) {
for (int i = 0; i < a.length / 2; i++) {
int temp = a[i];
a[i] = a[a.length - i - 1];
a[a.length - i - 1] = temp;
}
}
static void debug(int[] arr) {
for (int i = 0; i < arr.length; ++i)
System.out.print(arr[i] + " ");
System.out.println();
}
static void debug(int[][] arr) {
for (int i = 0; i < arr.length; ++i) {
for (int j = 0; j < arr[0].length; ++j)
System.out.print(arr[i][j] + " ");
System.out.println();
}
System.out.println();
}
static void debug(long[] arr) {
for (int i = 0; i < arr.length; ++i)
System.out.print(arr[i] + " ");
System.out.println();
}
static void debug(long[][] arr) {
for (int i = 0; i < arr.length; ++i) {
for (int j = 0; j < arr[0].length; ++j)
System.out.print(arr[i][j] + " ");
System.out.println();
}
System.out.println();
}
static int MOD = (int) (1e9 + 7);
static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static void debug(ArrayList<Integer> arr) {
for (int i = 0; i < arr.size(); i++) {
System.out.print(arr.get(i) + " ");
}
System.out.println();
}
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] == true) {
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++) {
// if (prime[i] == true)
// add it to HashSet
}
}
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 boolean isPalindrome(String s) {
for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
if (s.charAt(i) != s.charAt(j))
return false;
}
return true;
}
/*
*
* if RTE inside for loop: - check for increment/decrement in loop, (i++,i--) if
* after custom sorting an array using lamda , numbers are messed up - check the
* order in which input is taken
*
*/
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 983fb4fde74d5fe787ac3e6dfecac2de | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//--------------------------INPUT READER---------------------------------//
static class fs {
public BufferedReader br;
StringTokenizer st = new StringTokenizer("");
public fs() { this(System.in); }
public fs(InputStream is) {
br = new BufferedReader(new InputStreamReader(is));
}
String next() {
while (!st.hasMoreTokens()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String ns() { return next(); }
int[] na(long nn) {
int n = (int) nn;
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
long[] nal(long nn) {
int n = (int) nn;
long[] l = new long[n];
for(int i = 0; i < n; i++) l[i] = nl();
return l;
}
}
//-----------------------------------------------------------------------//
//---------------------------PRINTER-------------------------------------//
static class Printer {
static PrintWriter w;
public Printer() {this(System.out);}
public Printer(OutputStream os) {
w = new PrintWriter(os);
}
public void p(int i) {w.println(i);}
public void p(long l) {w.println(l);}
public void p(double d) {w.println(d);}
public void p(String s) { w.println(s);}
public void pr(int i) {w.print(i);}
public void pr(long l) {w.print(l);}
public void pr(double d) {w.print(d);}
public void pr(String s) { w.print(s);}
public void pl() {w.println();}
public void close() {w.close();}
}
//-----------------------------------------------------------------------//
//--------------------------VARIABLES------------------------------------//
static fs sc = new fs();
static OutputStream outputStream = System.out;
static Printer w = new Printer(outputStream);
static long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
static int ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
static long mod = 1000000007;
//-----------------------------------------------------------------------//
//--------------------------ADMIN_MODE-----------------------------------//
private static void ADMIN_MODE() throws IOException {
if (System.getProperty("ONLINE_JUDGE") == null) {
w = new Printer(new FileOutputStream("output.txt"));
sc = new fs(new FileInputStream("input.txt"));
}
}
//-----------------------------------------------------------------------//
//----------------------------START--------------------------------------//
public static void main(String[] args)
throws IOException {
ADMIN_MODE();
int t = sc.ni();while(t-->0)
solve();
w.close();
}
static void solve() throws IOException {
int n = sc.ni();
long a = sc.ni(), b = sc.ni();
long[] arr = new long[n+1];
for(int i = 1; i <= n; i++) {
arr[i] = sc.nl();
}
long[] suff = new long[n+1];
suff[n] = arr[n];
for(int i = n-1; i > 0; i--) {
suff[i] = suff[i+1]+arr[i];
}
int at = 0, target = 1, check = 0;
long cost = 0;
while(target <= n) {
int numberLeft = n - target + 1;
long withoutDistance, withoutCost;
// without
withoutDistance = suff[target] - (numberLeft * arr[at]);
withoutCost = withoutDistance * b;
// with
long withDistance, withCost;
withDistance = suff[target] - (numberLeft * arr[check]);
withCost = withDistance * b + ((arr[check]-arr[at]) * a);
if(withoutCost > withCost) {
cost += (arr[check]-arr[at]) * a;
at = check;
}
long distance = arr[target] - arr[at];
long currCost = distance * b;
cost += currCost;
check = target;
target++;
}
w.p(cost);
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 74b929ae2c1b8de6742f1327acf6bf29 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.math.BigInteger;
import static java.lang.System.out;
import static java.lang.Math.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
Read in = new Read(System.in);
int tN = in.nextInt();
while(tN > 0) {
tN--;
int n=in.nextInt();
long a = in.nextInt();
long b = in.nextInt();
int[] w = new int [n+1];
for(int i = 1; i <= n;i++) {
w[i] = in.nextInt();
}
long ans = b*w[1];
boolean f = true;
int now = 0;
for(int i = 2;i <= n;i++) {
if( a*(w[i-1]-now) >= b*(w[i-1]-now)*(n-i+1)) {
f=false;
}
if(f) {
ans+=a*(w[i-1]-now) ;
now = w[i-1];
}
ans +=b*(w[i]-now);
}
out.println(ans);
}
}
static class Read {//自定义快读 Read
public BufferedReader reader;
public StringTokenizer tokenizer;
public Read(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 String nextLine() {
String str = null;
try {
str = reader.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public Double nextDouble() {
return Double.parseDouble(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long gcd(long a, long b) {
return (a % b == 0) ? b : gcd(b, a % b);
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | f19323530b43ea9999ca357a3563098e | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String args[])
{
FastReader input=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int T=input.nextInt();
while(T-->0)
{
int n=input.nextInt();
int a=input.nextInt();
int b=input.nextInt();
int arr[]=new int[n+1];
for(int i=1;i<=n;i++)
{
int x=input.nextInt();
arr[i]=x;
}
long sufSum[]=new long[n+1];
long sum=0;
for(int i=n;i>=0;i--)
{
sum+=arr[i];
sufSum[i]=sum;
}
long min=Long.MAX_VALUE;
for(int i=0;i<=n-1;i++)
{
long val=0;
val+=(long)(a+b)*(long)arr[i];
long v1=sufSum[i+1];
v1-=(long)(n-i)*(long)(arr[i]);
v1*=b;
val+=v1;
min=Math.min(min,val);
}
out.println(min);
}
out.close();
}
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 | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 0cb73e2cbfa13d2827cedd09ac601dfa | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
/**
* @author Naitik
*
*/
public class A
{
static FastReader sc=new FastReader();
static long dp[][][];
static int mod=998244353;//1000000007;
// static int mod=1000000009;
static int max;
static int bit[];
static long ans;
static long seg[];
static long A[];
static HashMap<Integer,Integer> map;
static ArrayList<Integer> l=new ArrayList<Integer>();
public static void main(String[] args)
{
//CHECK FOR N=1
//CHECK FOR N=1
PrintWriter out=new PrintWriter(System.out);
//StringBuffer sb=new StringBuffer("");
int ttt=1;
ttt =i();
outer :while (ttt-- > 0)
{
int n=i();
long a=i(),b=i();
long A[]=inputL(n);
long S[]=suffix(A);
long ans=0;
ArrayList<Long> l=new ArrayList<Long>();
long pv=0;
long curr=0;
long left=0;
for(int i=0;i<n;i++) {
long len=n-i;
long y=b*(S[i]-len*pv)+curr+left;
l.add(y);
pv=A[i];
left+=b*(A[i]-(i>0?A[i-1]:0));
curr=a*A[i];
}
l.sort(null);
System.out.println(l.get(0));
}
out.close();
// System.out.println(sb.toString());
//CHECK FOR N=1 //CHECK FOR M=0
//CHECK FOR N=1 //CHECK FOR M=0
}
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) {
if(this.x>o.x)
return 1;
else if(this.x<o.x)
return -1;
else {
if(this.y>o.y)
return -1;
else if(this.y<o.y)
return 1;
else
return 0;
}
}
/* FOR TREE MAP PAIR USE */
// public int compareTo(Pair o) {
// if (x > o.x) {
// return 1;
// }
// if (x < o.x) {
// return -1;
// }
// if (y > o.y) {
// return 1;
// }
// if (y < o.y) {
// return -1;
// }
// return 0;
// }
// public int hashCode()
// {
// final int temp = 14;
// int ans = 1;
// ans =x*31+y*13;
// return ans;
// }
//
// // Equal objects must produce the same
// // hash code as long as they are equal
// @Override
// public boolean equals(Object o)
// {
// if (this == o) {
// return true;
// }
// if (o == null) {
// return false;
// }
// if (this.getClass() != o.getClass()) {
// return false;
// }
// Pair other = (Pair)o;
// if (this.x != other.x || this.y!=other.y) {
// return false;
// }
// return true;
// }
}
//FRENWICK TREE
static void update(int i, int x){
for(; i < bit.length; i += (i&-i))
bit[i] += x;
}
static int sum(int i){
int ans = 0;
for(; i > 0; i -= (i&-i))
ans += bit[i];
return ans;
}
//END
static void add(int v) {
if(!map.containsKey(v)) {
map.put(v, 1);
}
else {
map.put(v, map.get(v)+1);
}
}
static void remove(int v) {
if(map.containsKey(v)) {
map.put(v, map.get(v)-1);
if(map.get(v)==0)
map.remove(v);
}
}
static int[] copy(int A[]) {
int B[]=new int[A.length];
for(int i=0;i<A.length;i++) {
B[i]=A[i];
}
return B;
}
static long[] copy(long A[]) {
long B[]=new long[A.length];
for(int i=0;i<A.length;i++) {
B[i]=A[i];
}
return B;
}
static int[] input(int n) {
int A[]=new int[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextInt();
}
return A;
}
static long[] inputL(int n) {
long A[]=new long[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextLong();
}
return A;
}
static String[] inputS(int n) {
String A[]=new String[n];
for(int i=0;i<n;i++) {
A[i]=sc.next();
}
return A;
}
static long sum(int A[]) {
long sum=0;
for(int i : A) {
sum+=i;
}
return sum;
}
static long sum(long A[]) {
long sum=0;
for(long i : A) {
sum+=i;
}
return sum;
}
static void reverse(long A[]) {
int n=A.length;
long B[]=new long[n];
for(int i=0;i<n;i++) {
B[i]=A[n-i-1];
}
for(int i=0;i<n;i++)
A[i]=B[i];
}
static void reverse(int A[]) {
int n=A.length;
int B[]=new int[n];
for(int i=0;i<n;i++) {
B[i]=A[n-i-1];
}
for(int i=0;i<n;i++)
A[i]=B[i];
}
static void input(int A[],int B[]) {
for(int i=0;i<A.length;i++) {
A[i]=sc.nextInt();
B[i]=sc.nextInt();
}
}
static int[][] input(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]=i();
}
}
return A;
}
static char[][] charinput(int n,int m){
char A[][]=new char[n][m];
for(int i=0;i<n;i++) {
String s=s();
for(int j=0;j<m;j++) {
A[i][j]=s.charAt(j);
}
}
return A;
}
static int find(int A[],int a) {
if(A[a]==a)
return a;
return A[a]=find(A, A[a]);
}
static int max(int A[]) {
int max=Integer.MIN_VALUE;
for(int i=0;i<A.length;i++) {
max=Math.max(max, A[i]);
}
return max;
}
static int min(int A[]) {
int min=Integer.MAX_VALUE;
for(int i=0;i<A.length;i++) {
min=Math.min(min, A[i]);
}
return min;
}
static long max(long A[]) {
long max=Long.MIN_VALUE;
for(int i=0;i<A.length;i++) {
max=Math.max(max, A[i]);
}
return max;
}
static long min(long A[]) {
long min=Long.MAX_VALUE;
for(int i=0;i<A.length;i++) {
min=Math.min(min, A[i]);
}
return min;
}
static long [] prefix(long A[]) {
long p[]=new long[A.length];
p[0]=A[0];
for(int i=1;i<A.length;i++)
p[i]=p[i-1]+A[i];
return p;
}
static long [] prefix(int A[]) {
long p[]=new long[A.length];
p[0]=A[0];
for(int i=1;i<A.length;i++)
p[i]=p[i-1]+A[i];
return p;
}
static long [] suffix(long A[]) {
long p[]=new long[A.length];
p[A.length-1]=A[A.length-1];
for(int i=A.length-2;i>=0;i--)
p[i]=p[i+1]+A[i];
return p;
}
static long [] suffix(int A[]) {
long p[]=new long[A.length];
p[A.length-1]=A[A.length-1];
for(int i=A.length-2;i>=0;i--)
p[i]=p[i+1]+A[i];
return p;
}
static void fill(int dp[]) {
Arrays.fill(dp, -1);
}
static void fill(int dp[][]) {
for(int i=0;i<dp.length;i++)
Arrays.fill(dp[i], -1);
}
static void fill(int dp[][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
Arrays.fill(dp[i][j],-1);
}
}
}
static void fill(int dp[][][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
for(int k=0;k<dp[0][0].length;k++) {
Arrays.fill(dp[i][j][k],-1);
}
}
}
}
static void fill(long dp[]) {
Arrays.fill(dp, -1);
}
static void fill(long dp[][]) {
for(int i=0;i<dp.length;i++)
Arrays.fill(dp[i], -1);
}
static void fill(long dp[][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
Arrays.fill(dp[i][j],-1);
}
}
}
static void fill(long dp[][][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
for(int k=0;k<dp[0][0].length;k++) {
Arrays.fill(dp[i][j][k],-1);
}
}
}
}
static int min(int a,int b) {
return Math.min(a, b);
}
static int min(int a,int b,int c) {
return Math.min(a, Math.min(b, c));
}
static int min(int a,int b,int c,int d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static int max(int a,int b) {
return Math.max(a, b);
}
static int max(int a,int b,int c) {
return Math.max(a, Math.max(b, c));
}
static int max(int a,int b,int c,int d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long min(long a,long b) {
return Math.min(a, b);
}
static long min(long a,long b,long c) {
return Math.min(a, Math.min(b, c));
}
static long min(long a,long b,long c,long d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static long max(long a,long b) {
return Math.max(a, b);
}
static long max(long a,long b,long c) {
return Math.max(a, Math.max(b, c));
}
static long max(long a,long b,long c,long d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long power(long x, long y, long p)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static void print(int A[]) {
for(int i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long A[]) {
for(long i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static long mod(long x) {
return ((x%mod + mod)%mod);
}
static String reverse(String s) {
StringBuffer p=new StringBuffer(s);
p.reverse();
return p.toString();
}
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
static void sort(int[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static void sort(long[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static String sort(String s) {
Character ch[]=new Character[s.length()];
for(int i=0;i<s.length();i++) {
ch[i]=s.charAt(i);
}
Arrays.sort(ch);
StringBuffer st=new StringBuffer("");
for(int i=0;i<s.length();i++) {
st.append(ch[i]);
}
return st.toString();
}
static HashMap<Integer,Integer> hash(int A[]){
HashMap<Integer,Integer> map=new HashMap<Integer, Integer>();
for(int i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static TreeMap<Integer,Integer> tree(int A[]){
TreeMap<Integer,Integer> map=new TreeMap<Integer, Integer>();
for(int i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static boolean prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean prime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, 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;
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | c37f88934b7d7db20c69761e0bca17d2 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long startTime = System.currentTimeMillis();
// for global initializations and methods starts here
// global initialisations and methods end here
static void run() {
boolean tc = true;
AdityaFastIO r = new AdityaFastIO();
//FastReader r = new FastReader();
try (OutputStream out = new BufferedOutputStream(System.out)) {
//long startTime = System.currentTimeMillis();
int testcases = tc ? r.ni() : 1;
int tcCounter = 1;
// Hold Here Sparky------------------->>>
// Solution Starts Here
start:
while (testcases-- > 0) {
int n = r.ni();
long a = r.nl();
long b = r.nl();
List<Long> al = new ArrayList<>();
al.add(0L);
for (int i = 0; i < n; i++) al.add(r.nl());
//out.write((al + " ").getBytes());
long sum = 0;
for (long ele : al) sum += ele;
long ans = Long.MAX_VALUE;
for (int i = 0; i < n + 1; i++) {
long get = al.get(i);
long ele = (a + b) * get;
sum -= get;
long can = sum - (n - i) * get;
ele += can * b;
ans = Math.min(ans, ele);
}
out.write((ans + " ").getBytes());
out.write(("\n").getBytes());
}
// Solution Ends Here
} catch (IOException e) {
e.printStackTrace();
}
}
static class AdityaFastIO {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public BufferedReader br;
public StringTokenizer st;
public AdityaFastIO() {
br = new BufferedReader(new InputStreamReader(System.in));
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public AdityaFastIO(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public String readLine() throws IOException {
byte[] buf = new byte[100000001]; // 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 ni() 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 nl() 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 nd() 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 {
run();
}
static int[] readIntArr(int n, AdityaFastIO r) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = r.ni();
return arr;
}
static long[] readLongArr(int n, AdityaFastIO r) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = r.nl();
return arr;
}
static List<Integer> readIntList(int n, AdityaFastIO r) throws IOException {
List<Integer> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.ni());
return al;
}
static List<Long> readLongList(int n, AdityaFastIO r) throws IOException {
List<Long> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.nl());
return al;
}
static long mod = 998244353;
static long modInv(long base, long e) {
long result = 1;
base %= mod;
while (e > 0) {
if ((e & 1) > 0) result = result * base % mod;
base = base * base % mod;
e >>= 1;
}
return result;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int ni() {
return Integer.parseInt(word());
}
long nl() {
return Long.parseLong(word());
}
double nd() {
return Double.parseDouble(word());
}
}
static int MOD = (int) (1e9 + 7);
static long powerLL(long x, long n) {
long result = 1;
while (n > 0) {
if (n % 2 == 1) result = result * x % MOD;
n = n / 2;
x = x * x % MOD;
}
return result;
}
static long powerStrings(int i1, int i2) {
String sa = String.valueOf(i1);
String sb = String.valueOf(i2);
long a = 0, b = 0;
for (int i = 0; i < sa.length(); i++) a = (a * 10 + (sa.charAt(i) - '0')) % MOD;
for (int i = 0; i < sb.length(); i++) b = (b * 10 + (sb.charAt(i) - '0')) % (MOD - 1);
return powerLL(a, b);
}
static long gcd(long a, long b) {
if (a == 0) return b;
else return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long lower_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] >= x) r = m;
else l = m;
}
return r;
}
static int upper_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] <= x) l = m;
else r = m;
}
return l + 1;
}
static void addEdge(ArrayList<ArrayList<Integer>> graph, int edge1, int edge2) {
graph.get(edge1).add(edge2);
graph.get(edge2).add(edge1);
}
public static class Pair implements Comparable<Pair> {
int first;
int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(Pair o) {
// TODO Auto-generated method stub
if (this.first != o.first)
return (int) (this.first - o.first);
else return (int) (this.second - o.second);
}
}
public static class PairC<X, Y> implements Comparable<PairC> {
X first;
Y second;
public PairC(X first, Y second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(PairC o) {
// TODO Auto-generated method stub
return o.compareTo((PairC) first);
}
}
static boolean isCollectionsSorted(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) <= list.get(i - 1)) return false;
return true;
}
static boolean isCollectionsSortedReverseOrder(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) >= list.get(i - 1)) return false;
return true;
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 7d582f191994872b66f1c1c367ba62c5 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) {
InputReader in = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
for (int ca = 1; ca <= t; ca++) {
int n = in.nextInt(), a = in.nextInt(), b = in.nextInt();
int[] kingdoms = new int[n + 10];
long sum = 0;
for (int i = 1; i <= n; i++) {
kingdoms[i] = in.nextInt();
sum += ((long) kingdoms[i] * b);
}
long result = sum;
for (int i = 1; i <= n; i++) {
int diff = kingdoms[i] - kingdoms[i - 1];
sum -= (long) diff * (n - i) * b;
sum += (long) diff * a;
result = Math.min(result, sum);
}
out.println(result);
}
out.close();
}
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());
}
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 8df491eec3870bdc75623b25f4b0cf17 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 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.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);
CLineEmpire solver = new CLineEmpire();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class CLineEmpire {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
long a = in.nextLong();
long b = in.nextLong();
long ar[] = new long[n + 1];
for (int i = 1; i <= n; i++) {
ar[i] = in.nextInt();
}
long min = Long.MAX_VALUE;
long cost = 0;
for (int i = n; i >= 1; i--) {
long ans = (a + b) * ar[i];
min = Math.min(min, ans + cost * b);
cost = cost + (ar[i] - ar[i - 1]) * (n - i + 1);
}
min = Math.min(min, cost * b);
out.println(min);
}
}
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 long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
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 String next() {
return readString();
}
public boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 73856c38a5476169bf030f10dde811fb | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import org.omg.PortableInterceptor.INACTIVE;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.StringTokenizer;
public class C {
static StringTokenizer st;
static PrintWriter pw;
static BufferedReader br;
static String nextToken() {
try {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(br.readLine());
}
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
static int nextInt() {
return Integer.parseInt(nextToken());
}
static long nextLong() {
return Long.parseLong(nextToken());
}
public static void main(String[] args) {
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(System.out);
run();
pw.close();
}
private static void run() {
int t = nextInt();
for (int i = 0; i < t; i++) {
int n = nextInt();
long a = nextLong();
long b = nextLong();
long x[] = new long[n];
for (int j = 0; j < n; j++) {
x[j] = nextLong();
}
long ans = 0;
long place = 0;
for (int j = 0; j < n; j++) {
ans += (x[j] - place) * b;
if ((n - j - 1) * b > a) {
ans += a * (x[j] - place);
place = x[j];
}
}
pw.println(ans);
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 8836aabb49d3b58c1be0a330cfce4be5 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class C {
FastScanner in;
PrintWriter out;
boolean systemIO = true;
public class DSU {
int[] sz;
int[] p;
public DSU(int n) {
sz = new int[n];
p = new int[n];
for (int i = 0; i < p.length; i++) {
p[i] = i;
sz[i] = 1;
}
}
public int get(int x) {
if (x == p[x]) {
return x;
}
int par = get(p[x]);
p[x] = par;
return par;
}
public boolean unite(int a, int b) {
int pa = get(a);
int pb = get(b);
if (pa == pb) {
return false;
}
if (sz[pa] < sz[pb]) {
p[pa] = pb;
sz[pb] += sz[pa];
} else {
p[pb] = pa;
sz[pa] += sz[pb];
}
return true;
}
}
public class SegmentTreeAdd {
int pow;
long[] max;
long[] delta;
boolean[] flag;
public SegmentTreeAdd(long[] a) {
pow = 1;
while (pow < a.length) {
pow *= 2;
}
flag = new boolean[2 * pow];
max = new long[2 * pow];
delta = new long[2 * pow];
for (int i = 0; i < max.length; i++) {
max[i] = Long.MIN_VALUE / 2;
}
for (int i = 0; i < a.length; i++) {
max[pow + i] = a[i];
}
for (int i = pow - 1; i > 0; i--) {
max[i] = f(max[2 * i], max[2 * i + 1]);
}
}
public long get(int v, int tl, int tr, int l, int r) {
push(v, tl, tr);
if (l > r) {
return Long.MIN_VALUE / 2;
}
if (l == tl && r == tr) {
return max[v];
}
int tm = (tl + tr) / 2;
return f(get(2 * v, tl, tm, l, Math.min(r, tm)), get(2 * v + 1, tm + 1, tr, Math.max(l, tm + 1), r));
}
public void set(int v, int tl, int tr, int l, int r, long x) {
push(v, tl, tr);
if (l > tr || r < tl) {
return;
}
if (l <= tl && r >= tr) {
delta[v] += x;
flag[v] = true;
push(v, tl, tr);
return;
}
int tm = (tl + tr) / 2;
set(2 * v, tl, tm, l, r, x);
set(2 * v + 1, tm + 1, tr, l, r, x);
max[v] = f(max[2 * v], max[2 * v + 1]);
}
public void push(int v, int tl, int tr) {
if (flag[v]) {
if (v < pow) {
flag[2 * v] = true;
flag[2 * v + 1] = true;
delta[2 * v] += delta[v];
delta[2 * v + 1] += delta[v];
}
flag[v] = false;
max[v] += delta[v];
delta[v] = 0;
}
}
public long f(long a, long b) {
return Math.max(a, b);
}
}
public class SegmentTreeSet {
int pow;
int[] sum;
int[] delta;
boolean[] flag;
public SegmentTreeSet(int[] a) {
pow = 1;
while (pow < a.length) {
pow *= 2;
}
flag = new boolean[2 * pow];
sum = new int[2 * pow];
delta = new int[2 * pow];
for (int i = 0; i < a.length; i++) {
sum[pow + i] = a[i];
}
for (int i = pow - 1; i > 0; i--) {
sum[i] = f(sum[2 * i], sum[2 * i + 1]);
}
}
public int get(int v, int tl, int tr, int l, int r) {
push(v, tl, tr);
if (l > r) {
return 0;
}
if (l == tl && r == tr) {
return sum[v];
}
int tm = (tl + tr) / 2;
return f(get(2 * v, tl, tm, l, Math.min(r, tm)), get(2 * v + 1, tm + 1, tr, Math.max(l, tm + 1), r));
}
public void set(int v, int tl, int tr, int l, int r, int x) {
push(v, tl, tr);
if (l > tr || r < tl) {
return;
}
if (l <= tl && r >= tr) {
delta[v] = x;
flag[v] = true;
push(v, tl, tr);
return;
}
int tm = (tl + tr) / 2;
set(2 * v, tl, tm, l, r, x);
set(2 * v + 1, tm + 1, tr, l, r, x);
sum[v] = f(sum[2 * v], sum[2 * v + 1]);
}
public void push(int v, int tl, int tr) {
if (flag[v]) {
if (v < pow) {
flag[2 * v] = true;
flag[2 * v + 1] = true;
delta[2 * v] = delta[v];
delta[2 * v + 1] = delta[v];
}
flag[v] = false;
sum[v] = delta[v] * (tr - tl + 1);
}
}
public int f(int a, int b) {
return a + b;
}
}
public class Pair implements Comparable<Pair> {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
public Pair clone() {
return new Pair(x, y);
}
public String toString() {
return x + " " + y;
}
@Override
public int compareTo(Pair o) {
if (x > o.x) {
return 1;
}
if (x < o.x) {
return -1;
}
if (y > o.y) {
return 1;
}
if (y < o.y) {
return -1;
}
return 0;
}
}
long mod = 1000000007;
Random random = new Random();
public void shuffle(Pair[] a) {
for (int i = 0; i < a.length; i++) {
int x = random.nextInt(i + 1);
Pair t = a[x];
a[x] = a[i];
a[i] = t;
}
}
public void sort(int[][] a) {
for (int i = 0; i < a.length; i++) {
Arrays.sort(a[i]);
}
}
public void add(Map<Long, Integer> map, long l) {
if (map.containsKey(l)) {
map.put(l, map.get(l) + 1);
} else {
map.put(l, 1);
}
}
public void remove(Map<Integer, Integer> map, Integer s) {
if (map.get(s) > 1) {
map.put(s, map.get(s) - 1);
} else {
map.remove(s);
}
}
long max = Long.MAX_VALUE / 2;
double eps = 1e-10;
public int signum(double x) {
if (x > eps) {
return 1;
}
if (x < -eps) {
return -1;
}
return 0;
}
public long abs(long x) {
return x < 0 ? -x : x;
}
public long min(long x, long y) {
return x < y ? x : y;
}
public long max(long x, long y) {
return x > y ? x : y;
}
public long gcd(long x, long y) {
while (y > 0) {
long c = y;
y = x % y;
x = c;
}
return x;
}
public final Vector ZERO = new Vector(0, 0);
// public class Vector implements Comparable<Vector> {
// long x;
// long y;
// int type;
// int number;
//
// public Vector() {
// x = 0;
// y = 0;
// }
//
// public Vector(long x, long y, int type, int number) {
// this.x = x;
// this.y = y;
// this.type = type;
// this.number = number;
// }
//
// public Vector(long x, long y) {
//
// }
//
// public Vector(Point begin, Point end) {
// this(end.x - begin.x, end.y - begin.y);
// }
//
// public void orient() {
// if (x < 0) {
// x = -x;
// y = -y;
// }
// if (x == 0 && y < 0) {
// y = -y;
// }
// }
//
// public void normalize() {
// long gcd = gcd(abs(x), abs(y));
// x /= gcd;
// y /= gcd;
// }
//
// public String toString() {
// return x + " " + y;
// }
//
// public boolean equals(Vector v) {
// return x == v.x && y == v.y;
// }
//
// public boolean collinear(Vector v) {
// return cp(this, v) == 0;
// }
//
// public boolean orthogonal(Vector v) {
// return dp(this, v) == 0;
// }
//
// public Vector ort(Vector v) {
// return new Vector(-y, x);
// }
//
// public Vector add(Vector v) {
// return new Vector(x + v.x, y + v.y);
// }
//
// public Vector multiply(long c) {
// return new Vector(c * x, c * y);
// }
//
// public int quater() {
// if (x > 0 && y >= 0) {
// return 1;
// }
// if (x <= 0 && y > 0) {
// return 2;
// }
// if (x < 0) {
// return 3;
// }
// return 0;
// }
//
// public long len2() {
// return x * x + y * y;
// }
//
// @Override
// public int compareTo(Vector o) {
// if (quater() != o.quater()) {
// return quater() - o.quater();
// }
// return signum(cp(o, this));
// }
// }
// public long dp(Vector v1, Vector v2) {
// return v1.x * v2.x + v1.y * v2.y;
// }
//
// public long cp(Vector v1, Vector v2) {
// return v1.x * v2.y - v1.y * v2.x;
// }
// public class Line implements Comparable<Line> {
// Point p;
// Vector v;
//
// public Line(Point p, Vector v) {
// this.p = p;
// this.v = v;
// }
//
// public Line(Point p1, Point p2) {
// if (p1.compareTo(p2) < 0) {
// p = p1;
// v = new Vector(p1, p2);
// } else {
// p = p2;
// v = new Vector();
// }
// }
//
// public boolean collinear(Line l) {
// return v.collinear(l.v);
// }
//
// public boolean inLine(Point p) {
// return hv(p) == 0;
// }
//
// public boolean inSegment(Point p) {
// if (!inLine(p)) {
// return false;
// }
// Point p1 = p;
// Point p2 = p.add(v);
// return p1.x <= p.x && p.x <= p2.x && min(p1.y, p2.y) <= p.y && p.y <=
// max(p1.y, p2.y);
// }
//
// public boolean equalsSegment(Line l) {
// return p.equals(l.p) && v.equals(l.v);
// }
//
// public boolean equalsLine(Line l) {
// return collinear(l) && inLine(l.p);
// }
//
// public long hv(Point p) {
// Vector v1 = new Vector(this.p, p);
// return cp(v, v1);
// }
//
// public double h(Point p) {
// Vector v1 = new Vector(this.p, p);
// return cp(v, v1) / Math.sqrt(v.len2());
// }
//
// public long[] intersectLines(Line l) {
// if (collinear(l)) {
// return null;
// }
// long[] ans = new long[4];
//
// return ans;
// }
//
// public long[] intersectSegment(Line l) {
// long[] ans = intersectLines(l);
// if (ans == null) {
// return null;
// }
// Point p1 = p;
// Point p2 = p.add(v);
// boolean f1 = p1.x * ans[1] <= ans[0] && ans[0] <= p2.x * ans[1] && min(p1.y,
// p2.y) * ans[3] <= ans[2]
// && ans[2] <= max(p1.y, p2.y) * ans[3];
// p1 = l.p;
// p2 = l.p.add(v);
// boolean f2 = p1.x * ans[1] <= ans[0] && ans[0] <= p2.x * ans[1] && min(p1.y,
// p2.y) * ans[3] <= ans[2]
// && ans[2] <= max(p1.y, p2.y) * ans[3];
// if (!f1 || !f2) {
// return null;
// }
// return ans;
// }
//
// @Override
// public int compareTo(Line o) {
// return v.compareTo(o.v);
// }
// }
public class Rect {
long x1;
long x2;
long y1;
long y2;
int number;
public Rect(long x1, long x2, long y1, long y2, int number) {
this.x1 = x1;
this.x2 = x2;
this.y1 = y1;
this.y2 = y2;
this.number = number;
}
}
public static class Fenvik {
int[] t;
public Fenvik(int n) {
t = new int[n];
}
public void add(int x, int delta) {
for (int i = x; i < t.length; i = (i | (i + 1))) {
t[i] += delta;
}
}
private int sum(int r) {
int ans = 0;
int x = r;
while (x >= 0) {
ans += t[x];
x = (x & (x + 1)) - 1;
}
return ans;
}
public int sum(int l, int r) {
return sum(r) - sum(l - 1);
}
}
public class SegmentTreeMaxSum {
int pow;
int[] sum;
int[] maxPrefSum;
int[] maxSufSum;
int[] maxSum;
public SegmentTreeMaxSum(int[] a) {
pow = 1;
while (pow < a.length) {
pow *= 2;
}
sum = new int[2 * pow];
maxPrefSum = new int[2 * pow];
maxSum = new int[2 * pow];
maxSufSum = new int[2 * pow];
for (int i = 0; i < a.length; i++) {
sum[pow + i] = a[i];
maxSum[pow + i] = Math.max(a[i], 0);
maxPrefSum[pow + i] = maxSum[pow + i];
maxSufSum[pow + i] = maxSum[pow + i];
}
for (int i = pow - 1; i > 0; i--) {
update(i);
}
}
public int[] get(int v, int tl, int tr, int l, int r) {
if (r <= tl || l >= tr) {
int[] ans = { 0, 0, 0, 0 };
return ans;
}
if (l <= tl && r >= tr) {
int[] ans = { maxPrefSum[v], maxSum[v], maxSufSum[v], sum[v] };
return ans;
}
int tm = (tl + tr) / 2;
int[] left = get(2 * v, tl, tm, l, r);
int[] right = get(2 * v + 1, tm, tr, l, r);
int[] ans = { Math.max(left[0], right[0] + left[3]),
Math.max(left[1], Math.max(right[1], left[2] + right[0])), Math.max(right[2], left[2] + right[3]),
left[3] + right[3] };
return ans;
}
public void set(int v, int tl, int tr, int x, int value) {
if (v >= pow) {
sum[v] = value;
maxSum[v] = Math.max(value, 0);
maxPrefSum[v] = maxSum[v];
maxSufSum[v] = maxSum[v];
return;
}
int tm = (tl + tr) / 2;
if (x < tm) {
set(2 * v, tl, tm, x, value);
} else {
set(2 * v + 1, tm, tr, x, value);
}
update(v);
}
public void update(int i) {
sum[i] = f(sum[2 * i], sum[2 * i + 1]);
maxSum[i] = Math.max(maxSum[2 * i], Math.max(maxSum[2 * i + 1], maxSufSum[2 * i] + maxPrefSum[2 * i + 1]));
maxPrefSum[i] = Math.max(maxPrefSum[2 * i], maxPrefSum[2 * i + 1] + sum[2 * i]);
maxSufSum[i] = Math.max(maxSufSum[2 * i + 1], maxSufSum[2 * i] + sum[2 * i + 1]);
}
public int f(int a, int b) {
return a + b;
}
}
public class Point implements Comparable<Point> {
double x;
double y;
public Point() {
x = 0;
y = 0;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Point p) {
return x == p.x && y == p.y;
}
public double dist2() {
return x * x + y * y;
}
public Point add(Point v) {
return new Point(x + v.x, y + v.y);
}
@Override
public int compareTo(Point o) {
int z = signum(x + y - o.x - o.y);
if (z != 0) {
return z;
}
return signum(x - o.x) != 0 ? signum(x - o.x) : signum(y - o.y);
}
}
public class Circle implements Comparable<Circle> {
Point p;
int r;
public Circle(Point p, int r) {
this.p = p;
this.r = r;
}
public Point angle() {
double z = r / sq2;
z -= z % 1e-5;
return new Point(p.x - z, p.y - z);
}
public boolean inside(Point p) {
return hypot2(p.x - this.p.x, p.y - this.p.y) <= sq(r);
}
@Override
public int compareTo(Circle o) {
Point a = angle();
Point oa = o.angle();
int z = signum(a.x + a.y - oa.x - oa.y);
if (z != 0) {
return z;
}
return signum(a.y - oa.y);
}
}
public class Fraction implements Comparable<Fraction> {
long x;
long y;
public Fraction(long x, long y, boolean needNorm) {
this.x = x;
this.y = y;
if (y < 0) {
this.x *= -1;
this.y *= -1;
}
if (needNorm) {
long gcd = gcd(this.x, this.y);
this.x /= gcd;
this.y /= gcd;
}
}
public Fraction clone() {
return new Fraction(x, y, false);
}
public String toString() {
return x + "/" + y;
}
@Override
public int compareTo(Fraction o) {
long res = x * o.y - y * o.x;
if (res > 0) {
return 1;
}
if (res < 0) {
return -1;
}
return 0;
}
}
public double sq(double x) {
return x * x;
}
public long sq(long x) {
return x * x;
}
public double hypot2(double x, double y) {
return sq(x) + sq(y);
}
public long hypot2(long x, long y) {
return sq(x) + sq(y);
}
public boolean kuhn(int v, int[][] edge, boolean[] used, int[] mt) {
used[v] = true;
for (int u : edge[v]) {
if (mt[u] < 0 || (!used[mt[u]] && kuhn(mt[u], edge, used, mt))) {
mt[u] = v;
return true;
}
}
return false;
}
public int matching(int[][] edge) {
int n = edge.length;
int[] mt = new int[n];
Arrays.fill(mt, -1);
boolean[] used = new boolean[n];
int ans = 0;
for (int i = 0; i < n; i++) {
if (!used[i] && kuhn(i, edge, used, mt)) {
Arrays.fill(used, false);
ans++;
}
}
return ans;
}
double sq2 = Math.sqrt(2);
int small = 20;
public class MyStack {
int[] st;
int sz;
public MyStack(int n) {
this.st = new int[n];
sz = 0;
}
public boolean isEmpty() {
return sz == 0;
}
public int peek() {
return st[sz - 1];
}
public int pop() {
sz--;
return st[sz];
}
public void clear() {
sz = 0;
}
public void add(int x) {
st[sz++] = x;
}
public int get(int x) {
return st[x];
}
}
public int[][] readGraph(int n, int m) {
int[][] to = new int[n][];
int[] sz = new int[n];
int[] x = new int[m];
int[] y = new int[m];
for (int i = 0; i < m; i++) {
x[i] = in.nextInt() - 1;
y[i] = in.nextInt() - 1;
sz[x[i]]++;
sz[y[i]]++;
}
for (int i = 0; i < to.length; i++) {
to[i] = new int[sz[i]];
sz[i] = 0;
}
for (int i = 0; i < x.length; i++) {
to[x[i]][sz[x[i]]++] = y[i];
to[y[i]][sz[y[i]]++] = x[i];
}
return to;
}
public class Pol {
double[] coeff;
public Pol(double[] coeff) {
this.coeff = coeff;
}
public Pol mult(Pol p) {
double[] ans = new double[coeff.length + p.coeff.length - 1];
for (int i = 0; i < ans.length; i++) {
for (int j = Math.max(0, i - p.coeff.length + 1); j < coeff.length && j <= i; j++) {
ans[i] += coeff[j] * p.coeff[i - j];
}
}
return new Pol(ans);
}
public String toString() {
String ans = "";
for (int i = 0; i < coeff.length; i++) {
ans += coeff[i] + " ";
}
return ans;
}
public double value(double x) {
double ans = 0;
double p = 1;
for (int i = 0; i < coeff.length; i++) {
ans += coeff[i] * p;
p *= x;
}
return ans;
}
public double integrate(double r) {
Pol p = new Pol(new double[coeff.length + 1]);
for (int i = 0; i < coeff.length; i++) {
p.coeff[i + 1] = coeff[i] / fact[i + 1];
}
return p.value(r);
}
public double integrate(double l, double r) {
return integrate(r) - integrate(l);
}
}
double[] fact = new double[100];
public class SparseTable {
int pow;
int[] lessPow;
int[][] min;
public SparseTable(int[] a) {
pow = 0;
while ((1 << pow) <= a.length) {
pow++;
}
min = new int[pow][a.length];
for (int i = 0; i < a.length; i++) {
min[0][i] = a[i];
}
for (int i = 1; i < pow; i++) {
for (int j = 0; j < a.length; j++) {
min[i][j] = min[i - 1][j];
if (j + (1 << (i - 1)) < a.length) {
min[i][j] = func(min[i][j], min[i - 1][j + (1 << (i - 1))]);
}
}
}
lessPow = new int[a.length + 1];
for (int i = 1; i < lessPow.length; i++) {
if (i < (1 << (lessPow[i - 1]) + 1)) {
lessPow[i] = lessPow[i - 1];
} else {
lessPow[i] = lessPow[i - 1] + 1;
}
}
}
public int get(int l, int r) { // [l, r)
int p = lessPow[r - l];
return func(min[p][l], min[p][r - (1 << p)]);
}
public int func(int a, int b) {
if (a < b) {
return a;
}
return b;
}
}
public double check(int n, ArrayList<Integer> masks) {
int good = 0;
for (int colorMask = 0; colorMask < (1 << n); ++colorMask) {
int best = 2 << n;
int cnt = 0;
for (int curMask : masks) {
int curScore = 0;
for (int i = 0; i < n; ++i) {
if (((curMask >> i) & 1) == 1) {
if (((colorMask >> i) & 1) == 0) {
curScore += 1;
} else {
curScore += 2;
}
}
}
if (curScore < best) {
best = curScore;
cnt = 1;
} else if (curScore == best) {
++cnt;
}
}
if (cnt == 1) {
++good;
}
}
return (double) good / (double) (1 << n);
}
public int builtin_popcount(int x) {
int ans = 0;
for (int i = 0; i < 14; i++) {
if (((x >> i) & 1) > 0) {
ans++;
}
}
return ans;
}
public int number(int[] x) {
int ans = 0;
for (int i = 0; i < x.length; i++) {
ans *= 3;
ans += x[i];
}
return ans;
}
public int[] rotate(int[] x) {
int[] ans = { x[2], x[0], x[3], x[1] };
return ans;
}
int MAX = 100000;
// boolean[] b = new boolean[MAX];
int[][] max0 = new int[MAX][2];
int[][] max1 = new int[MAX][2];
int[][] max2 = new int[MAX][2];
int[] index0 = new int[MAX];
int[] index1 = new int[MAX];
int[] p = new int[MAX];
public int place(String s) {
if (s.charAt(s.length() - 1) == '1') {
return 1;
}
int number = 16;
boolean w = true;
boolean a = true;
for (int i = 0; i < s.length(); i++) {
if (number == 1) {
return 2;
}
if (s.charAt(i) == '1') {
if (w) {
number /= 2;
} else {
if (a) {
a = false;
} else {
number /= 2;
a = true;
}
}
} else {
if (w) {
if (number == 16) {
w = false;
number /= 2;
} else {
w = false;
a = false;
}
} else {
if (number == 8) {
if (a) {
return 13;
} else {
return 9;
}
} else if (number == 4) {
if (a) {
return 7;
} else {
return 5;
}
} else if (a) {
return 4;
} else {
return 3;
}
}
}
}
return 0;
}
public class P implements Comparable<P> {
Integer x;
String s;
public P(Integer x, String s) {
this.x = x;
this.s = s;
}
@Override
public String toString() {
return x + " " + s;
}
@Override
public int compareTo(P o) {
if (x != o.x) {
return x - o.x;
}
return s.compareTo(o.s);
}
}
public BigInteger prod(int l, int r) {
if (l + 1 == r) {
return BigInteger.valueOf(l);
}
int m = (l + r) >> 1;
return prod(l, m).multiply(prod(m, r));
}
public class Frac {
BigInteger p;
BigInteger q;
public Frac(BigInteger p, BigInteger q) {
BigInteger gcd = p.gcd(q);
this.p = p.divide(gcd);
this.q = q.divide(gcd);
}
public String toString() {
return p + "\n" + q;
}
public Frac(long p, long q) {
this(BigInteger.valueOf(p), BigInteger.valueOf(q));
}
public Frac mul(Frac o) {
return new Frac(p.multiply(o.p), q.multiply(o.q));
}
public Frac sum(Frac o) {
return new Frac(p.multiply(o.q).add(q.multiply(o.p)), q.multiply(o.q));
}
public Frac diff(Frac o) {
return new Frac(p.multiply(o.q).subtract(q.multiply(o.p)), q.multiply(o.q));
}
}
public int[] transform(int[] x, int k, int step) {
int n = x.length;
int[] a = new int[n];
int[][] prefsum = new int[n][];
int[] elements = new int[n];
int[] start = new int[n];
int[] id = new int[n];
for (int i = 0; i < n; i++) {
if (elements[i] > 0) {
continue;
}
int cur = i;
do {
elements[i]++;
cur += step;
cur %= n;
} while (cur != i);
for (int j = 0; j < elements[i]; j++) {
start[cur] = i;
id[cur] = j;
elements[cur] = elements[i];
cur += step;
cur %= n;
}
prefsum[i] = new int[3 * elements[i] + 1];
for (int j = 0; j < 3 * elements[i]; j++) {
prefsum[i][j + 1] = prefsum[i][j] ^ x[cur];
cur += step;
cur %= n;
}
}
for (int i = 0; i < x.length; i++) {
int curlen = k % (2 * elements[i]);
a[i] = prefsum[start[i]][id[i] + curlen] ^ prefsum[start[i]][id[i]];
}
return a;
}
public int[][] readTree(int n) {
int m = n - 1;
int[][] to = new int[n][];
int[] sz = new int[n];
int[] x = new int[m];
int[] y = new int[m];
for (int i = 0; i < m; i++) {
x[i] = in.nextInt() - 1;
y[i] = i + 1;
sz[x[i]]++;
}
for (int i = 0; i < to.length; i++) {
to[i] = new int[sz[i]];
sz[i] = 0;
}
for (int i = 0; i < x.length; i++) {
to[x[i]][sz[x[i]]++] = y[i];
}
return to;
}
int[][] to;
int[] b;
public double dfs(int v) {
if (b[v] > 0) {
return b[v];
}
double max = 0;
double sum = 0;
for (int i : to[v]) {
double cur = dfs(i);
max = Math.max(max, cur);
sum += cur;
}
if (b[v] < 0) {
return max;
}
return sum / to[v].length;
}
public void solve() {
for (int qwerty = in.nextInt(); qwerty > 0; --qwerty) {
int n = in.nextInt();
long a = in.nextLong();
long b = in.nextLong();
long[] x = new long[n];
for (int i = 0; i < x.length; i++) {
x[i] = in.nextLong();
}
long[] sufSumm = new long[n];
sufSumm[n - 1] = x[n - 1];
for (int i = n - 2; i >= 0; --i) {
sufSumm[i] = sufSumm[i + 1] + x[i];
}
long ans = sufSumm[0] * b;
for (int i = 0; i < n - 1; i++) {
ans = Math.min(ans, (a + b) * x[i] + (sufSumm[i + 1] - (n - i - 1) * x[i]) * b);
}
out.println(ans);
}
}
public void add(HashMap<Integer, Integer> map, int x) {
if (map.containsKey(x)) {
map.put(x, map.get(x) + 1);
} else {
map.put(x, 1);
}
}
public void run() {
try {
if (systemIO) {
in = new FastScanner(System.in);
out = new PrintWriter(System.out);
} else {
in = new FastScanner(new File("input.txt"));
out = new PrintWriter(new File("output.txt"));
}
solve();
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
return null;
}
}
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());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
// AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
public static void main(String[] arg) {
new C().run();
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 64b3c100dd5eca41acd79632f4fad616 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.StringTokenizer;
import java.util.Vector;
public class template {
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 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;
}
}
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 ignored) {}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
private QuickReader sc;
private PrintWriter ptk;
public template(QuickReader sc, PrintWriter ptk) {
this.sc = sc;
this.ptk = ptk;
}
public static void main(String[] args) {
QuickReader in = new QuickReader(System.in);
try(PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));)
{
new template(in, out).solve();
}
}
public static String sortString(String inputString)
{
// Converting input string to character array
char tempArray[] = inputString.toCharArray();
// Sorting temp array using
Arrays.sort(tempArray);
// Returning new sorted string
return new String(tempArray);
}
public void solve() {
int t =sc.nextInt();
while (t-- > 0) {
int n=sc.nextInt();
n++;
int a=sc.nextInt();
int b=sc.nextInt();
long [] arr=new long[n];
for (int i = 1; i <n ; i++) {
arr[i]=sc.nextLong();
}
long sum=0;
for (int i = 1; i <n ; i++) {
sum+=arr[i];
}
long min=Long.MAX_VALUE;
for (int i = 0; i <n ; i++) {
long now=arr[i]*(a+b);
sum=sum-arr[i];
now=now+b*(sum-(n-i-1)*(arr[i]));
min=Math.min(now,min);
}
System.out.println(min);
}
}
static long lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
private static long gcd(long a, long b)
{
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | a8cdd0765df1ce07e8ce328b70902ed2 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
import java.math.BigInteger;
public final class B
{
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans=new StringBuilder();
static FastReader in=new FastReader();
// static int g[][];
static ArrayList<Integer> g[];
static long mod=(long)998244353,INF=Long.MAX_VALUE;
static boolean set[];
static int max=0;
static int lca[][];
static int par[],col[],D[];
static long fact[];
static int size[],N;
static long dp[][],sum[][],f[];
static int seg[];
public static void main(String args[])throws IOException
{
/*
* star,rope,TPST
* BS,LST,MS,MQ
*/
int T=i();
outer:while(T-->0)
{
int N=i();
long a=l(),b=l();
long c=a+b;
long A[]=new long[N+1];
for(int i=1; i<=N; i++)A[i]=l();
long pre[]=new long[N+1];
for(int i=1; i<=N; i++)
{
pre[i]=b*A[i];
pre[i]+=pre[i-1];
}
long min=pre[N];//cost with 0 as captial
long move[]=new long[N+1];
for(int i=1; i<=N; i++)
{
move[i]=c*(A[i]-A[i-1]);
move[i]+=move[i-1];
}
long left=N-1;
for(int i=1; i<N; i++)
{
long temp=move[i];//min cost to make it capital
long s=pre[N]-pre[i];//summation b*A[i]
long t=b*A[i];
t*=left;
temp+=s-t;
min=Math.min(min, temp);
left--;
}
out.println(min);
}
out.print(ans);
out.close();
}
static void dfs(int n,int p,long A[],HashMap<String,Integer> mp)
{
for(int c:g[n])
{
if(c!=p)
{
long x=mp.get(n+" "+c);
long y=mp.get(c+" "+n);
if(A[n]*y!=A[c]*x)
{
}
dfs(c,n,A,mp);
}
}
}
static int count(long N)
{
int cnt=0;
long p=1L;
while(p<=N)
{
if((p&N)!=0)cnt++;
p<<=1;
}
return cnt;
}
static long kadane(long A[])
{
long lsum=A[0],gsum=0;
gsum=Math.max(gsum, lsum);
for(int i=1; i<A.length; i++)
{
lsum=Math.max(lsum+A[i],A[i]);
gsum=Math.max(gsum,lsum);
}
return gsum;
}
public static boolean pal(int i)
{
StringBuilder sb=new StringBuilder();
StringBuilder rev=new StringBuilder();
int p=1;
while(p<=i)
{
if((i&p)!=0)
{
sb.append("1");
}
else sb.append("0");
p<<=1;
}
rev=new StringBuilder(sb.toString());
rev.reverse();
if(i==8)System.out.println(sb+" "+rev);
return (sb.toString()).equals(rev.toString());
}
public static void reverse(int i,int j,int A[])
{
while(i<j)
{
int t=A[i];
A[i]=A[j];
A[j]=t;
i++;
j--;
}
}
public static int ask(int a,int b,int c)
{
System.out.println("? "+a+" "+b+" "+c);
return i();
}
static int[] reverse(int A[],int N)
{
int B[]=new int[N];
for(int i=N-1; i>=0; i--)
{
B[N-i-1]=A[i];
}
return B;
}
static boolean isPalin(char X[])
{
int i=0,j=X.length-1;
while(i<=j)
{
if(X[i]!=X[j])return false;
i++;
j--;
}
return true;
}
static int distance(int a,int b)
{
int d=D[a]+D[b];
int l=LCA(a,b);
l=2*D[l];
return d-l;
}
static int LCA(int a,int b)
{
if(D[a]<D[b])
{
int t=a;
a=b;
b=t;
}
int d=D[a]-D[b];
int p=1;
for(int i=0; i>=0 && p<=d; i++)
{
if((p&d)!=0)
{
a=lca[a][i];
}
p<<=1;
}
if(a==b)return a;
for(int i=max-1; i>=0; i--)
{
if(lca[a][i]!=-1 && lca[a][i]!=lca[b][i])
{
a=lca[a][i];
b=lca[b][i];
}
}
return lca[a][0];
}
static void dfs(int n,int p)
{
lca[n][0]=p;
if(p!=-1)D[n]=D[p]+1;
for(int c:g[n])
{
if(c!=p)
{
dfs(c,n);
}
}
}
static int[] prefix_function(char X[])//returns pi(i) array
{
int N=X.length;
int pre[]=new int[N];
for(int i=1; i<N; i++)
{
int j=pre[i-1];
while(j>0 && X[i]!=X[j])
j=pre[j-1];
if(X[i]==X[j])j++;
pre[i]=j;
}
return pre;
}
// static TreeNode start;
// public static void f(TreeNode root,TreeNode p,int r)
// {
// if(root==null)return;
// if(p!=null)
// {
// root.par=p;
// }
// if(root.val==r)start=root;
// f(root.left,root,r);
// f(root.right,root,r);
// }
//
static int right(int A[],int Limit,int l,int r)
{
while(r-l>1)
{
int m=(l+r)/2;
if(A[m]<Limit)l=m;
else r=m;
}
return l;
}
static int left(int A[],int a,int l,int r)
{
while(r-l>1)
{
int m=(l+r)/2;
if(A[m]<a)l=m;
else r=m;
}
return l;
}
static void build(int v,int tl,int tr,int A[])
{
if(tl==tr)
{
seg[v]=A[tl];
return;
}
int tm=(tl+tr)/2;
build(v*2,tl,tm,A);
build(v*2+1,tm+1,tr,A);
seg[v]=seg[v*2]+seg[v*2+1];
}
static void update(int v,int tl,int tr,int index)
{
if(index==tl && index==tr)
{
seg[v]--;
}
else
{
int tm=(tl+tr)/2;
if(index<=tm)update(v*2,tl,tm,index);
else update(v*2+1,tm+1,tr,index);
seg[v]=seg[v*2]+seg[v*2+1];
}
}
static int ask(int v,int tl,int tr,int l,int r)
{
if(l>r)return 0;
if(tl==l && r==tr)
{
return seg[v];
}
int tm=(tl+tr)/2;
return ask(v*2,tl,tm,l,Math.min(tm, r))+ask(v*2+1,tm+1,tr,Math.max(tm+1, l),r);
}
static boolean f(long A[],long m,int N)
{
long B[]=new long[N];
for(int i=0; i<N; i++)
{
B[i]=A[i];
}
for(int i=N-1; i>=0; i--)
{
if(B[i]<m)return false;
if(i>=2)
{
long extra=Math.min(B[i]-m, A[i]);
long x=extra/3L;
B[i-2]+=2L*x;
B[i-1]+=x;
}
}
return true;
}
static int f(int l,int r,long A[],long x)
{
while(r-l>1)
{
int m=(l+r)/2;
if(A[m]>=x)l=m;
else r=m;
}
return r;
}
static boolean f(long m,long H,long A[],int N)
{
long s=m;
for(int i=0; i<N-1;i++)
{
s+=Math.min(m, A[i+1]-A[i]);
}
return s>=H;
}
static long ask(long l,long r)
{
System.out.println("? "+l+" "+r);
return l();
}
static long f(long N,long M)
{
long s=0;
if(N%3==0)
{
N/=3;
s=N*M;
}
else
{
long b=N%3;
N/=3;
N++;
s=N*M;
N--;
long a=N*M;
if(M%3==0)
{
M/=3;
a+=(b*M);
}
else
{
M/=3;
M++;
a+=(b*M);
}
s=Math.min(s, a);
}
return s;
}
static int ask(StringBuilder sb,int a)
{
System.out.println(sb+""+a);
return i();
}
static void swap(char X[],int i,int j)
{
char x=X[i];
X[i]=X[j];
X[j]=x;
}
static int min(int a,int b,int c)
{
return Math.min(Math.min(a, b), c);
}
static long and(int i,int j)
{
System.out.println("and "+i+" "+j);
return l();
}
static long or(int i,int j)
{
System.out.println("or "+i+" "+j);
return l();
}
static int len=0,number=0;
static void f(char X[],int i,int num,int l)
{
if(i==X.length)
{
if(num==0)return;
//update our num
if(isPrime(num))return;
if(l<len)
{
len=l;
number=num;
}
return;
}
int a=X[i]-'0';
f(X,i+1,num*10+a,l+1);
f(X,i+1,num,l);
}
static boolean is_Sorted(int A[])
{
int N=A.length;
for(int i=1; i<=N; i++)if(A[i-1]!=i)return false;
return true;
}
static boolean f(StringBuilder sb,String Y,String order)
{
StringBuilder res=new StringBuilder(sb.toString());
HashSet<Character> set=new HashSet<>();
for(char ch:order.toCharArray())
{
set.add(ch);
for(int i=0; i<sb.length(); i++)
{
char x=sb.charAt(i);
if(set.contains(x))continue;
res.append(x);
}
}
String str=res.toString();
return str.equals(Y);
}
static boolean all_Zero(int f[])
{
for(int a:f)if(a!=0)return false;
return true;
}
static long form(int a,int l)
{
long x=0;
while(l-->0)
{
x*=10;
x+=a;
}
return x;
}
static int count(String X)
{
HashSet<Integer> set=new HashSet<>();
for(char x:X.toCharArray())set.add(x-'0');
return set.size();
}
static int f(long K)
{
long l=0,r=K;
while(r-l>1)
{
long m=(l+r)/2;
if(m*m<K)l=m;
else r=m;
}
return (int)l;
}
// static void build(int v,int tl,int tr,long A[])
// {
// if(tl==tr)
// {
// seg[v]=A[tl];
// }
// else
// {
// int tm=(tl+tr)/2;
// build(v*2,tl,tm,A);
// build(v*2+1,tm+1,tr,A);
// seg[v]=Math.min(seg[v*2], seg[v*2+1]);
// }
// }
static int [] sub(int A[],int B[])
{
int N=A.length;
int f[]=new int[N];
for(int i=N-1; i>=0; i--)
{
if(B[i]<A[i])
{
B[i]+=26;
B[i-1]-=1;
}
f[i]=B[i]-A[i];
}
for(int i=0; i<N; i++)
{
if(f[i]%2!=0)f[i+1]+=26;
f[i]/=2;
}
return f;
}
static int[] f(int N)
{
char X[]=in.next().toCharArray();
int A[]=new int[N];
for(int i=0; i<N; i++)A[i]=X[i]-'a';
return A;
}
static int max(int a ,int b,int c,int d)
{
a=Math.max(a, b);
c=Math.max(c,d);
return Math.max(a, c);
}
static int min(int a ,int b,int c,int d)
{
a=Math.min(a, b);
c=Math.min(c,d);
return Math.min(a, c);
}
static HashMap<Integer,Integer> Hash(int A[])
{
HashMap<Integer,Integer> mp=new HashMap<>();
for(int a:A)
{
int f=mp.getOrDefault(a,0)+1;
mp.put(a, f);
}
return mp;
}
static long mul(long a, long b)
{
return ( a %mod * 1L * b%mod )%mod;
}
static void swap(int A[],int a,int b)
{
int t=A[a];
A[a]=A[b];
A[b]=t;
}
static int find(int a)
{
if(par[a]<0)return a;
return par[a]=find(par[a]);
}
static void union(int a,int b)
{
a=find(a);
b=find(b);
if(a!=b)
{
if(par[a]>par[b]) //this means size of a is less than that of b
{
int t=b;
b=a;
a=t;
}
par[a]+=par[b];
par[b]=a;
}
}
static boolean isSorted(int A[])
{
for(int i=1; i<A.length; i++)
{
if(A[i]<A[i-1])return false;
}
return true;
}
static boolean isDivisible(StringBuilder X,int i,long num)
{
long r=0;
for(; i<X.length(); i++)
{
r=r*10+(X.charAt(i)-'0');
r=r%num;
}
return r==0;
}
static int lower_Bound(int A[],int low,int high, int x)
{
if (low > high)
if (x >= A[high])
return A[high];
int mid = (low + high) / 2;
if (A[mid] == x)
return A[mid];
if (mid > 0 && A[mid - 1] <= x && x < A[mid])
return A[mid - 1];
if (x < A[mid])
return lower_Bound( A, low, mid - 1, x);
return lower_Bound(A, mid + 1, high, x);
}
static String f(String A)
{
String X="";
for(int i=A.length()-1; i>=0; i--)
{
int c=A.charAt(i)-'0';
X+=(c+1)%2;
}
return X;
}
static void sort(long[] a) //check for long
{
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 String swap(String X,int i,int j)
{
char ch[]=X.toCharArray();
char a=ch[i];
ch[i]=ch[j];
ch[j]=a;
return new String(ch);
}
static int sD(long n)
{
if (n % 2 == 0 )
return 2;
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0 )
return i;
}
return (int)n;
}
// static void setGraph(int N,int nodes)
// {
//// size=new int[N+1];
// par=new int[N+1];
// col=new int[N+1];
//// g=new int[N+1][];
// D=new int[N+1];
// int deg[]=new int[N+1];
// int A[][]=new int[nodes][2];
// for(int i=0; i<nodes; i++)
// {
// int a=i(),b=i();
// A[i][0]=a;
// A[i][1]=b;
// deg[a]++;
// deg[b]++;
// }
// for(int i=0; i<=N; i++)
// {
// g[i]=new int[deg[i]];
// deg[i]=0;
// }
// for(int a[]:A)
// {
// int x=a[0],y=a[1];
// g[x][deg[x]++]=y;
// g[y][deg[y]++]=x;
// }
// }
static long pow(long a,long b)
{
//long mod=1000000007;
long pow=1;
long x=a;
while(b!=0)
{
if((b&1)!=0)pow=(pow*x)%mod;
x=(x*x)%mod;
b/=2;
}
return pow;
}
static long toggleBits(long x)//one's complement || Toggle bits
{
int n=(int)(Math.floor(Math.log(x)/Math.log(2)))+1;
return ((1<<n)-1)^x;
}
static int countBits(long a)
{
return (int)(Math.log(a)/Math.log(2)+1);
}
static long fact(long N)
{
long n=2;
if(N<=1)return 1;
else
{
for(int i=3; i<=N; i++)n=(n*i)%mod;
}
return n;
}
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 boolean isPrime(long N)
{
if (N<=1) return false;
if (N<=3) return true;
if (N%2L == 0 || N%3L == 0) return false;
for (long i=5; i*i<=N; i+=2)
if (N%i==0)
return false;
return true;
}
static void print(char A[])
{
for(char c:A)System.out.print(c+" ");
System.out.println();
}
static void print(boolean A[])
{
for(boolean c:A)System.out.print(c+" ");
System.out.println();
}
static void print(int A[])
{
for(int a:A)System.out.print(a+" ");
System.out.println();
}
static void print(long A[])
{
for(long i:A)System.out.print(i+ " ");
System.out.println();
}
static void print(boolean A[][])
{
for(boolean a[]:A)print(a);
}
static void print(long A[][])
{
for(long a[]:A)print(a);
}
static void print(int A[][])
{
for(int a[]:A)print(a);
}
static void print(ArrayList<Integer> A)
{
for(int a:A)System.out.print(a+" ");
System.out.println();
}
static int i()
{
return in.nextInt();
}
static long l()
{
return in.nextLong();
}
static int[] input(int N){
int A[]=new int[N];
for(int i=0; i<N; i++)
{
A[i]=in.nextInt();
}
return A;
}
static long[] inputLong(int N) {
long A[]=new long[N];
for(int i=0; i<A.length; i++)A[i]=in.nextLong();
return A;
}
static long GCD(long a,long b)
{
if(b==0)
{
return a;
}
else return GCD(b,a%b );
}
}
//Code For FastReader
//Code For FastReader
//Code For FastReader
//Code For FastReader
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 | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 26d3f6f55d2e1c4328c2842ffa22a128 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.function.Function;
public class C {
public static void main(String[] args) throws NumberFormatException, IOException {
MyReader reader = new MyReader();
int t = reader.nextInt();
for(int testcase=0; testcase<t; testcase++) {
int n = reader.nextInt();
int a = reader.nextInt();
int b = reader.nextInt();
//0 at front
long[] x = new long[n+1];
for(int i=0; i<n; i++) {
x[i+1] = reader.nextInt();
}
PrefixSums prefSums = new PrefixSums(x);
long[] d = new long[n+1];
d[0] = 0;
for(int i=1; i<n+1; i++) {
final int iFinal = i;
int argj = binarySearch(
j -> (d[j] - d[j+1] + (a + 1L*b*(iFinal-j+1))*(x[j+1]-x[j])),
iFinal-1);
//System.out.println(i + " " + argj);
d[i] = d[argj] + cost(argj, i, x,prefSums,a,b) + 1L*a*(x[i]-x[argj]);
}
long ans = Long.MAX_VALUE;
for(int j=0; j<n+1; j++) {
ans = Math.min(ans, d[j] + cost(j,n, x,prefSums,a,b));
}
System.out.println(ans);
//System.out.println("---------");
}
}
//cost to conquer j+1...i from j, without including capital moving cost
public static long cost(int j, int i, long[] x, PrefixSums xPrefSums, int a, int b) {
return b*xPrefSums.query(j+1,i) - b*(i-j)*x[j];
}
//return min of f, compare tells us f(i) - f(i+1) for a input i
public static int binarySearch(Function<Integer, Long> compare, int initR) {
int l = 0;
int r = initR;
while(r>l) {
Integer m = (l+r)/2;
if(compare.apply(m) > 0) l = m+1;
else r = m;
}
return l;
}
static class PrefixSums {
long[] prefixSums;
public PrefixSums(long[] arr) {
prefixSums = new long[arr.length];
prefixSums[0] = arr[0];
for(int i=1; i<arr.length; i++) {
prefixSums[i] = prefixSums[i-1] + arr[i];
}
}
public long query(int l, int r) {
return prefixSums[r] - (l==0 ? 0 : prefixSums[l-1]);
}
}
static class MyReader {
BufferedReader br;
StringTokenizer st;
public MyReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() throws IOException {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(next());
}
long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(next());
}
double nextDouble() throws NumberFormatException, IOException {
return Double.parseDouble(next());
}
String nextLine() throws IOException {
return br.readLine();
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | 2a1de3d4f6f6b3baaaa1f5d00c35e85b | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | //make sure to make new file!
import java.io.*;
import java.util.*;
public class C782{
public static void main(String[] args)throws IOException{
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(f.readLine());
for(int q = 1; q <= t; q++){
StringTokenizer st = new StringTokenizer(f.readLine());
int n = Integer.parseInt(st.nextToken());
long b = Long.parseLong(st.nextToken());
long a = Long.parseLong(st.nextToken());
long[] array = new long[n+1];
array[0] = 0;
st = new StringTokenizer(f.readLine());
for(int k = 1; k <= n; k++){
array[k] = Long.parseLong(st.nextToken());
}
long[] sufsum = new long[n+1];
sufsum[n] = 0;
for(int k = n-1; k >= 0; k--){
sufsum[k] = sufsum[k+1] + array[k+1];
}
//cost to get the capital to point array[k]
long curcost = 0;
long answer = Long.MAX_VALUE;
for(int k = 0; k < n; k++){
answer = Math.min(answer,a*(sufsum[k]-array[k]*(long)(n-k))+curcost);
if(k < n-1) curcost += (a+b)*(array[k+1]-array[k]);
}
if(n > 1) answer = Math.min(answer,curcost+a*(array[n]-array[n-1]));
out.println(answer);
}
out.close();
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | a455f1d56e2a329b043659e007780a19 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes | import java.util.*;
import java.io.*;
public class LineEmpire {
public static void main(String[] args) throws IOException {
Reader in = new Reader();
PrintWriter out = new PrintWriter(System.out);
int T = in.nextInt();
for (int t = 0; t < T; t++) {
int N = in.nextInt() + 1, A = in.nextInt(), B = in.nextInt();
int[] arr = new int[N];
for (int i = 1; i < N; i++) {
arr[i] = in.nextInt();
}
long res = 0;
for (int i = 1; i < N; i++) {
res += (long)arr[i] * B;
}
int current = 0;
for (int i = 1; i < N; i++) {
long cost = (long)(arr[i] - arr[current]) * A;
long save = (long)(arr[i] - arr[current]) * B * (N - 1 - i);
if (save > cost) {
current = i;
res += cost;
res -= save;
}
}
out.println(res);
}
out.close();
}
static class Reader {
BufferedReader in;
StringTokenizer st;
public Reader() {
in = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer("");
}
public String nextLine() throws IOException {
st = new StringTokenizer("");
return in.readLine();
}
public String next() throws IOException {
while (!st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
}
public static void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i : arr) {
list.add(i);
}
Collections.sort(list);
for (int i = 0; i < arr.length; i++) {
arr[i] = list.get(i);
}
}
} | Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | f60c7a31fe65011980399a977f6b2f6c | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 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;
/*
X X X X
------- B
--------- B
------- A
------- B
Distance between doesn't change move strategy at all
$1 to conquere, $1 to move
X X X X X
-
- -
- - -
- - - -
$10
-
-
-
-
-
-
Each thing is min of (timesAcross*b, b+a)
*/
public class C {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int T=fs.nextInt();
for (int tt=0; tt<T; tt++) {
int n=fs.nextInt();
long c1=fs.nextInt(), c2=fs.nextInt();
int[] a=fs.readArray(n);
long ans=0;
for (int i=0; i<n; i++) {
int nToRight=n-i;
int last=i==0?0:a[i-1];
long dist=a[i]-last;
ans+=dist*Math.min(nToRight*c2, c1+c2);
}
System.out.println(ans);
}
}
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\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 8 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | f2c0f4993061d3c1fb1966acd41d2dc4 | train_110.jsonl | 1650206100 | You are an ambitious king who wants to be the Emperor of The Reals. But to do that, you must first become Emperor of The Integers.Consider a number axis. The capital of your empire is initially at $$$0$$$. There are $$$n$$$ unconquered kingdoms at positions $$$0<x_1<x_2<\ldots<x_n$$$. You want to conquer all other kingdoms.There are two actions available to you: You can change the location of your capital (let its current position be $$$c_1$$$) to any other conquered kingdom (let its position be $$$c_2$$$) at a cost of $$$a\cdot |c_1-c_2|$$$. From the current capital (let its current position be $$$c_1$$$) you can conquer an unconquered kingdom (let its position be $$$c_2$$$) at a cost of $$$b\cdot |c_1-c_2|$$$. You cannot conquer a kingdom if there is an unconquered kingdom between the target and your capital. Note that you cannot place the capital at a point without a kingdom. In other words, at any point, your capital can only be at $$$0$$$ or one of $$$x_1,x_2,\ldots,x_n$$$. Also note that conquering a kingdom does not change the position of your capital.Find the minimum total cost to conquer all kingdoms. Your capital can be anywhere at the end. | 256 megabytes |
import java.util.*;
import java.io.*;
public class B {
public static void main(String[] args) throws IOException {
Soumit sc = new Soumit();
int tc = sc.nextInt();
StringBuilder sb = new StringBuilder();
while (tc-->0){
int n = sc.nextInt();
long a = sc.nextLong();
long b = sc.nextLong();
long[] x = sc.nextLongArray(n);
if(a <= b){
if(n == 1){
long tot = x[0] * b;
sb.append(tot).append("\n");
continue;
}
long tot = (x[n-2]) * a;
long last = 0;
for(int i=0;i<n;i++){
tot += (x[i] - last) * b;
last = x[i];
}
sb.append(tot).append("\n");
continue;
}
long tot_bi = 0;
for(int i=0;i<n;i++)
tot_bi += x[i];
long tot = (tot_bi * b);
long cur_bi = 0;
long done_bi = 0;
for(int i=0;i<n;i++){
long ad = x[i] * a;
cur_bi += x[i];
if(i>0)
done_bi += (x[i] - x[i-1]);
else done_bi += x[i];
long bi = cur_bi + (x[i] * (n - i - 1));
long bd = (tot_bi - bi + done_bi) * b;
long cur = ad + bd;
tot = Math.min(tot, cur);
}
sb.append(tot).append("\n");
}
System.out.println(sb);
sc.close();
}
static class Soumit {
final private int BUFFER_SIZE = 1 << 18;
final private DataInputStream din;
final private byte[] buffer;
private PrintWriter pw;
private int bufferPointer, bytesRead;
StringTokenizer st;
public Soumit() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Soumit(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public void streamOutput(String file) throws IOException {
FileWriter fw = new FileWriter(file);
BufferedWriter bw = new BufferedWriter(fw);
pw = new PrintWriter(bw);
}
public void println(String a) {
pw.println(a);
}
public void print(String a) {
pw.print(a);
}
public String readLine() throws IOException {
byte[] buf = new byte[3000064]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public void sort(int[] arr) {
ArrayList<Integer> arlist = new ArrayList<>();
for (int i : arr)
arlist.add(i);
Collections.sort(arlist);
for (int i = 0; i < arr.length; i++)
arr[i] = arlist.get(i);
}
public void sort(long[] arr) {
ArrayList<Long> arlist = new ArrayList<>();
for (long i : arr)
arlist.add(i);
Collections.sort(arlist);
for (int i = 0; i < arr.length; i++)
arr[i] = arlist.get(i);
}
public int[] nextIntArray(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextLong();
}
return arr;
}
public double[] nextDoubleArray(int n) throws IOException {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = nextDouble();
}
return arr;
}
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;*/
if (din != null) din.close();
if (pw != null) pw.close();
}
}
}
| Java | ["4\n5 2 7\n3 5 12 13 21\n5 6 3\n1 5 6 21 30\n2 9 3\n10 15\n11 27182 31415\n16 18 33 98 874 989 4848 20458 34365 38117 72030"] | 1 second | ["173\n171\n75\n3298918744"] | NoteHere is an optimal sequence of moves for the second test case: Conquer the kingdom at position $$$1$$$ with cost $$$3\cdot(1-0)=3$$$. Move the capital to the kingdom at position $$$1$$$ with cost $$$6\cdot(1-0)=6$$$. Conquer the kingdom at position $$$5$$$ with cost $$$3\cdot(5-1)=12$$$. Move the capital to the kingdom at position $$$5$$$ with cost $$$6\cdot(5-1)=24$$$. Conquer the kingdom at position $$$6$$$ with cost $$$3\cdot(6-5)=3$$$. Conquer the kingdom at position $$$21$$$ with cost $$$3\cdot(21-5)=48$$$. Conquer the kingdom at position $$$30$$$ with cost $$$3\cdot(30-5)=75$$$. The total cost is $$$3+6+12+24+3+48+75=171$$$. You cannot get a lower cost than this. | Java 17 | standard input | [
"binary search",
"brute force",
"dp",
"greedy",
"implementation",
"math"
] | ce2b12f1d7c0388c39fee52f5410b94b | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of each test case follows. The first line of each test case contains $$$3$$$ integers $$$n$$$, $$$a$$$, and $$$b$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$1 \leq a,b \leq 10^5$$$). The second line of each test case contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_1 < x_2 < \ldots < x_n \leq 10^8$$$). The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,500 | For each test case, output a single integer — the minimum cost to conquer all kingdoms. | standard output | |
PASSED | b10d67acdff8588280ff7dfe84d4f001 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.util.regex.*;
import java.util.stream.*;
public class B {
static FastScanner fs = new FastScanner();
public static void main(String[] args) throws IOException {
int T = fs.nextInt();
int N;
int K;
int k;
while (T-- > 0) {
N = fs.nextInt();
K = fs.nextInt();
k = K;
String bitVect = fs.nextLine();
List<List<Integer>> bits = new ArrayList<>(2);
bits.add(new LinkedList<Integer>());
bits.add(new LinkedList<Integer>());
for (int i = 0; i < N; i++) {
if (bitVect.charAt(i) == '1')
bits.get(1).add(i);
else
bits.get(0).add(i);
}
int badBits = (K % 2 == 0)?0:1;
int good = (1 + badBits) % 2;
// touch as many bad bits as possible
int n = Math.min(bits.get(badBits).size(), K);
int flipped;
int[] result = new int[N];
while(n-- > 0) {
flipped = bits.get(badBits).remove(0);
result[flipped]++;
bits.get(good).add(flipped);
K--;
}
result[N-1] += K;
// Final number;
char[] binNum = new char[N];
Arrays.fill(binNum, '1');
for (Integer idx: bits.get(badBits))
binNum[idx] = '0';
// Value of last bit
if ((k - result[N-1]) % 2 == 0)
binNum[N-1] = bitVect.charAt(N-1);
else
binNum[N-1] = (bitVect.charAt(N-1)=='0')?'1':'0';
System.out.println(String.valueOf(binNum));
System.out.println(Arrays.stream(result)
.mapToObj(Integer::toString)
.collect(Collectors.joining(" "))
);
}
fs.close();
}
}
class FastScanner {
BufferedReader br;
StringTokenizer stk;
FastScanner() {
this(System.in);
}
FastScanner(InputStream stream) {
br = new BufferedReader(new InputStreamReader(stream));
}
String nextLine() throws IOException {
return br.readLine().trim();
}
String next() throws IOException {
if (stk == null || !stk.hasMoreTokens())
stk = new StringTokenizer(this.nextLine());
return stk.nextToken();
}
int nextInt() throws IOException {
return Integer.parseInt(this.next());
}
long nextLong() throws IOException {
return Long.parseLong(this.next());
}
double nextDouble() throws IOException {
return Double.parseDouble(this.next());
}
char nextChar() throws IOException {
return this.next().charAt(0);
}
int[] readArray(int count) throws IOException {
int[] ret = new int[count];
for (int i = 0; i < count; i++)
ret[i] = this.nextInt();
return ret;
}
long[] readArrayLong(int count) throws IOException {
long[] ret = new long[count];
for (int i = 0; i < count; i++)
ret[i] = this.nextLong();
return ret;
}
boolean hasNextLine() throws IOException {
br.mark(2);
try {
int ret = br.read();
br.reset();
if (ret != -1)
return true;
return false;
}
catch(IOException e) {
return false;
}
}
boolean hasNext() throws IOException {
return (stk != null && stk.hasMoreTokens()) || this.hasNextLine();
}
void close() throws IOException {
br.close();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 17 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 6c60ae978662b0fc63430ae32ea9814b | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class practice {
public static void solve() {
Reader sc = new Reader();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
int k = sc.nextInt();
char[] s = sc.next().toCharArray();
int[] a = new int[n];
int temp_k = k;
for(int i = 0;i < n && temp_k > 0;i++) {
if(k % 2 == s[i] - '0') {
a[i] = 1;
temp_k--;
}
}
a[n - 1] += temp_k;
for(int i = 0;i < n;i++)
if((k - a[i]) % 2 == 1)
s[i] = (char) ('1' - (s[i] - '0'));
for(char c : s)
out.print(c);
out.println();
for(int i : a)
out.print(i + " ");
out.println();
}
out.flush();
}
public static void main(String[] args) throws IOException {
solve();
}
public 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 {
if(st.hasMoreTokens())
str = st.nextToken("\n");
else
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 17 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | fcbccf4bf9225938515ee6493f220611 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
public class pb2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0 ){
int n = sc.nextInt();
int k = sc.nextInt();
char[] str = sc.next().toCharArray();
int tt = k;
HashSet<Integer> set = new HashSet<>();
if (k % 2 != 0){
boolean flag = true;
for (int i = 0 ; i < n ; i++){
if (flag && str[i] == '1'){
flag = false;
str[i] = '1';
set.add(i);
}
else{
if (str[i] == '0'){
str[i] = '1';
}
else{
str[i] = '0';
}
}
}
if (flag){
str[n-1] = '0';
set.add(n-1);
}
tt--;
}
for (int i = 0; i < n && tt > 0 ; i++){
if (str[i] == '0'){
str[i] = '1';
set.add(i);
tt--;
}
}
if (tt > 0 && tt%2 != 0){
str[n-1] = '0';
}
System.out.println(new String(str));
for (int i = 0; i < n-1 ; i++){
if (set.contains(i)){
System.out.print(1 + " ");
k--;
}
else{
System.out.print(0 + " ");
}
}
System.out.println(k);
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 17 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | a61ca0f11bb9ddef044cf24e73dc46e2 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
//new Main().run();
//}
//void run() {
Scanner sc=new Scanner(System.in);
PrintWriter pw=new PrintWriter(System.out);
int t=sc.nextInt();
while (t-->0) {
int n=sc.nextInt();
int k=sc.nextInt();
char[] cs=sc.next().toCharArray();
int[] a=new int[cs.length];
Arrays.setAll(a, i->cs[i]-'0');
for (int i=0;i<a.length;++i) {
a[i]^=k%2;
}
int[] ans=new int[a.length];
for (int i=0;i<a.length;++i) {
if (a[i]==0&&k>0) {
ans[i]=1;
a[i]^=1;
--k;
}
}
ans[a.length-1]+=k;
a[a.length-1]^=k%2;
for (int i=0;i<a.length;++i) {
pw.print(a[i]+(i==a.length-1?"\n":""));
}
for (int i=0;i<ans.length;++i) {
pw.print(ans[i]+(i==ans.length-1?"\n":" "));
}
}
pw.close();
}
//void tr(Object...o) {System.out.println(Arrays.deepToString(o));}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 17 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | d1eff8db1e6478bc3806c72c535324b8 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
String s = br.readLine();
char arr2[] = s.toCharArray();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = arr2[i] - '0';
int res[] = new int[n];
if(k == 0)
{
output.write(s + "\n");
for(int i = 0; i < n; i++)
output.write("0 ");
output.write("\n");
continue;
}
if(k % 2 == 1)
{
if(arr[0] == 1)
{
res[0]++;
k--;
for(int i = 1; i < n; i++)
arr[i] = 1 - arr[i];
}
else
{
int flag = 0;
arr[0] = 1;
for(int i = 1; i < n; i++)
{
if(flag == 1)
{
arr[i] = 1 - arr[i];
continue;
}
if(arr[i] == 1)
{
flag = 1;
k--;
res[i]++;
}
arr[i] = 1;
}
if(flag == 0)
{
k--;
res[n-1]++;
for(int i = 0; i < n; i++)
{
if(i == n - 1)
arr[i] = 0;
else
arr[i] = 1;
}
}
}
}
for(int i = 0; i < n - 1; i++)
{
if(k == 0)
break;
if(arr[i] == 1)
continue;
else
{
res[i]++;
arr[i] = 1;
i++;
k-=2;
int flag = 0;
while(arr[i]!=0)
{
arr[i] = 1;
i++;
if(i == n)
{
flag = 1;
break;
}
}
if(flag == 1)
{res[n-1]++;
arr[n-1] = 1 - arr[n-1];
break;}
else
{
res[i]++;
arr[i] = 1-arr[i];
i--;
continue;
}
}
}
if(k > 0)
res[0]+=k;
for(int i : arr)
output.write(i + "");
output.write("\n");
for(int i : res)
output.write(i + " ");
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// 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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f64f792b25430d6a8fb798bd23973d83 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
String s = br.readLine();
char arr2[] = s.toCharArray();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = arr2[i] - '0';
int res[] = new int[n];
if(k == 0)
{
output.write(s + "\n");
for(int i = 0; i < n; i++)
output.write("0 ");
output.write("\n");
continue;
}
if(k % 2 == 1)
{
int ind = -1;
for(int i = 0; i < n; i++)
{
if(arr[i] == 1)
{
ind = i;
break;
}
}
if(ind == -1)
ind = n - 1;
for(int i = 0; i < n; i++)
{
if(i == ind)
continue;
else
arr[i] = 1 - arr[i];
}
k--;
res[ind]++;
}
for(int i = 0; i < n - 1; i++)
{
if(k == 0)
break;
if(arr[i] == 1)
continue;
else
{
res[i]++;
arr[i] = 1;
i++;
k-=2;
int flag = 0;
while(arr[i]!=0)
{
arr[i] = 1;
i++;
if(i == n)
{
flag = 1;
break;
}
}
if(flag == 1)
{res[n-1]++;
arr[n-1] = 1 - arr[n-1];
break;}
else
{
res[i]++;
arr[i] = 1-arr[i];
i--;
continue;
}
}
}
if(k > 0)
res[0]+=k;
for(int i : arr)
output.write(i + "");
output.write("\n");
for(int i : res)
output.write(i + " ");
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// 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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f3e1508c3b193071559de01cf6074955 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
String s = br.readLine();
char arr2[] = s.toCharArray();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = arr2[i] - '0';
int res[] = new int[n];
if(k == 0)
{
output.write(s + "\n");
for(int i = 0; i < n; i++)
output.write("0 ");
output.write("\n");
continue;
}
if(k % 2 == 1)
{
int ind = -1;
for(int i = 0; i < n; i++)
{
if(arr[i] == 1)
{
ind = i;
break;
}
}
if(ind == -1)
ind = n - 1;
for(int i = 0; i < n; i++)
{
if(i == ind)
continue;
else
arr[i] = 1 - arr[i];
}
k--;
res[ind]++;
}
for(int i = 0; i < n - 1; i++)
{
if(k == 0)
break;
if(arr[i] == 1)
continue;
else
{
res[i]++;
arr[i] = 1;
i++;
k-=2;
int flag = 0;
while(arr[i]!=0)
{
arr[i] = 1;
i++;
if(i == n)
{
flag = 1;
break;
}
}
if(flag == 1)
{res[n-1]++;
arr[n-1] = 1 - arr[n-1];
break;}
else
{
res[i]++;
arr[i] = 1;
i--;
continue;
}
}
}
if(k > 0)
res[0]+=k;
for(int i : arr)
output.write(i + "");
output.write("\n");
for(int i : res)
output.write(i + " ");
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// 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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 67401b50b718e5a5bfd5cb60839488ae | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
String s = br.readLine();
char arr2[] = s.toCharArray();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = arr2[i] - '0';
int res[] = new int[n];
if(k == 0)
{
output.write(s + "\n");
for(int i = 0; i < n; i++)
output.write("0 ");
output.write("\n");
continue;
}
if(k % 2 == 1)
{
int ind = -1;
for(int i = 0; i < n; i++)
{
if(arr[i] == 1)
{
ind = i;
break;
}
}
if(ind == -1)
ind = n - 1;
for(int i = 0; i < n; i++)
{
if(i == ind)
continue;
else
arr[i] = 1 - arr[i];
}
k--;
res[ind]++;
}
for(int i = 0; i < n - 1; i++)
{
if(k == 0)
break;
if(arr[i] == 1)
continue;
else
{
res[i]++;
arr[i] = 1;
i++;
k-=2;
int flag = 0;
while(arr[i]!=0)
{
arr[i] = 1;
i++;
if(i == n)
{
flag = 1;
break;
}
}
if(flag == 1)
{res[n-1]++;
arr[n-1] = 1 - arr[n-1];
break;}
else
{
res[i]++;
arr[i] = 1;
// i--;
continue;
}
}
}
if(k > 0)
res[0]+=k;
for(int i : arr)
output.write(i + "");
output.write("\n");
for(int i : res)
output.write(i + " ");
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// 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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | dd501300ad5ebdad64ffcc9799e6b9c4 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
String s = br.readLine();
char arr2[] = s.toCharArray();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = arr2[i] - '0';
int res[] = new int[n];
if(k == 0)
{
output.write(s + "\n");
for(int i = 0; i < n; i++)
output.write("0 ");
output.write("\n");
continue;
}
if(k % 2 == 1)
{
int ind = -1;
for(int i = 0; i < n; i++)
{
if(arr[i] == 1)
{
ind = i;
break;
}
}
if(ind == -1)
ind = n - 1;
for(int i = 0; i < n; i++)
{
if(i == ind)
continue;
else
arr[i] = 1 - arr[i];
}
k--;
res[ind]++;
}
for(int i = 0; i < n - 1; i++)
{
if(k == 0)
break;
if(arr[i] == 1)
continue;
else
{
res[i]++;
arr[i] = 1;
i++;
k-=2;
int flag = 0;
while(arr[i]!=0)
{
i++;
if(i == n)
{
flag = 1;
break;
}
}
if(flag == 1)
{res[n-1]++;
arr[n-1] = 1 - arr[n-1];
break;}
else
{
res[i]++;
arr[i] = 1-arr[i];
// i--;
continue;
}
}
}
if(k > 0)
res[0]+=k;
for(int i : arr)
output.write(i + "");
output.write("\n");
for(int i : res)
output.write(i + " ");
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// 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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 4eb3f1db39bb2521277916f95044ab42 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
String s = br.readLine();
char arr2[] = s.toCharArray();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = arr2[i] - '0';
int res[] = new int[n];
int k2 = k;
for(int i = 0; i < n; i++)
{
if(k2 == 0)
break;
if((k%2 == 0 && arr[i] == 0) || (k%2 == 1 && arr[i] == 1))
{
res[i] = 1;
k2--;
continue;
}
}
res[n-1]+=k2;
for(int i = 0; i < n; i++)
arr[i] = ((k-res[i])%2==0)?arr[i]:1-arr[i];
for(int i : arr)
output.write(i + "");
output.write("\n");
for(int i : res)
output.write(i + " ");
output.write("\n");
}
output.flush();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | fc79f55b348faa96df46659d0e19225d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int not = Integer.parseInt(st.nextToken());
int knot = Integer.parseInt(st.nextToken());
int Onot = knot;
char sot[] = br.readLine().toCharArray();
int Sot[] = new int[not];
for(int i = 0; i < sot.length; i++)
Sot[i] = sot[i] - '0';
int fin[] = new int[not];
int onot = 0;
for(int i = 0; i < not && knot > 0; i++)
{
int stemp = (sot[i] - '0') ^ onot;
sot[i] = (stemp==0)?'0':'1';
if(knot%2!=stemp || i==not-1)
continue;
else
{
fin[i]++;
onot = onot ^ 1;
knot--;
}
}
fin[fin.length - 1]+=knot;
for(int i=0;i<not;i++)
{
int e = Onot-fin[i];
if(e%2 == 1)
Sot[i]^=1;
}
for(int i : Sot)
output.write(i + "");
output.write("\n");
for(int i : fin)
output.write(i + " ");
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// 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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 83d628c92b43e91b6a05d387b5fd0449 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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;
import java.util.*;
public class spidername {
static FastScanner sc = new FastScanner();
static PrintWriter out = new PrintWriter(System.out);
static int nxt() {
int x = sc.nextInt();
return x;
}
static void solve() {
int n = nxt();
int k = nxt();
String s = sc.next();
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = (int)s.charAt(i) - '0';
}
int ans[] = new int[n];
int cnt[] = new int[n];
int need = 1, flip = 1;
if (k % 2 == 0) {
need = 0;
flip = 0;
}
for (int i = 0; i < n ; i++) {
ans[i] = a[i] ^ flip;
if (a[i] == need && k > 0) {
cnt[i]++;
k--;
}
}
cnt[n - 1] += k;
for (int i = 0; i < n; i++) {
if (cnt[i] % 2 == 1) {
ans[i] ^= 1;
}
out.print(ans[i]);
}
out.println();
for (int i = 0; i < n; i++) {
out.print(cnt[i] + " ");
}
out.println();
}
public static void main(String args[]) {
int t = nxt();
while (t-- > 0) {
solve();
}
out.close();
}
}
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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 210c58ec5a4555e891d760678e797a10 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
// when can't think of anything -->>
// 1. In sorting questions try to think about all possibilities like starting from start, end, middle.
// 2. Two pointers, brute force.
// 3. In graph query questions try to solve it reversely or try to process all the queries in a single parse.
// 4. If order does not matter then just sort the data if constraints allow. It'll never harm.
// 5. In greedy problems if you are just overwhelmed by the possibilities and stuck, try to code whatever you come up with.
// 6. Try to solve it from back or reversely.
// 7. When can't prove something take help of contradiction.
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
PrintWriter writer = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int k = sc.nextInt();
int ok = k;
String s = sc.next();
int ans[] = new int[n];
int arr[] = new int[n];
int onecount = 0;
for(char c : s.toCharArray())if(c=='1')onecount++;
int zerocount = n-onecount;
if(k%2==0) {
int val = 0;
int count = 0;
for(char c : s.toCharArray()) {
if(c=='0' && count < k) {
arr[val]++;
count++;
}
val++;
}
arr[n-1]+=k-count;
for(int i = 0; i < n; i++) {
int val1 = arr[i];
int val2 = s.charAt(i) - '0';
ans[i] = val1^val2;
}
if(arr[n-1]>1) {
if(zerocount%2==0)ans[n-1]=1;
else ans[n-1]=0;
}
}else {
int val = 0;
int count = 0;
for(char c : s.toCharArray()) {
if(c=='1' && count < k) {
arr[val]++;
count++;
}
val++;
}
arr[n-1]+=k-count;
for(int i = 0; i < n; i++) {
int val1 = arr[i];
int val2 = s.charAt(i) - '0';
if(val1==val2)ans[i]=1;
else ans[i]=0;
}
if(arr[n-1]>1) {
if(onecount%2==1)ans[n-1]=1;
else ans[n-1]=0;
}
}
//writer.println(zerocount + " " + onecount);
for(int i : ans)writer.print(i);
writer.println();
for(int i : arr) {
writer.print(i + " ");
}
writer.println();
}
writer.flush();
writer.close();
}
private static int findVal(String a, String b, int n) {
int diff1 = 0;
for(int i = 0; i < n; i++) {
if(a.charAt(i) != b.charAt(i))diff1++;
}
return diff1;
}
private static long power (long a, long n, long p) {
long res = 1;
while(n!=0) {
if(n%2==1) {
res=(res*a)%p;
n--;
}else {
a= (a*a)%p;
n/=2;
}
}
return res;
}
private static boolean isPrime(int c) {
for (int i = 2; i*i <= c; i++) {
if(c%i==0)return false;
}
return true;
}
private static int find(int a , int arr[]) {
if(arr[a] != a) return arr[a] = find(arr[a], arr);
return arr[a];
}
private static void union(int a, int b, int arr[]) {
int aa = find(a,arr);
int bb = find(b, arr);
arr[aa] = bb;
}
private static int gcd(int a, int b) {
if(a==0)return b;
return gcd(b%a, a);
}
public static int[] readIntArray(int size, FastReader s) {
int[] array = new int[size];
for (int i = 0; i < size; i++) {
array[i] = s.nextInt();
}
return array;
}
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;
}
}
}
class Pair implements Comparable<Pair>{
int a;
int b;
Pair(int a, int b){
this.a = a;
this.b = b;
}
@Override
public boolean equals(Object obj) {
if(this == obj) return true;
if(obj == null || obj.getClass()!= this.getClass()) return false;
Pair pair = (Pair) obj;
return (pair.a == this.a && pair.b == this.b);
}
@Override
public int hashCode()
{
return Objects.hash(a,b);
}
@Override
public int compareTo(Pair o) {
if(this.a == o.a) {
return Integer.compare(this.b, o.b);
}else {
return Integer.compare(this.a, o.a);
}
}
@Override
public String toString() {
return this.a + " " + this.b;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | c5e7c79b5d58e360bff528ddf2e29240 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Main{
public static FastReader cin;
public static PrintWriter out;
public static void main(String[] args) throws Exception{
out = new PrintWriter(new BufferedOutputStream(System.out));
cin = new FastReader();
int t=cin.nextInt();
for(int z=0;z<t;z++) {
int n=cin.nextInt();
int k=cin.nextInt();
char []arr=cin.next().toCharArray();
int tmpk=k;
int []f=new int [n];
for(int i=0;i<n&&tmpk>0;i++) {
if((k%2==1&&arr[i]=='1')||(k%2==0&&arr[i]=='0')) {
f[i]=1;
tmpk--;
}
/*if(k%2==(int)(arr[i]-'0')) {
f[i]=1;
tmpk--;
}*/
}
f[n-1]+=tmpk;
for(int i=0;i<n;i++) {
int change=(k-f[i])%2;//变换次数
if(change==1) {
if(arr[i]=='1')arr[i]='0';
else arr[i]='1';
}
}
out.println(String.valueOf(arr));
for(int i=0;i<n;i++)out.printf("%d ",f[i]);
out.println();
}
out.close();
}
static class FastReader
{
BufferedReader br;
StringTokenizer str;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (str == null || !str.hasMoreElements())
{
try
{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
}
return str.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 lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 3ce06d92864bd2ebcf62ded1731bddc8 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
public class codeforces {
public static void main(String[] args) throws IOException {
BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(scan.readLine());
while (t-->0) {
String[] s = scan.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int k = Integer.parseInt(s[1]);
char[] bitStr = scan.readLine().toCharArray();
taskB(n, k, bitStr);
}
}
// public static String compute(long games, long r, long b) {
// long rest = r%(b+1);
// long ratio = r/(b+1);
// String sol = "";
// long j = 0;
// while(j <= b) {
// for (int i = 0; i < ratio; i++) {
// sol += "R";
// }
// if (j < rest) sol += "R";
// if (j < b) sol = sol + "B";
// j++;
// }
//
// return sol;
// }
public static void taskB(long n, long k, char[] bitStr) {
StringBuilder sol = new StringBuilder();
StringBuilder f = new StringBuilder();
char[] switchStr = bitStr;
if (k%2 == 1) {
for (int i = 0; i < n; i++) {
if (switchStr[i] == "0".charAt(0)) switchStr[i] = "1".charAt(0);
else switchStr[i] = "0".charAt(0);
}
}
for (int i = 0; i < n; i++) {
if (i == n-1) {
f.append(k);
if (k%2 == 0) sol.append(switchStr[i]);
else if (switchStr[i] == "1".charAt(0)) sol.append("0");
else sol.append("1");
k=0;
}
else if (switchStr[i] == "0".charAt(0) && k > 0) {
f.append("1 ");
sol.append("1");
k--;
}
else {
f.append("0 ");
sol.append(switchStr[i]);
}
}
System.out.println(sol);
System.out.println(f);
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 12ecdb5ed0753783e4b82bcd1cd45cb7 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class codeforces_782_B {
private static void solve(FastIOAdapter in, PrintWriter out) {
int n = in.nextInt();
int k = in.nextInt();
char[] s = in.next().toCharArray();
int cur = 0;
char[] sans = new char[n];
int[] ans = new int[n];
boolean last = s[n - 1] == '1';
for (int i = 0; i < n - 1; i++) {
if (cur < k) {
int left = k - cur;
if (s[i] == '1' && cur % 2 == 0 || s[i] == '0' && cur % 2 == 1) {
sans[i] = '1';
if (left % 2 != 0) {
ans[i] = 1;
last = !last;
cur++;
}
} else {
sans[i] = '1';
if (left % 2 == 0) {
if (left > 1) {
ans[i] = 1;
last = !last;
cur++;
} else {
boolean g = false;
for (int j = i + 1; j < n - 1; j++) {
if (s[j] == '1' && cur % 2 == 0 || s[j] == '0' && cur % 2 == 1) {
g = true;
ans[j] = 1;
sans[j] = '1';
last = !last;
cur++;
break;
}
}
if (!g) {
ans[n - 1] = 1;
cur++;
}
}
}
}
} else if (s[i] == '1' && k % 2 == 0 || s[i] == '0' && k % 2 == 1) {
sans[i] = '1';
} else sans[i] = '0';
}
int left = k - cur;
ans[n - 1] += left;
if (last)
sans[n - 1] = '1';
else sans[n - 1] = '0';
for (int i = 0; i < n; i++) {
out.print(sans[i]);
}
out.println();
for (int i = 0; i < n; i++) {
out.print(ans[i] + " ");
}
out.println();
}
public static void main(String[] args) throws Exception {
try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
int count = 1;
count = ioAdapter.nextInt();
while (count-- > 0) {
solve(ioAdapter, ioAdapter.out);
}
}
}
static void ruffleSort(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 class FastIOAdapter implements AutoCloseable {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
e.printStackTrace();
return null;
}
}
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[] readArrayLong(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());
}
@Override
public void close() throws Exception {
out.flush();
out.close();
br.close();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 741db639cf685a16614710d37dd414e4 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
public class BitFlipping {
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while(t > 0)
{
int n = scan.nextInt();
int k = scan.nextInt();
scan.nextLine();
String str = scan.nextLine();
int[] arr = new int[n];
int sum = 0;
for(int i = 0; i < n; i++)
{
if(sum == k)
{
break;
}
if(str.charAt(i) == '0')
{
if(k % 2 == 0)
{
arr[i]++;
sum++;
}
}
else
{
if(k % 2 == 1)
{
arr[i]++;
sum++;
}
}
}
arr[n-1] += (k - sum);
StringBuilder str1 = new StringBuilder("");
StringBuilder str2 = new StringBuilder("");
for(int i = 0; i < n; i++)
{
str1.append((((str.charAt(i)-'0')+k-arr[i])%2));
str2.append(arr[i] + " ");
}
System.out.println(str1);
System.out.println(str2);
t--;
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 2d5c7633a795151909382e784e984eb1 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.io.*;
import java.util.*;
public class Solution {
static int mod=(int)998244353;
public static void main(String[] args) {
Copied io = new Copied(System.in, System.out);
// int k=1;
int t = 1;
t = io.nextInt();
for (int i = 0; i < t; i++) {
// io.print("Case #" + k + ": ");
solve(io);
// k++;
}
io.close();
}
public static void solve(Copied io) {
int n = io.nextInt(), k = io.nextInt();
String s = io.next();
char c[] = new char[n];
for(int i = 0; i < n; i++) {
c[i] = s.charAt(i);
}
int ans[] = new int[n];
if(k % 2 == 0) {
int nz = 0;
for(int i = 0; i < n; i++) {
if(c[i] == '0') {
nz++;
}
}
for(int i = 0; i < n && nz > 1 && k > 0; i++) {
if(c[i] == '0') {
ans[i]++;
c[i] = '1';
while(i < n && c[i] == '1') {
i++;
}
ans[i]++;
nz -= 2;
k -= 2;
c[i] = '1';
}
}
if(k > 0) {
for(int i = 0; i < n; i++) {
if(c[i] == '0') {
c[i] = '1';
c[n - 1] = '0';
ans[i]++;
ans[n - 1]++;
nz--;
k -= 2;
break;
}
}
ans[n - 1] += k;
}
}else {
int nz = 0;
for(int i = 0; i < n; i++) {
if(c[i] == '1') {
nz++;
}
}
for(int i = 0; i < n && nz > 1 && k > 1; i++) {
if(c[i] == '1') {
ans[i]++;
c[i] = '0';
while(i < n && c[i] == '0') {
i++;
}
ans[i]++;
nz -= 2;
k -= 2;
c[i] = '0';
}
}
if(k >= 1) {
if(k > 1) {
ans[n - 1] += (k - 1);
k = 1;
}
int pos = n - 1;
for(int i = 0; i < n; i++) {
if(c[i] == '1') {
pos = i;
break;
}
}
ans[pos]++;
for(int i = 0; i < pos; i++) {
c[i] = (c[i] == '0' ? '1' : '0');
}
for(int i = pos + 1; i < n; i++) {
c[i] = (c[i] == '0' ? '1' : '0');
}
}
}
for(char ch: c) {
io.print(ch);
}
io.println();
printArr(ans, io);
}
static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
static void binaryOutput(boolean ans, Copied io){
String a=new String("YES"), b=new String("NO");
if(ans){
io.println(a);
}
else{
io.println(b);
}
}
static int power(int x, int y, int p)
{
int res = 1;
x = x % p;
if (x == 0)
return 0;
while (y > 0)
{
if ((y & 1) != 0)
res = (res * x) % p;
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
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 printArr(int[] arr,Copied io) {
for (int x : arr)
io.print(x + " ");
io.println();
}
static void printArr(long[] arr,Copied io) {
for (long x : arr)
io.print(x + " ");
io.println();
}
static int[] listToInt(ArrayList<Integer> a){
int n=a.size();
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=a.get(i);
}
return arr;
}
static int mod_mul(int a, int b){ a = a % mod; b = b % mod; return (((a * b) % mod) + mod) % mod;}
static int mod_add(int a, int b){ a = a % mod; b = b % mod; return (((a + b) % mod) + mod) % mod;}
static int mod_sub(int a, int b){ a = a % mod; b = b % mod; return (((a - b) % mod) + mod) % mod;}
}
class Pair implements Cloneable, Comparable<Pair>
{
int x,y;
Pair(int a,int b)
{
this.x=a;
this.y=b;
}
@Override
public boolean equals(Object obj)
{
if(obj instanceof Pair)
{
Pair p=(Pair)obj;
return p.x==this.x && p.y==this.y;
}
return false;
}
@Override
public int hashCode()
{
return Math.abs(x)+101*Math.abs(y);
}
@Override
public String toString()
{
return "("+x+" "+y+")";
}
@Override
protected Pair clone() throws CloneNotSupportedException {
return new Pair(this.x,this.y);
}
@Override
public int compareTo(Pair a)
{
int t= this.x-a.x;
if(t!=0)
return t;
else
return this.y-a.y;
}
}
class byY implements Comparator<Pair>{
// @Override
public int compare(Pair o1,Pair o2){
// -1 if want to swap and (1,0) otherwise.
int t= o1.y-o2.y;
if(t!=0)
return t;
else
return o1.x-o2.x;
}
}
class byX implements Comparator<Pair>{
// @Override
public int compare(Pair o1,Pair o2){
// -1 if want to swap and (1,0) otherwise.
int t= o1.x-o2.x;
if(t!=0)
return t;
else
return o1.y-o2.y;
}
}
class DescendingOrder<T> implements Comparator<T>{
@Override
public int compare(Object o1,Object o2){
// -1 if want to swap and (1,0) otherwise.
int addingNumber=(Integer) o1,existingNumber=(Integer) o2;
if(addingNumber>existingNumber) return -1;
else if(addingNumber<existingNumber) return 1;
else return 0;
}
}
class Copied extends PrintWriter {
public Copied(InputStream i) {
super(new BufferedOutputStream(System.out));
r = new BufferedReader(new InputStreamReader(i));
}
public Copied(InputStream i, OutputStream o) {
super(new BufferedOutputStream(o));
r = new BufferedReader(new InputStreamReader(i));
}
public boolean hasMoreTokens() {
return peekToken() != null;
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public String next() {
return nextToken();
}
public double[] nextDoubles(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextLongs(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
private BufferedReader r;
private String line;
private StringTokenizer st;
private String token;
private String peekToken() {
if (token == null)
try {
while (st == null || !st.hasMoreTokens()) {
line = r.readLine();
if (line == null) return null;
st = new StringTokenizer(line);
}
token = st.nextToken();
} catch (IOException e) { }
return token;
}
private String nextToken() {
String ans = peekToken();
token = null;
return ans;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f6590aec452aafa108498f198ba12bbd | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Solution6 {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int t=fs.nextInt();
int l=0;
while(t-->0){
//long n=fs.nextLong();
//long k=fs.nextLong();
int n=fs.nextInt();
int k=fs.nextInt();
String s=fs.next();
char ch[]=s.toCharArray();
int fn[]=new int[n];
//char ch[]=s.toCharArray();
int m=k;
for(int i=0;i<n && m>0;i++){
if(k%2==ch[i]-'0'){
fn[i]=1;
m--;
}
}
fn[n-1]+=m;
String res="";
for(int i=0;i<n;i++){
if((k-fn[i])%2!=0)
ch[i]=(char)('1'-(ch[i]-'0'));
}
out.println(ch);
for(int i:fn)
out.print(i+" ");
out.println();
}
out.close();
}
// public static void check(int n,int k,char ch[]){
// PrintWriter out=new PrintWriter(System.out);
//
//
// }
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | adbf55ca2e306182201399194e432123 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Solution6 {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int t=fs.nextInt();
int l=0;
while(t-->0){
//long n=fs.nextLong();
//long k=fs.nextLong();
int n=fs.nextInt();
int k=fs.nextInt();
String s=fs.next();
char ch[]=s.toCharArray();
check(n,k,ch);
}
}
public static void check(int n,int k,char ch[]){
int fn[]=new int[n];
//char ch[]=s.toCharArray();
int m=k;
for(int i=0;i<n && m>0;i++){
if(k%2==ch[i]-'0'){
fn[i]=1;
m--;
}
}
fn[n-1]+=m;
String res="";
for(int i=0;i<n;i++){
if((k-fn[i])%2!=0)
ch[i]=(char)('1'-(ch[i]-'0'));
}
System.out.println(ch);
for(int i:fn)
System.out.print(i+" ");
System.out.println();
}
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | c07a0d17e86f52bc17b4f68c4c67a7e5 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Solution6 {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int t=fs.nextInt();
int l=0;
while(t-->0){
//long n=fs.nextLong();
//long k=fs.nextLong();
int n=fs.nextInt();
int k=fs.nextInt();
String s=fs.next();
char ch[]=s.toCharArray();
check(n,k,ch);
}
}
public static void check(int n,int k,char ch[]){
int fn[]=new int[n];
//char ch[]=s.toCharArray();
int m=k;
for(int i=0;i<n && m>0;i++){
if(k%2==ch[i]-'0'){
fn[i]=1;
m--;
}
}
fn[n-1]+=m;
String res="";
for(int i=0;i<n;i++){
if((k-fn[i])%2!=0)
ch[i]=(char)('1'-(ch[i]-'0'));
}
System.out.println(String.valueOf(ch));
for(int i:fn)
System.out.print(i+" ");
System.out.println();
}
static long binpow(long a, long b) {
a %= mod;
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
static long setbit(long x){
long ind=0;int i=0;
while(x>0){
if((x & 1)==1)
ind =i;
i++;
x>>=1;
}
return ind;
}
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 94d4129f24c65255bfe475c7ea744e9e | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
public class Solution{
public static void main(String[] args) {
TaskA solver = new TaskA();
int t = in.nextInt();
for (int i = 1; i <= t ; i++) {
solver.solve(i, in, out);
}
// solver.solve(1, in, out);
out.flush();
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n= in.nextInt();
int k= in.nextInt();
char[]c=in.next().toCharArray();
if(n==1) {
println((char)c[0]-'0');
println(k);
return;
}
int []ans=new int[n];int k1=k;
for(int i=0;i<n&&k1>0;i++) {
if(k%2==c[i]-'0') {
ans[i]=1;k1--;
}
}
ans[n-1]+=k1;
for(int i=0;i<n;i++) {
if((k-ans[i])%2==1) {
if(c[i]=='0') {
c[i]='1';
}
else {
c[i]='0';
}
}
}
println(String.valueOf(c));
println(ans);
}
}
static int ceil(int a,int b) {
int ans=a/b;if(a%b!=0) {
ans++;
}
return ans;
}
static int query(int l,int r) {
System.out.println("? "+l+" "+r);
System.out.print ("\n");
int x=in.nextInt();
return x;
}
static boolean check(Pair[]arr,int mid,int n) {
int c=0;
for(int i=0;i<n;i++) {
if(mid-1-arr[i].first<=c&&c<=arr[i].second) {
c++;
}
}
return c>=mid;
}
static long sum(int[]arr) {
long s=0;
for(int x:arr) {
s+=x;
}
return s;
}
static long sum(long[]arr) {
long s=0;
for(long x:arr) {
s+=x;
}
return s;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
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);
}
static void println(int[][]arr) {
for(int i=0;i<arr.length;i++) {
for(int j=0;j<arr[0].length;j++) {
print(arr[i][j]+" ");
}
print("\n");
}
}
static void println(long[][]arr) {
for(int i=0;i<arr.length;i++) {
for(int j=0;j<arr[0].length;j++) {
print(arr[i][j]+" ");
}
print("\n");
}
}
static void println(int[]arr){
for(int i=0;i<arr.length;i++) {
print(arr[i]+" ");
}
print("\n");
}
static void println(long[]arr){
for(int i=0;i<arr.length;i++) {
print(arr[i]+" ");
}
print("\n");
}
static long[]input(int n){
long[]arr=new long[n];
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
return arr;
}
static long[]input(){
long n= in.nextInt();
long[]arr=new long[(int)n];
for(int i=0;i<n;i++) {
arr[i]=in.nextLong();
}
return arr;
}
////////////////////////////////////////////////////////
static class Pair {
long first;
long second;
Pair(long x, long y)
{
this.first = x;
this.second = y;
}
}
static void sortS(Pair arr[])
{
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return (int)(p1.second - p2.second);
}
});
}
static void sortF(Pair arr[])
{
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return (int)(p1.first - p2.first);
}
});
}
/////////////////////////////////////////////////////////////
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
static void println(long c) {
out.println(c);
}
static void print(long c) {
out.print(c);
}
static void print(int c) {
out.print(c);
}
static void println(int x) {
out.println(x);
}
static void print(String s) {
out.print(s);
}
static void println(String s) {
out.println(s);
}
static void println(boolean b) {
out.println(b);
}
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());
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 9cdc742df730ff1b78d8c1d3eff56b11 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
// static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
while(t-->0)
{
int n = sc.nextInt();
int k = sc.nextInt();
char arr[] = sc.next().toCharArray();
int ans[] = new int[n];
if(k % 2 == 0)
{
for(int i=0;i<n;i++)
{
if(k > 0 && arr[i] == '0')
{
arr[i] = '1';
k--;
ans[i]++;
}
}
if(k % 2 == 1)
{
ans[n - 1]++;
arr[n - 1] = (arr[n - 1] == '0') ? '1' : '0';
k--;
}
ans[0] += k;
System.out.println(arr);
for(int x:ans)
{
System.out.print(x+" ");
}
System.out.println();
}
else
{
k -= 1;
for(int i=0;i<n-1;i++)
{
if(k > 0 && arr[i] == '1')
{
arr[i] = '0';
k--;
ans[i]++;
}
}
if(k % 2 == 1)
{
if(arr[n - 1] == '0')
{
arr[n - 1] = '1';
ans[n - 1]++;
}
else
{
arr[n - 2] = '1';
ans[n - 2]++;
}
k--;
}
ans[0] += k;
int index = n - 1;
for(int i=0;i<n;i++)
{
if(arr[i] == '1')
{
index = i;
break;
}
}
ans[index]++;
for(int i=0;i<n;i++)
{
if(i != index)
{
arr[i] = (arr[i] == '1') ? '0' : '1';
}
}
System.out.println(arr);
for(int x:ans)
{
System.out.print(x+" ");
}
System.out.println();
}
}
}
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());
}
float nextFloat()
{
return Float.parseFloat(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;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
static class FenwickTree
{
//Binary Indexed Tree
//1 indexed
public int[] tree;
public int size;
public FenwickTree(int size)
{
this.size = size;
tree = new int[size+5];
}
public void add(int i, int v)
{
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
private static long 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;long 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 long mergeSortAndCount(int[] arr, int l,
int r)
{
// Keeps track of the inversion count at a
// particular node of the recursion tree
long 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 void my_sort(long[] arr)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list);
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
static void reverse_sorted(int[] arr)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
list.add(arr[i]);
}
Collections.sort(list , Collections.reverseOrder());
for(int i=0;i<arr.length;i++)
{
arr[i] = list.get(i);
}
}
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(ArrayList<Integer> list, int x) {// x is the key or target value
int l=-1,r=list.size();
while(l+1<r) {
int m=(l+r)>>>1;
if(list.get(m)<=x) l=m;
else r=m;
}
return l+1;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static class Queue_Pair implements Comparable<Queue_Pair> {
int first , second;
public Queue_Pair(int first, int second) {
this.first=first;
this.second=second;
}
public int compareTo(Queue_Pair o) {
return Integer.compare(o.first, first);
}
}
static void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
static boolean isPalindrome(String str)
{
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
static boolean palindrome_array(char arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
public static void reverse_Long(long[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
long temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
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 (long i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static long getSum(long n)
{
long sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
if (b == 0)
return a;
return gcdLong(b, a % b);
}
static void swap(int i, int j)
{
int temp = i;
i = j;
j = temp;
}
static int countDigit(int n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
}
class Pair
{
int first;
int second;
Pair(int first , int second)
{
this.first = first;
this.second = second;
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 24ccd5e885660d288b3eece3b36ce368 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.util.*;
import java.io.*;
public class Main {
public static Scanner obj=new Scanner(System.in);
public static PrintWriter out=new PrintWriter(System.out);
public static void main(String[] args)
{
int len=obj.nextInt();
while(len--!=0)
{
int n=obj.nextInt();
int k=obj.nextInt();
String s=obj.next();
char[] c=s.toCharArray();
int[] ans=new int[n];
int op=k;
int i=0;
for(i=0;i<n && op>0 ;i++)
{
if(c[i]=='1' )
{
if(k%2!=0) {
ans[i]+=1;
op--;
}
}
if(c[i]=='0') {
if(k%2==0)
{
ans[i]+=1;
op--;
}
c[i]='1';
}
}
if(i==n)
{
if(op%2!=0) c[n-1]='0';
ans[n-1]+=op;
}
else {
if(k%2!=0)
for(;i<n;i++)
{
if(c[i]=='1')c[i]='0';
else c[i]='1';
}
}
out.println(c);
for(i=0;i<n;i++)out.print(ans[i]+" ");
out.println();
}
out.flush();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 7c030e98d82ee264eb037ecc85671e65 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
public class E {
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;
}
}
static class Pair<T extends Comparable<T>, U extends Comparable<U>>
implements Comparable<Pair<T, U>>{
// note first and second must not change
T first;
U second;
private int hashCode;
public Pair(T first, U second) {
this.first = first;
this.second = second;
hashCode = Objects.hash(first, second);
}
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair p = (Pair) o;
return first.equals(p.first) && second.equals(p.second);
}
public int hashCode() {
return hashCode;
}
public int compareTo(Pair<T, U> o) {
return first.compareTo(o.first);
}
}
public static void main(String args[]) {
Reader rd = new Reader();
int tcs = rd.nextInt();
for (int tc = 1; tc <= tcs; tc++) {
int n = rd.nextInt();
long kk = rd.nextLong();
long k = kk;
String s = rd.next();
/*
if (k % 2 != 0) {
StringBuilder rev = new StringBuilder();
for (int i = 0; i < n; i++) {
rev.append(s.charAt(i) == '0' ? "1" : "0");
}
s = rev.toString();
}
*/
long[] o = new long[n];
for (int i = 0; i < n && k > 0; i++) {
if ((kk + s.charAt(i) - '0') % 2 == 0) {
o[i] = 1;
k--;
}
}
o[n - 1] += k;
StringBuilder a = new StringBuilder();
for (int i = 0; i < n; i++) {
long add = (o[i] + kk + s.charAt(i) - '0') % 2;
a.append(Long.toString(add));
}
System.out.println(a.toString());
a.setLength(0);
for (int i = 0; i < n; i++) {
a.append(Long.toString(o[i]));
a.append(" ");
}
System.out.println(a.toString());
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | ae6a14244873cc6098941107c749ce03 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | // package div_2_782;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class B{
public static void main(String[] args){
FastReader sc = new FastReader();
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int k=sc.nextInt();
char c[]=sc.next().toCharArray();
int ans[]=new int [n];
if((k&1)==0) {
StringBuilder s= new StringBuilder();
for(int i=0;i<n;i++) {
int cnt=0;
if(i==n-1) {
if((k&1)==0) {
if(c[i]=='0')s.append('0');
else s.append('1');
}else {
if(c[i]=='0')s.append('1');
else s.append('0');
}
cnt=k;
}
else if(c[i]=='0') {
if(k>0) {
cnt=1;
s.append('1');
k--;
}else s.append('0');
}
else s.append('1');
ans[i]=cnt;
}
System.out.println(s);
}else {
StringBuilder s= new StringBuilder();
for(int i=0;i<n;i++) {
int cnt=0;
if(i==n-1) {
if((k&1)==0) {
if(c[i]=='0')s.append('1');
else s.append('0');
}else {
if(c[i]=='0')s.append('0');
else s.append('1');
}
cnt=k;
}
else if(c[i]=='1') {
if(k>0) {
cnt=1;
s.append('1');
k--;
}else s.append('0');
}
else s.append('1');;
ans[i]=cnt;
}
System.out.println(s);
}
print(ans);
}
}
static int pow(int a,int b) {
if(b==0)return 1;
if(b==1)return a;
return a*pow(a,b-1);
}
static class pair {
int x;int y;
pair(int x,int y){
this.x=x;
this.y=y;
}
}
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;
int 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]==1) {
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 int [] sieve_of_Eratosthenes(int n) {
int prime[]=new int [n];
Arrays.fill(prime, 1); // 1-prime | 0-not prime
prime[0]=prime[1]=0;
for(int i=2;i<n;i++) {
if(prime[i]==1) {
for(int j=i*i;j<n;j+=i) {
prime[j]=0;
}
}
}
return prime;
}
static long binpow(long a,long b) {
long res=1;
if(b==0)return 1;
if(a==0)return 0;
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 long rolling_hashcode(String s ,int st,int end,long Hashcode,int n) {
if(end>=s.length()) return -1;
int mod=1000000007;
Hashcode=Hashcode-(s.charAt(st-1)*(long)Math.pow(27,n-1));
Hashcode*=10;
Hashcode=(Hashcode+(long)s.charAt(end))%mod;
return Hashcode;
}
static long hashcode(String s,int n) {
long code=0;
for(int i=0;i<n;i++) {
code+=((long)s.charAt(i)*(long)Math.pow(27, n-i-1)%1000000007);
}
return code;
}
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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | fbede8b0f0635b6427abbaad5ded5a9c | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Main {
static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter pw = new PrintWriter(System.out);;
public static void main(String[] args) throws IOException {
int T = Integer.parseInt(in.readLine());
while (T-- > 0) {
solve();
}
pw.close();
}
private static void solve() throws IOException {
String input = in.readLine();
int n = Integer.parseInt(input.split(" ")[0]);
int k = Integer.parseInt(input.split(" ")[1]);
String str = in.readLine();
int[] ans = new int[n];
int time = k;
if(k%2==0){
for (int i = 0; i < n && time > 0; i++) {
if(str.charAt(i) == '0'){
ans[i]++;
time--;
}
}
}else {
for (int i = 0; i < n && time > 0; i++) {
if(str.charAt(i) == '1'){
ans[i]++;
time--;
}
}
}
if(time > 0){
ans[n - 1] += time;
}
if(k%2==0){
for (int i = 0; i < n - 1; i++) {
if(str.charAt(i) == '0' && ans[i] == 1){
pw.print("1");
}else {
pw.print(str.charAt(i));
}
}
}else {
for (int i = 0; i < n - 1; i++) {
if(str.charAt(i) == '1' && ans[i] == 0){
pw.print("0");
}else{
pw.print('1');
}
}
}
if((k - ans[n - 1]) % 2 == 0){
pw.println(str.charAt(n - 1));
}else {
pw.println(str.charAt(n - 1) == '1'?'0':'1');
}
for (int i = 0; i < n; i++){
pw.print(ans[i] + " ");
}
pw.println();
pw.flush();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 25a2cb3390e72821913d1b77fa45dfc9 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class D {
static FastScanner fs = new FastScanner();
static Scanner scn = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
static StringBuilder sb = new StringBuilder("");
public static void main(String[] args) {
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int k = fs.nextInt();
String s = fs.next();
int r = k;
int[] a = new int[n];
for (int i = 0; i < n; i++) {
if (s.charAt(i) == '1' && k % 2 == 1 && r > 0) {
a[i]++;
r--;
}
if (s.charAt(i) == '0' && k % 2 == 0 && r > 0) {
a[i]++;
r--;
}
}
a[n-1] += Math.max(r, 0);
for (int i = 0; i < n; i++) {
if (a[i] % 2 == 1 && s.charAt(i) == '1' && k % 2 == 1) sb.append("1");
if (a[i] % 2 == 0 && s.charAt(i) == '0' && k % 2 == 1) sb.append("1");
if (a[i] % 2 == 0 && s.charAt(i) == '1' && k % 2 == 0) sb.append("1");
if (a[i] % 2 == 1 && s.charAt(i) == '0' && k % 2 == 0) sb.append("1");
if (a[i] % 2 == 1 && s.charAt(i) == '1' && k % 2 == 0) sb.append("0");
if (a[i] % 2 == 0 && s.charAt(i) == '1' && k % 2 == 1) sb.append("0");
if (a[i] % 2 == 0 && s.charAt(i) == '0' && k % 2 == 0) sb.append("0");
if (a[i] % 2 == 1 && s.charAt(i) == '0' && k % 2 == 1) sb.append("0");
}
sb.append("\n");
for (int i = 0; i < n; i++) sb.append(a[i] + " ");
sb.append("\n");
}
pw.print(sb.toString());
pw.close();
}
static void sort(int[] a) {
ArrayList<Integer> al = new ArrayList<>();
for (int i : a)
al.add(i);
Collections.sort(al);
for (int i = 0; i < a.length; i++)
a[i] = al.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) {}return st.nextToken();}
int nextInt() {return Integer.parseInt(next());}
long nextLong() {return Long.parseLong(next());}
double nextDouble() {return Double.parseDouble(next());}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 2be4433caa06dcab5a95fe1a0d630ab5 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.lang.*;
import java.util.*;
public class ComdeFormces {
public static int cc2;
public static pair pr;
public static long sum;
public static void main(String[] args) throws Exception{
// TODO Auto-generated method stub
// Reader.init(System.in);
FastReader sc=new FastReader();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
// OutputStream out = new BufferedOutputStream ( System.out );
int t=sc.nextInt();
int tc=1;
while(t--!=0) {
int n=sc.nextInt();
int k=sc.nextInt();
char a[]=sc.next().toCharArray();
int cnt1=0;
ArrayList<Integer> ar=new ArrayList<>();
char b[]=a.clone();
int mov[]=new int[n];
int cnt=0;
int i=0;
int ptr=0;
if(k%2!=0) {
if(b[0]=='0') {
int ct=0;
for( i=0;i<b.length;i++) {
if(b[i]=='0') {
b[i]='1';
}
else {
if(ct==0) {
ptr=i;
mov[i]=1;
ct++;
continue;
}
b[i]='0';
}
}
if(ct==0) {
b[n-1]='0';
mov[n-1]=1;
}
}
else if(b[0]=='1') {
mov[0]=1;
for( i=1;i<b.length;i++) {
if(b[i]=='0') {
b[i]='1';
}
else {
b[i]='0';
}
}
}
k--;
}
cnt=0;
i=0;
for( i=0;i<b.length;i++) {
if(b[i]=='0')ar.add(i);
}
for(i=0;i<ar.size();i+=2) {
if(i+1<ar.size() && cnt+2<=k) {
mov[ar.get(i)]++;
mov[ar.get(i+1)]++;
b[ar.get(i)]='1';
b[ar.get(i+1)]='1';
cnt+=2;
}
else break;
}
if(k-cnt>0) {
int p2=n-1;
int p1=0;
int vl=k-cnt;
while(p1<p2) {
while(p1<n && b[p1]!='0')p1++;
while(p2>p1 && b[p2]!='1')p2--;
if(p1<p2 && vl>0) {
mov[p1]++;
mov[p2]++;
b[p1]='1';
b[p2]='0';
vl-=2;
}
else break;
}
mov[ptr]+=vl;
}
for(i=0;i<b.length;i++)log.write(b[i]+"");
log.write("\n");
for(i=0;i<n;i++)log.write(mov[i]+" ");
log.write("\n");
log.flush();
}
}
static long eval(ArrayList<ArrayList<Integer>> ar,int src,long f[], boolean vis[]) {
long mn=Integer.MAX_VALUE;
vis[src]=true;
for(int p:ar.get(src)) {
if(!vis[p]) {
long s=eval(ar,p,f,vis);
mn=Math.min(mn,s);
sum+=s;
}
}
if(src==0)return 0;
if(mn==Integer.MAX_VALUE)return f[src];
sum-=mn;
return Math.max(f[src], mn);
}
public static void radixSort(int a[]) {
int n=a.length;
int res[]=new int[n];
int p=1;
for(int i=0;i<=8;i++) {
int cnt[]=new int[10];
for(int j=0;j<n;j++) {
a[j]=res[j];
cnt[(a[j]/p)%10]++;
}
for(int j=1;j<=9;j++) {
cnt[j]+=cnt[j-1];
}
for(int j=n-1;j>=0;j--) {
res[cnt[(a[j]/p)%10]-1]=a[j];
cnt[(a[j]/p)%10]--;
}
p*=10;
}
}
static int bits(long n) {
int ans=0;
while(n!=0) {
if((n&1)==1)ans++;
n>>=1;
}
return ans;
}
static long flor(ArrayList<Long> ar,long el) {
int s=0;
int e=ar.size()-1;
while(s<=e) {
int m=s+(e-s)/2;
if(ar.get(m)==el)return ar.get(m);
else if(ar.get(m)<el)s=m+1;
else e=m-1;
}
return e>=0?e:-1;
}
public static int kadane(int a[]) {
int sum=0,mx=Integer.MIN_VALUE;
for(int i=0;i<a.length;i++) {
sum+=a[i];
mx=Math.max(mx, sum);
if(sum<0) sum=0;
}
return mx;
}
public static int m=1000000007;
public static int mul(int a, int b) {
return ((a%m)*(b%m))%m;
}
public static long mul(long a, long b) {
return ((a%m)*(b%m))%m;
}
public static int add(int a, int b) {
return ((a%m)+(b%m))%m;
}
public static long add(long a, long b) {
return ((a%m)+(b%m))%m;
}
//debug
public static <E> void p(E[][] a,String s) {
System.out.println(s);
for(int i=0;i<a.length;i++) {
for(int j=0;j<a[0].length;j++) {
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
public static void p(int[] a,String s) {
System.out.print(s+"=");
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
public static void p(long[] a,String s) {
System.out.print(s+"=");
for(int i=0;i<a.length;i++)System.out.print(a[i]+" ");
System.out.println();
}
public static <E> void p(E a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(ArrayList<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(LinkedList<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(HashSet<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(Stack<E> a,String s){
System.out.println(s+"="+a);
}
public static <E> void p(Queue<E> a,String s){
System.out.println(s+"="+a);
}
//utils
static ArrayList<Integer> divisors(int n){
ArrayList<Integer> ar=new ArrayList<>();
for (int i=2; i<=Math.sqrt(n); i++){
if (n%i == 0){
if (n/i == i) {
ar.add(i);
}
else {
ar.add(i);
ar.add(n/i);
}
}
}
return ar;
}
static int primeDivisor(int n){
ArrayList<Integer> ar=new ArrayList<>();
int cnt=0;
boolean pr=false;
while(n%2==0) {
pr=true;
ar.add(2);
n/=2;
}
for(int i=3;i*i<=n;i+=2) {
pr=false;
while(n%i==0) {
n/=i;
ar.add(i);
pr=true;
}
}
if(n>2) ar.add(n);
return ar.size();
}
static long gcd(long a,long b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static int gcd(int a,int b) {
if(b==0)return a;
else return gcd(b,a%b);
}
static long factmod(long n,long mod,long img) {
if(n==0)return 1;
long ans=1;
long temp=1;
while(n--!=0) {
if(temp!=img) {
ans=((ans%mod)*((temp)%mod))%mod;
}
temp++;
}
return ans%mod;
}
static int ncr(int n, int r){
if(r>n-r)r=n-r;
int ans=1;
for(int i=0;i<r;i++){
ans*=(n-i);
ans/=(i+1);
}
return ans;
}
public static class trip{
int a,b;
int c;
public trip(int a,int b,int c) {
this.a=a;
this.b=b;
this.c=c;
}
public int compareTo(trip q) {
return this.b-q.b;
}
}
static void mergesort(long[] a,int start,int end) {
if(start>=end)return ;
int mid=start+(end-start)/2;
mergesort(a,start,mid);
mergesort(a,mid+1,end);
merge(a,start,mid,end);
}
static void merge(long[] a, int start,int mid,int end) {
int ptr1=start;
int ptr2=mid+1;
long b[]=new long[end-start+1];
int i=0;
while(ptr1<=mid && ptr2<=end) {
if(a[ptr1]<=a[ptr2]) {
b[i]=a[ptr1];
ptr1++;
i++;
}
else {
b[i]=a[ptr2];
ptr2++;
i++;
}
}
while(ptr1<=mid) {
b[i]=a[ptr1];
ptr1++;
i++;
}
while(ptr2<=end) {
b[i]=a[ptr2];
ptr2++;
i++;
}
for(int j=start;j<=end;j++) {
a[j]=b[j-start];
}
}
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{
int a;
int b;
public pair(int a,int b) {
this.a=a;
this.b=b;
}
public int compareTo(pair b) {
return this.a-b.a;
}
// public int compareToo(pair b) {
// return this.b-b.b;
// }
@Override
public String toString() {
return "{"+this.a+" "+this.b+"}";
}
}
static long pow(long a, long pw) {
long temp;
if(pw==0)return 1;
temp=pow(a,pw/2);
if(pw%2==0)return temp*temp;
return a*temp*temp;
}
static int pow(int a, int pw) {
int temp;
if(pw==0)return 1;
temp=pow(a,pw/2);
if(pw%2==0)return temp*temp;
return a*temp*temp;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 523ca7e2d99a18997931d88a98b230ff | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
public class codeforces2 {
static List<Integer> primes;
static final long X = 10000000000L;
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter pw = new PrintWriter(System.out);
// primes = sieveOfEratosthenes(100_000);
// System.out.println(primes.toString());
int tc = sc.ni();
// int tc = 1;
for (int rep = 0; rep < tc; rep++) {
int N = sc.ni();
int K = sc.ni();
String s=sc.next();
// int[] arr = sc.intArray(N);
pw.println(solve(N,K, s));
}
pw.close();
}
public static String solve(final int N, int K, String s) {
if (K == 0) {
int[] bruh = new int[N];
StringBuilder sb = new StringBuilder();
sb.append(s + "\n");
for (int x : bruh) sb.append(x + " ");
return sb.substring(0,sb.length()-1);
}
final int orig = K;
char[] S = s.toCharArray();
int[] arr = new int[N];
StringBuilder newstr = new StringBuilder();
for (int i = 0; i < N; i++) {
if ((S[i] == '1' && orig % 2 == 1) || (S[i] == '0' && orig % 2 == 0)) {
K--;
arr[i]++;
newstr.append('1');
if (K == 0) {
if (orig % 2 == 1) newstr.append(flip(s.substring(i+1, s.length())));
else newstr.append(s.substring(i+1, s.length()));
break;
}
}
else newstr.append('1');
}
// if (K > 0) {
// if (K % 2 == 1) {
// arr[N-1]++;
// }
if (K % 2==1) {
newstr.deleteCharAt(newstr.length() - 1);
newstr.append(0);
}
arr[N-1] += K;
// }
StringBuilder sb = new StringBuilder();
sb.append(newstr);
sb.append("\n");
for (int x : arr) sb.append(x + " ");
return sb.substring(0, sb.length() - 1);
}
static String flip(String s) {
StringBuilder sb = new StringBuilder();
for (char c : s.toCharArray()) {
if (c == '0') sb.append('1');
else sb.append('0');
}
return sb.toString();
}
static int charint(char c) {
return (int)(c) - (int)('a');
}
static long stringHash(String s) {
switch (s.length()) {
case 1:
return ((long)((int)(s.charAt(0)) - (int)('a')));
case 2:
return (getHash(charint(s.charAt(0)), charint(s.charAt(1))));
default:
return (getHash( charint(s.charAt(0)) , charint(s.charAt(1)) , charint(s.charAt(2)) ));
}
}
static long getHash(int a,int b, int c) {
long fault = 1_00000;
return ((a*fault)+b)*fault + c;
}
static long getHash(int a, int b) {
long fault = 1_00000;
return a*fault+b;
}
static boolean isSet(long n, int bit) {
if (((1 << bit) & n) > 0) return true;
return false;
}
static long nextPrime(long input){
long counter;
input++;
while(true){
int l = (int) Math.sqrt(input);
counter = 0;
for(long i = 2; i <= l; i ++){
if(input % i == 0) counter++;
}
if(counter == 0)
return input;
else{
input++;
continue;
}
}
}
public static List<Integer> sieveOfEratosthenes(int n) {
boolean prime[] = new boolean[n + 1];
Arrays.fill(prime, true);
for (int p = 2; p * p <= n; p++) {
if (prime[p]) {
for (int i = p * 2; i <= n; i += p) {
prime[i] = false;
}
}
}
List<Integer> primeNumbers = new LinkedList<>();
for (int i = 2; i <= n; i++) {
if (prime[i]) {
primeNumbers.add(i);
}
}
return primeNumbers;
}
public static boolean perfectSquare(long n) {
long lo = 0;
long hi = n;
while (lo < hi) {
long k = (lo + hi) / 2;
if (k * k < n)
lo = k + 1;
else
hi = k;
}
return (lo * lo == n);
}
static Set<Integer> divisors(int n) {
Set<Integer> set = new HashSet();
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
// If divisors are equal, print only one
if (n/i == i)
set.add(i);
else {// Otherwise print both
set.add(i);
set.add(n / i);
}
}
}
return set;
}
static Map<Integer, Integer> primeFactorization(int x) {
//first divide by 2
Map<Integer, Integer> map = new HashMap();
if (x == 0) return map;
int count = 0;
while (x % 2 == 0) {
x /=2;
count++;
}
//insert 2
if (count > 0) map.put(2, count);
for (int divisor = 3; divisor * divisor <= x; divisor += 2) {
int cnt = 0;
while (x % divisor == 0) {
x /= divisor;
cnt++;
}
if (cnt > 0) map.put(divisor, cnt);
}
if (x > 1) {
map.put(x, 1);
}
return map;
}
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;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
// method to return LCM of two numbers
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
public static void sort(int[] arr) {
Random rgen = new Random();
for (int i = 0; i < arr.length; i++) {
int r = rgen.nextInt(arr.length);
int temp = arr[i];
arr[i] = arr[r];
arr[r] = temp;
}
Arrays.sort(arr);
}
public static void sort(long[] arr) {
Random rgen = new Random();
for (int i = 0; i < arr.length; i++) {
int r = rgen.nextInt(arr.length);
long temp = arr[i];
arr[i] = arr[r];
arr[r] = temp;
}
Arrays.sort(arr);
}
/* */
//printing methods
/* */
//WOW!
/* */
public static void printArr(PrintWriter pw, int[] arr) {
StringBuilder sb = new StringBuilder();
for (int x : arr) {
sb.append(x + "");
}
sb.setLength(sb.length() - 1);
pw.println(sb.toString());
}
public static void printArr2d(PrintWriter pw, int[][] arr) {
StringBuilder sb = new StringBuilder();
for (int[] row : arr) {
for (int x : row) {
sb.append(x + " ");
}
sb.setLength(sb.length() - 1);
sb.append("\n");
}
sb.setLength(sb.length() - 1);
pw.println(sb.toString());
}
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in), 32768);
st = null;
}
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());
}
int[] intArray(int N) {
int[] ret = new int[N];
for (int i = 0; i < N; i++)
ret[i] = ni();
return ret;
}
long nl() {
return Long.parseLong(next());
}
long[] longArray(int N) {
long[] ret = new long[N];
for (int i = 0; i < N; i++)
ret[i] = nl();
return ret;
}
double nd() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
class UnionFind {
int size;
private int[] id;
public UnionFind(int size) {
this.size = size;
id = new int[size];
for (int i = 0; i < id.length; i++) {
id[i] = i;
}
}
public int find(int a) {
if (id[a] != a) {
id[a] = find(id[a]);
}
return id[a];
}
public void union(int a, int b) {
int r1 = find(a);
int r2 = find(b);
if (r1 == r2) return;
size--;
id[r1] = r2;
}
@Override
public String toString() {
return Arrays.toString(id);
}
}
class isSame {
private long[] arr;
TreeMap<Integer, Integer> map = new TreeMap();
public isSame(long[] in) {
arr = in;
int prev = 0;
for (int i = 1; i < arr.length; i++) {
if (arr[i] == arr[i-1]) {
}
else {
map.put(prev, i-1);
prev = i;
}
}
map.put(prev, arr.length - 1);
}
public boolean query(int l, int r) {
Integer lower = map.floorKey(l);
//should never happen i think
if (lower == null) return false;
return map.get(lower) >= r;
}
@Override
public String toString() {
return map.toString();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | bf1f2687d6e3c7f25f466af04b8fa999 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.PrintWriter;
import java.util.*;
public class B_Bit_Flipping {
static Scanner s = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
static void solve() {
int n = s.nextInt();
int k = s.nextInt();
String str = s.next();
int[] arr = new int[n];
StringBuilder temp = new StringBuilder();
int c1 = 0;
int c0 = 0;
for (int i = 0; i < str.length() - 1; i++) {
if (str.charAt(i) == '1')
c1++;
else
c0++;
}
if (k % 2 == 0) {
for (int i = 0; i < n; i++) {
if (k == 0) {
temp.append(str.charAt(i));
continue;
}
if (k > 0 && i == n - 1) {
// System.out.println("hello" + k);
arr[i] = k;
if (c0 % 2 == 0)
temp.append(str.charAt(i));
else
temp.append(1 - (str.charAt(i) - '0'));
continue;
}
if (str.charAt(i) == '0') {
k--;
arr[i] = 1;
}
temp.append("1");
}
} else {
for (int i = 0; i < n; i++) {
if (k == 0) {
temp.append(1 - (str.charAt(i) - '0'));
continue;
}
if (k > 0 && i == n - 1) {
arr[i] = k;
if (c1 % 2 == 0)
temp.append(str.charAt(i));
else
temp.append(1 - (str.charAt(i) - '0'));
continue;
}
if (str.charAt(i) == '1') {
k--;
arr[i] = 1;
}
temp.append("1");
}
}
out.println(temp);
for (int i = 0; i < arr.length; i++) {
out.print(arr[i] + " ");
}
out.println();
out.flush();
}
public static void main(String[] args) {
int t = s.nextInt();
while (t-- > 0) {
solve();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 708f8ded38dcfaff1c11bf2e428ccf05 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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 B_Bit_Flipping {
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 k = f.nextInt();
int ops = k;
StringBuilder sb = new StringBuilder(f.nextLine());
int arr[] = new int[n];
if(k%2 == 0) {
for(int i = 0; i < n && k > 0; i++) {
if(sb.charAt(i) == '0') {
arr[i] = 1;
k--;
}
}
arr[n-1] += k;
for(int i = 0; i < n; i++) {
if((ops-arr[i])%2 == 1) {
sb.setCharAt(i, (char)(sb.charAt(i)^'1'^'0'));
}
}
} else {
for(int i = 0; i < n && k > 0; i++) {
if(sb.charAt(i) == '1') {
arr[i] = 1;
k--;
}
}
arr[n-1] += k;
for(int i = 0; i < n; i++) {
if((ops-arr[i])%2 == 1) {
sb.setCharAt(i, (char)(sb.charAt(i)^'1'^'0'));
}
}
}
out.println(sb);
for(int i = 0; i < n; i++) {
out.print(arr[i] + " ");
}
out.println();
}
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);
}
}
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 + " ");
}
}
}
}
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;
}
public static int gcd(int a, int b) {
int dividend = a > b ? a : b;
int divisor = a < b ? a : b;
while(divisor > 0) {
int reminder = dividend % divisor;
dividend = divisor;
divisor = reminder;
}
return dividend;
}
public static int lcm(int a, int b) {
int lcm = gcd(a, b);
int hcf = (a * b) / lcm;
return hcf;
}
public static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
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());
}
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
*/
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 0cd6f075476b962bfdf4d7fa24dd590e | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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 B_Bit_Flipping {
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 k = f.nextInt();
StringBuilder sb = new StringBuilder(f.nextLine());
int arr[] = new int[n];
if(k%2 == 1) {
boolean flag = true;
for(int i = 0; i < n; i++) {
if(sb.charAt(i) == '1' && flag) {
arr[i]++;
flag = false;
} else {
sb.setCharAt(i, (char)(sb.charAt(i)^'0'^'1'));
}
}
if(flag) {
arr[n-1]++;
sb.setCharAt(n-1, '0');
}
k--;
}
for(int i = 0; i < n && k > 0; i++) {
if(sb.charAt(i) == '0') {
sb.setCharAt(i, '1');
arr[i]++;
k--;
}
}
if(k%2 == 1) {
sb.setCharAt(n-1, '0');
}
arr[n-1] += max(k, 0);
out.println(sb);
for(int i = 0; i < n; i++) {
out.print(arr[i] + " ");
}
out.println();
}
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);
}
}
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 + " ");
}
}
}
}
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;
}
public static int gcd(int a, int b) {
int dividend = a > b ? a : b;
int divisor = a < b ? a : b;
while(divisor > 0) {
int reminder = dividend % divisor;
dividend = divisor;
divisor = reminder;
}
return dividend;
}
public static int lcm(int a, int b) {
int lcm = gcd(a, b);
int hcf = (a * b) / lcm;
return hcf;
}
public static String sortString(String inputString) {
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
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());
}
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
*/
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 225aa2d4827ae4dc663693ebdb3c29ff | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
public class Solution{
public static void main(String[] args) {
TaskA solver = new TaskA();
int t = in.nextInt();
for (int i = 1; i <= t ; i++) {
solver.solve(i, in, out);
}
// solver.solve(1, in, out);
out.flush();
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n= in.nextInt();
int k= in.nextInt();
char[]c=in.next().toCharArray();
if(n==1) {
println((char)c[0]-'0');
println(k);
return;
}
int []ans=new int[n];int k1=k;
for(int i=0;i<n&&k1>0;i++) {
if(k%2==c[i]-'0') {
ans[i]=1;k1--;
}
}
ans[n-1]+=k1;
for(int i=0;i<n;i++) {
if((k-ans[i])%2==1) {
if(c[i]=='0') {
c[i]='1';
}
else {
c[i]='0';
}
}
}
println(String.valueOf(c));
println(ans);
}
}
static int ceil(int a,int b) {
int ans=a/b;if(a%b!=0) {
ans++;
}
return ans;
}
static int query(int l,int r) {
System.out.println("? "+l+" "+r);
System.out.print ("\n");
int x=in.nextInt();
return x;
}
static boolean check(Pair[]arr,int mid,int n) {
int c=0;
for(int i=0;i<n;i++) {
if(mid-1-arr[i].first<=c&&c<=arr[i].second) {
c++;
}
}
return c>=mid;
}
static long sum(int[]arr) {
long s=0;
for(int x:arr) {
s+=x;
}
return s;
}
static long sum(long[]arr) {
long s=0;
for(long x:arr) {
s+=x;
}
return s;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
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);
}
static void println(int[][]arr) {
for(int i=0;i<arr.length;i++) {
for(int j=0;j<arr[0].length;j++) {
print(arr[i][j]+" ");
}
print("\n");
}
}
static void println(long[][]arr) {
for(int i=0;i<arr.length;i++) {
for(int j=0;j<arr[0].length;j++) {
print(arr[i][j]+" ");
}
print("\n");
}
}
static void println(int[]arr){
for(int i=0;i<arr.length;i++) {
print(arr[i]+" ");
}
print("\n");
}
static void println(long[]arr){
for(int i=0;i<arr.length;i++) {
print(arr[i]+" ");
}
print("\n");
}
static long[]input(int n){
long[]arr=new long[n];
for(int i=0;i<n;i++) {
arr[i]=in.nextInt();
}
return arr;
}
static long[]input(){
long n= in.nextInt();
long[]arr=new long[(int)n];
for(int i=0;i<n;i++) {
arr[i]=in.nextLong();
}
return arr;
}
////////////////////////////////////////////////////////
static class Pair {
long first;
long second;
Pair(long x, long y)
{
this.first = x;
this.second = y;
}
}
static void sortS(Pair arr[])
{
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return (int)(p1.second - p2.second);
}
});
}
static void sortF(Pair arr[])
{
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
return (int)(p1.first - p2.first);
}
});
}
/////////////////////////////////////////////////////////////
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
static void println(long c) {
out.println(c);
}
static void print(long c) {
out.print(c);
}
static void print(int c) {
out.print(c);
}
static void println(int x) {
out.println(x);
}
static void print(String s) {
out.print(s);
}
static void println(String s) {
out.println(s);
}
static void println(boolean b) {
out.println(b);
}
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());
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 0b4208b39d359b018f16c6fb779c384c | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class JavaApplication {
static BufferedReader in;
static StringTokenizer st;
static String token;
String getLine() throws IOException {
return in.readLine();
}
String getToken() throws IOException {
if (st == null || !st.hasMoreTokens())
st = new StringTokenizer(getLine());
return st.nextToken();
}
int getInt() throws IOException {
return Integer.parseInt(getToken());
}
long getLong() throws IOException {
return Long.parseLong(getToken());
}
char getChar() throws IOException {
if (token == null || token.length() == 0)
token = getToken();
char r = token.charAt(0);
token = token.substring(1);
return r;
}
public void Solve() throws IOException {
int t = getInt();
while (t-- > 0) {
int n = getInt();
int k = getInt();
String s = getLine();
char str[] = s.toCharArray();
int f[] = new int[n];
if (k % 2 == 1) {
int i = 0;
while (i < n && (str[i] != '1'))
i++;
if (i == n)
i--;
for (int j = 0; j < n; j++)
if (j != i) {
str[j] = (char) ('1' + '0' - str[j]);
}
f[i]++;
k--;
}
int i = 0;
while (k > 0 && i < n) {
if (str[i] == '0') {
str[i] = '1';
f[i]++;
k--;
}
i++;
}
if (k % 2 == 1) {
i--;
int j = n - 1;
while (j >= 0 && str[j] == '0')
j--;
if (i < n && j > i - 1) {
str[j] = '0';
f[j]++;
k--;
}
}
f[n - 1] += k;
StringBuilder ans = new StringBuilder("");
for (int w = 0; w < n; w++) {
ans.append(str[w]);
}
System.out.println(ans);
for (int w = 0; w < n; w++) {
System.out.print(f[w] + " ");
}
System.out.println();
}
}
public static void main(String[] args) throws java.lang.Exception {
in = new BufferedReader(new InputStreamReader(System.in));
new JavaApplication().Solve();
return;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 05e32a52087084dc05d1c8ac42695143 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | // import static java.lang.System.out;
import static java.lang.Math.*;
import static java.lang.System.*;
import java.lang.reflect.Array;
import java.net.CookieHandler;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static FastReader sc;
static long mod=((long)1e9)+7;
// static FastWriter out;
static FastWriter out;
public static void main(String args[]) throws IOException {
// initializeIO();
out=new FastWriter();
sc=new FastReader();
long startTimeProg=System.currentTimeMillis();
long endTimeProg=Long.MAX_VALUE;
int t=sc.nextInt();
// int t=1;
// boolean[] seave=sieveOfEratosthenes((int)(1e5));
// int[] seave2=sieveOfEratosthenesInt((int)(1e5));
while(t--!=0) {
int n=sc.nextInt();
long k=sc.nextLong();
String s=sc.next();
long[] arr=new long[n];
chk(arr,s,n,k);
}
endTimeProg=System.currentTimeMillis();
debug("[finished : "+(endTimeProg-startTimeProg)+".ms ]");
// System.out.println(String.format("%.9f", max));
}
private static void chk(long[] arr, String s, int n, long k) throws IOException {
List<Integer> one=new ArrayList<>();
List<Integer> zero=new ArrayList<>();
for(int i=0;i<n;i++){
if(s.charAt(i)=='0'){
zero.add(i);
}else{
one.add(i);
}
}
int count=0;
if(k%2==0){
for(int i=0;i<min(zero.size(),k);i++){
arr[zero.get(i)]=1;
count++;
}
}else{
for(int i=0;i<min(one.size(),k);i++){
arr[one.get(i)]=1;
count++;
}
}
arr[n-1]+=k-count;
// debug(count);
// debug(arr);
StringBuilder sb=new StringBuilder(s);
for(int i=0;i<n;i++){
if(k%2!=0){
if(arr[i]%2==0){
sb.setCharAt(i,sb.charAt(i)=='0'?'1':'0');
}
}else{
if(arr[i]%2!=0){
sb.setCharAt(i,sb.charAt(i)=='0'?'1':'0');
}
}
}
print(sb.toString());
out.print(arr);
}
public static boolean isNumeric(String strNum) {
if (strNum == null) {
return false;
}
try {
double d = Double.parseDouble(strNum);
} catch (NumberFormatException nfe) {
return false;
}
return true;
}
private static boolean isSorted(List<Integer> li){
int n=li.size();
if(n<=1)return true;
for(int i=0;i<n-1;i++){
if(li.get(i)>li.get(i+1))return false;
}
return true;
}
public static long log2(long N)
{
// calculate log2 N indirectly
// using log() method
long result = (long)(Math.log(N) / Math.log(2));
return result;
}
static boolean isPowerOfTwo(long x)
{
return x != 0 && ((x & (x - 1)) == 0);
}
private static boolean ispallindromeList(List<Integer> res){
int l=0,r=res.size()-1;
while(l<r){
if(res.get(l)!=res.get(r))return false;
l++;
r--;
}
return true;
}
private static class Pair{
int first=0;
int sec=0;
int[] arr;
char ch;
Map<Integer,Integer> map;
Pair(int first,int sec){
this.first=first;
this.sec=sec;
}
Pair(int[] arr){
this.map=new HashMap<>();
for(int x:arr) this.map.put(x,map.getOrDefault(x,0)+1);
this.arr=arr;
}
Pair(char ch,int first){
this.ch=ch;
this.first=first;
}
}
private static int sizeOfSubstring(int st,int e){
int s=e-st+1;
return (s*(s+1))/2;
}
private static Set<Long> factors(long n){
Set<Long> res=new HashSet<>();
// res.add(n);
for (long i=1; i*i<=(n); i++){
if (n%i==0){
res.add(i);
if (n/i != i){
res.add(n/i);
}
}
}
return res;
}
private static long fact(long n){
if(n<=2)return n;
return n*fact(n-1);
}
private static long ncr(long n,long r){
return fact(n)/(fact(r)*fact(n-r));
}
private static int lessThen(long[] nums,long val,boolean work,int l,int r){
int i=-1;
if(work)i=0;
while(l<=r){
int mid=l+((r-l)/2);
if(nums[mid]<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int lessThen(List<Long> nums,long val,boolean work,int l,int r){
int i=-1;
if(work)i=0;
while(l<=r){
int mid=l+((r-l)/2);
if(nums.get(mid)<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int lessThen(List<Integer> nums,int val,boolean work,int l,int r){
int i=-1;
if(work)i=0;
while(l<=r){
int mid=l+((r-l)/2);
if(nums.get(mid)<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int greaterThen(List<Long> nums,long val,boolean work,int l,int r){
int i=-1;
if(work)i=r;
while(l<=r){
int mid=l+((r-l)/2);
if(nums.get(mid)>=val){
i=mid;
r=mid-1;
}else{
l=mid+1;
}
}
return i;
}
private static int greaterThen(List<Integer> nums,int val,boolean work){
int i=-1,l=0,r=nums.size()-1;
if(work)i=r;
while(l<=r){
int mid=l+((r-l)/2);
if(nums.get(mid)>=val){
i=mid;
r=mid-1;
}else{
l=mid+1;
}
}
return i;
}
private static int greaterThen(long[] nums,long val,boolean work,int l,int r){
int i=-1;
if(work)i=r;
while(l<=r){
int mid=l+((r-l)/2);
if(nums[mid]>=val){
i=mid;
r=mid-1;
}else{
l=mid+1;
}
}
return i;
}
private static long gcd(long[] arr){
long ans=0;
for(long x:arr){
ans=gcd(x,ans);
}
return ans;
}
private static int gcd(int[] arr){
int ans=0;
for(int x:arr){
ans=gcd(x,ans);
}
return ans;
}
private static long sumOfAp(long a,long n,long d){
long val=(n*( 2*a+((n-1)*d)));
return val/2;
}
//geometrics
private static double areaOftriangle(double x1,double y1,double x2,double y2,double x3,double y3){
double[] mid_point=midOfaLine(x1,y1,x2,y2);
// debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" "+y3);
double height=distanceBetweenPoints(mid_point[0],mid_point[1],x3,y3);
double wight=distanceBetweenPoints(x1,y1,x2,y2);
// debug(height+" "+wight);
return (height*wight)/2;
}
private static double distanceBetweenPoints(double x1,double y1,double x2,double y2){
double x=x2-x1;
double y=y2-y1;
return (Math.pow(x,2)+Math.pow(y,2));
}
public static boolean isPerfectSquareByUsingSqrt(long n) {
if (n <= 0) {
return false;
}
double squareRoot = Math.sqrt(n);
long tst = (long)(squareRoot + 0.5);
return tst*tst == n;
}
private static double[] midOfaLine(double x1,double y1,double x2,double y2){
double[] mid=new double[2];
mid[0]=(x1+x2)/2;
mid[1]=(y1+y2)/2;
return mid;
}
private static long sumOfN(long n){
return (n*(n+1))/2;
}
private static long power(long x,long y,long p){
long res = 1; // Initialize result
x = x % p; // 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) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
/* Function to calculate x raised to the power y in O(logn)*/
static long power(long x, long y){
long temp;
if( y == 0)
return 1l;
temp = power(x, y / 2);
if (y % 2 == 0)
return (temp*temp);
else
return (x*temp*temp);
}
private static StringBuilder reverseString(String s){
StringBuilder sb=new StringBuilder(s);
int l=0,r=sb.length()-1;
while(l<=r){
char ch=sb.charAt(l);
sb.setCharAt(l,sb.charAt(r));
sb.setCharAt(r,ch);
l++;
r--;
}
return sb;
}
private static void swap(List<Integer> li,int i,int j){
int t=li.get(i);
li.set(i,li.get(j));
li.set(j,t);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b){
return (a / gcd(a, b)) * b;
}
private static String decimalToString(int x){
return Integer.toBinaryString(x);
}
private static String decimalToString(long x){
return Long.toBinaryString(x);
}
private static boolean isPallindrome(String s,int l,int r){
while(l<r){
if(s.charAt(l)!=s.charAt(r))return false;
l++;
r--;
}
return true;
}
private static boolean isSubsequence(String s, String t) {
if (s == null || t == null) return false;
Map<Character, List<Integer>> map = new HashMap<>(); //<character, index>
//preprocess t
for (int i = 0; i < t.length(); i++) {
char curr = t.charAt(i);
if (!map.containsKey(curr)) {
map.put(curr, new ArrayList<Integer>());
}
map.get(curr).add(i);
}
int prev = -1; //index of previous character
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (map.get(c) == null) {
return false;
} else {
List<Integer> list = map.get(c);
prev = lessThen( list, prev,false,0,list.size()-1);
if (prev == -1) {
return false;
}
prev++;
}
}
return true;
}
private static StringBuilder removeLeadingZero(StringBuilder sb){
int i=0;
while(i<sb.length()&&sb.charAt(i)=='0')i++;
// debug("remove "+i);
if(i==sb.length())return new StringBuilder();
return new StringBuilder(sb.substring(i,sb.length()));
}
private static StringBuilder removeEndingZero(StringBuilder sb){
int i=sb.length()-1;
while(i>=0&&sb.charAt(i)=='0')i--;
// debug("remove "+i);
if(i<0)return new StringBuilder();
return new StringBuilder(sb.substring(0,i+1));
}
private static int stringToDecimal(String binaryString){
// debug(decimalToString(n<<1));
return Integer.parseInt(binaryString,2);
}
private static int stringToInt(String s){
return Integer.parseInt(s);
}
private static String toString(long val){
return String.valueOf(val);
}
private static void debug(int[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(long[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(List<int[]> arr){
for(int[] a:arr){
err.println(Arrays.toString(a));
}
}
private static void debug(float[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(double[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(boolean[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void print(String s)throws IOException {
out.println(s);
}
private static void debug(String s)throws IOException {
err.println(s);
}
private static int charToIntS(char c){
return ((((int)(c-'0'))%48));
}
private static void print(double s)throws IOException {
out.println(s);
}
private static void print(float s)throws IOException {
out.println(s);
}
private static void print(long s)throws IOException {
out.println(s);
}
private static void print(int s)throws IOException {
out.println(s);
}
private static void debug(double s)throws IOException {
err.println(s);
}
private static void debug(float s)throws IOException {
err.println(s);
}
private static void debug(long s){
err.println(s);
}
private static void debug(int s){
err.println(s);
}
private 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;
}
private static List<List<Integer>> readUndirectedGraph(int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
int x=sc.nextInt();
int y=sc.nextInt();
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readUndirectedGraph(int[][] intervals,int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<intervals.length;i++){
int x=intervals[i][0];
int y=intervals[i][1];
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int[][] intervals,int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<intervals.length;i++){
int x=intervals[i][0];
int y=intervals[i][1];
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
int x=sc.nextInt();
int y=sc.nextInt();
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
static String[] readStringArray(int n){
String[] arr=new String[n];
for(int i=0;i<n;i++){
arr[i]=sc.next();
}
return arr;
}
private static Map<Character,Integer> freq(String s){
Map<Character,Integer> map=new HashMap<>();
for(char c:s.toCharArray()){
map.put(c,map.getOrDefault(c,0)+1);
}
return map;
}
private static Map<Long,Integer> freq(long[] arr){
Map<Long,Integer> map=new HashMap<>();
for(long x:arr){
map.put(x,map.getOrDefault(x,0)+1);
}
return map;
}
private static Map<Integer,Integer> freq(int[] arr){
Map<Integer,Integer> map=new HashMap<>();
for(int x:arr){
map.put(x,map.getOrDefault(x,0)+1);
}
return map;
}
static boolean[] sieveOfEratosthenes(long n){
boolean prime[] = new boolean[(int)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;
}
}
return prime;
}
static int[] sieveOfEratosthenesInt(long n){
boolean prime[] = new boolean[(int)n + 1];
Set<Integer> li=new HashSet<>();
for (int i = 1; i <= n; i++){
prime[i] = true;
li.add(i);
}
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true){
for (int i = p * p; i <= n; i += p){
li.remove(i);
prime[i] = false;
}
}
}
int[] arr=new int[li.size()];
int i=0;
for(int x:li){
arr[i++]=x;
}
return arr;
}
public static long Kadens(List<Long> prices) {
long sofar=0;
long max_v=0;
for(int i=0;i<prices.size();i++){
sofar+=prices.get(i);
if (sofar<0) {
sofar=0;
}
max_v=Math.max(max_v,sofar);
}
return max_v;
}
public static int Kadens(int[] prices) {
int sofar=0;
int max_v=0;
for(int i=0;i<prices.length;i++){
sofar+=prices[i];
if (sofar<0) {
sofar=0;
}
max_v=Math.max(max_v,sofar);
}
return max_v;
}
static boolean isMemberAC(int a, int d, int x){
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
static boolean isMemberAC(long a, long d, long x){
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
private static void sort(int[] arr){
int n=arr.length;
List<Integer> li=new ArrayList<>();
for(int x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(int[] arr){
int n=arr.length;
List<Integer> li=new ArrayList<>();
for(int x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sort(double[] arr){
int n=arr.length;
List<Double> li=new ArrayList<>();
for(double x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(double[] arr){
int n=arr.length;
List<Double> li=new ArrayList<>();
for(double x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(long[] arr){
int n=arr.length;
List<Long> li=new ArrayList<>();
for(long x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sort(long[] arr){
int n=arr.length;
List<Long> li=new ArrayList<>();
for(long x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static long sum(int[] arr){
long sum=0;
for(int x:arr){
sum+=x;
}
return sum;
}
private static long sum(long[] arr){
long sum=0;
for(long x:arr){
sum+=x;
}
return sum;
}
private static long evenSumFibo(long n){
long l1=0,l2=2;
long sum=0;
while (l2<n) {
long l3=(4*l2)+l1;
sum+=l2;
if(l3>n)break;
l1=l2;
l2=l3;
}
return sum;
}
private static void initializeIO(){
try {
System.setIn(new FileInputStream("input"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
System.setErr(new PrintStream(new FileOutputStream("error.txt")));
} catch (Exception e) {
System.err.println(e.getMessage());
}
}
private static int maxOfArray(int[] arr){
int max=Integer.MIN_VALUE;
for(int x:arr){
max=Math.max(max,x);
}
return max;
}
private static long maxOfArray(long[] arr){
long max=Long.MIN_VALUE;
for(long x:arr){
max=Math.max(max,x);
}
return max;
}
private static int[][] readIntIntervals(int n,int m){
int[][] arr=new int[n][m];
for(int j=0;j<n;j++){
for(int i=0;i<m;i++){
arr[j][i]=sc.nextInt();
}
}
return arr;
}
private static long gcd(long a,long b){
if(b==0)return a;
return gcd(b,a%b);
}
private static int gcd(int a,int b){
if(b==0)return a;
return gcd(b,a%b);
}
private static int[] readIntArray(int n){
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
return arr;
}
private static double[] readDoubleArray(int n){
double[] arr=new double[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextDouble();
}
return arr;
}
private static long[] readLongArray(int n){
long[] arr=new long[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextLong();
}
return arr;
}
private static void print(int[] arr)throws IOException {
out.println(Arrays.toString(arr));
}
private static void print(long[] arr)throws IOException {
out.println(Arrays.toString(arr));
}
private static void print(String[] arr)throws IOException {
out.println(Arrays.toString(arr));
}
private static void print(double[] arr)throws IOException {
out.println(Arrays.toString(arr));
}
private static void debug(String[] arr){
err.println(Arrays.toString(arr));
}
private static void debug(Boolean[][] arr){
for(int i=0;i<arr.length;i++)err.println(Arrays.toString(arr[i]));
}
private static void debug(int[] arr){
err.println(Arrays.toString(arr));
}
private static void debug(long[] arr){
err.println(Arrays.toString(arr));
}
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
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());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
BufferedWriter bw;
List<String> list = new ArrayList<>();
Set<String> set = new HashSet<>();
FastWriter() {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
<T> void print(T obj) throws IOException {
bw.write(obj.toString());
bw.flush();
}
void println() throws IOException {
print("\n");
}
<T> void println(T obj) throws IOException {
print(obj.toString() + "\n");
}
void print(int[] arr)throws IOException {
for(int x:arr){
print(x+" ");
}
println();
}
void print(long[] arr)throws IOException {
for(long x:arr){
print(x+" ");
}
println();
}
void print(double[] arr)throws IOException {
for(double x:arr){
print(x+" ");
}
println();
}
void printCharN(char c, int n) throws IOException {
for (int i = 0; i < n; i++) {
print(c);
}
}
}
static class Dsu {
int[] parent, size;
Dsu(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
private int findParent(int u) {
if (parent[u] == u) return u;
return parent[u] = findParent(parent[u]);
}
private boolean union(int u, int v) {
// System.out.println("uf "+u+" "+v);
int pu = findParent(u);
// System.out.println("uf2 "+pu+" "+v);
int pv = findParent(v);
// System.out.println("uf3 " + u + " " + pv);
if (pu == pv) return false;
if (size[pu] <= size[pv]) {
parent[pu] = pv;
size[pv] += size[pu];
} else {
parent[pv] = pu;
size[pu] += size[pv];
}
return true;
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 5cdf9779266fe7d31a7aab70e2cd9658 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class B {
static class RealScanner {
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());
}
}
public static void main(String[] args) {
RealScanner sc = new RealScanner();
int t = sc.nextInt();
while (t-- > 0) {
int n, k;
n = sc.nextInt();
k = sc.nextInt();
char[] s = sc.next().toCharArray();
if (k % 2 != 0) {
for (int i = 0; i < n; i++) {
if (s[i] == '1') {
s[i] = '0';
} else {
s[i] = '1';
}
}
}
int[] res = new int[n];
for (int i = 0; i < n; i++) {
if (s[i] == '0' && k > 0) {
s[i] = '1';
k--;
res[i]++;
}
// if (k == 0) {
// break;
// }
}
if (k % 2 != 0) {
s[n - 1] = '0';
}
System.out.println(String.valueOf(s));
for (int i = 0; i < n - 1; i++) {
System.out.print(res[i] + " ");
}
System.out.println(res[n - 1] + k);
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 5574ee3cbde697359ae8d44ae4dc3065 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Question4 {
static boolean multipleTC = true;
final static int Mod = 1000000007;
final static int Mod2 = 998244353;
final double PI = 3.14159265358979323846;
int MAX = 1000000007;
void pre() throws Exception {
}
long knapSack(long W, long wt[], long val[], int n){
long []dp = new long[(int) (W + 1)];
for (int i = 1; i < n + 1; i++) {
for (long w = W; w >= 0; w--) {
if (wt[i - 1] <= w)
dp[(int) w] = Math.max(dp[(int) w], dp[(int) (w - wt[i - 1])] + val[i - 1]);
}
}
return dp[(int) W];
}
void solve(int t) throws Exception {
int n = ni();
long k = nl();
long temp = k;
char arr[] = n().toCharArray();
long is1 = 0;
for(int i=0;i<n;i++)
if(arr[i] == '1')
is1++;
int ind = -1;
if(k%2 != 0) {
if(is1 > 0) {
boolean flag = false;
for(int i=0;i<n;i++) {
if(arr[i] == '1' && !flag) {
ind = i;
flag = true;
continue;
}
else {
if(arr[i] == '1')
arr[i] = '0';
else
arr[i] = '1';
}
}
}
else {
for(int i=0;i<n;i++)
arr[i] = '1';
ind = (n-1);
arr[n-1] = '0';
}
k--;
temp--;
}
long cnt1 = 0, cnt0 = 0;
for(int i=0;i<n;i++) {
if(arr[i] == '1')
cnt1++;
else
cnt0++;
}
StringBuilder ans = new StringBuilder();
StringBuilder op = new StringBuilder();
for(int i=0;i<n;i++) {
if(arr[i] == '0') {
if(temp>0) {
ans.append('1');
temp--;
}
else
ans.append('0');
}
else
ans.append('1');
}
temp = k;
if(k>=cnt0) {
if(cnt0%2 != 0) {
ans.deleteCharAt(n-1);
ans.append(0);
}
}
long oparr[] = new long[n];
if(ind != -1)
oparr[ind]++;
for(int i=0;i<n;i++) {
if(temp > 0) {
if(arr[i] == '0') {
oparr[i]++;
temp--;
}
}
}
if(k >= cnt0) {
if(cnt0%2 != 0) {
oparr[n-1]++;
temp--;
}
}
oparr[0] += temp;
pn(ans);
for(int i=0;i<n;i++)
op.append(oparr[i] + " ");
pn(op);
}
double dist(int x1, int y1, int x2, int y2) {
double a = x1 - x2, b = y1 - y2;
return Math.sqrt((a * a) + (b * b));
}
int[] readArr(int n) throws Exception {
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = ni();
}
return arr;
}
void sort(int arr[], int left, int right) {
ArrayList<Integer> list = new ArrayList<>();
for (int i = left; i <= right; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = left; i <= right; i++)
arr[i] = list.get(i - left);
}
void sort(int arr[]) {
ArrayList<Integer> list = new ArrayList<>();
for (int i = 0; i < arr.length; i++)
list.add(arr[i]);
Collections.sort(list);
for (int i = 0; i < arr.length; i++)
arr[i] = list.get(i);
}
public long max(long... arr) {
long max = arr[0];
for (long itr : arr)
max = Math.max(max, itr);
return max;
}
public int max(int... arr) {
int max = arr[0];
for (int itr : arr)
max = Math.max(max, itr);
return max;
}
public long min(long... arr) {
long min = arr[0];
for (long itr : arr)
min = Math.min(min, itr);
return min;
}
public int min(int... arr) {
int min = arr[0];
for (int itr : arr)
min = Math.min(min, itr);
return min;
}
public long sum(long... arr) {
long sum = 0;
for (long itr : arr)
sum += itr;
return sum;
}
public long sum(int... arr) {
long sum = 0;
for (int itr : arr)
sum += itr;
return sum;
}
String bin(long n) {
return Long.toBinaryString(n);
}
String bin(int n) {
return Integer.toBinaryString(n);
}
static int bitCount(int x) {
return x == 0 ? 0 : (1 + bitCount(x & (x - 1)));
}
static void dbg(Object... o) {
System.err.println(Arrays.deepToString(o));
}
int bit(long n) {
return (n == 0) ? 0 : (1 + bit(n & (n - 1)));
}
int abs(int a) {
return (a < 0) ? -a : a;
}
long abs(long a) {
return (a < 0) ? -a : a;
}
void p(Object o) {
out.print(o);
}
void pn(Object o) {
out.println(o);
}
void pni(Object o) {
out.println(o);
out.flush();
}
void pn(int[] arr) {
int n = arr.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
sb.append(arr[i] + " ");
}
pn(sb);
}
void pn(long[] arr) {
int n = arr.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
sb.append(arr[i] + " ");
}
pn(sb);
}
String n() throws Exception {
return in.next();
}
String nln() throws Exception {
return in.nextLine();
}
int ni() throws Exception {
return Integer.parseInt(in.next());
}
long nl() throws Exception {
return Long.parseLong(in.next());
}
double nd() throws Exception {
return Double.parseDouble(in.next());
}
public static void main(String[] args) throws Exception {
new Question4().run();
}
FastReader in;
PrintWriter out;
void run() throws Exception {
in = new FastReader();
out = new PrintWriter(System.out);
int T = (multipleTC) ? ni() : 1;
pre();
for (int t = 1; t <= T; t++)
solve(t);
out.flush();
out.close();
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception {
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
throw new Exception(e.toString());
}
return str;
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 2ff9f36d597b82dd0c3d5931909d8478 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
static FastReader sc = new FastReader();
public static void main (String[] args) throws java.lang.Exception
{
int t = sc.nextInt();
while(t-->0){
solve(sc);
}
}
public static void solve(FastReader sc) throws IOException{
int n = i();int k = i();String s = s();
if(k==0 ){
out.println(s);
for(int i = 0;i<n;++i){
out.print(0 + " ");
}
out.println();
out.flush();return;
}
if(n==1){
out.println(s.charAt(0));out.println(k);out.flush();return;
}
char[] str = s.toCharArray();
int[] arr = new int[n];
int deter = 0;
int z = 0;
for(;z<n&&k>0;++z){
char ch = str[z];
if(deter%2!=0){
if(ch == '0') ch = '1';
else ch = '0';
}
if(ch == '0'){
if(k%2==0){
arr[z]=1;
k--;deter++;
}
}else{
if(k%2!=0){
arr[z]=1;
k--;deter++;
}
}
str[z]='1';
}
if(k!=0){
arr[n-1]+=k;
if(k%2!=0){
str[n-1]='0';
}
}
if(z!=n && deter%2!=0){
for(int j = z;j<n;++j){
if(str[j]=='0'){
str[j]='1';
}else{
str[j] = '0';
}
}
}
for(int i = 0;i<n;++i){
out.print(str[i]);
}out.println();
for(int i = 0;i<n;++i){
out.print(arr[i] + " ");
}
out.println();
out.flush();
}
/*
int [] arr = new int[n];
for(int i = 0;i<n;++i){
arr[i] = sc.nextInt();
}
*/
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 class Pair implements Comparable<Pair>{
int ind;int val;
Pair(int ind, int val){
this.ind=ind;
this.val=val;
}
public int compareTo(Pair o){
return this.val-o.val;
}
}
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | da8bd8e2dc27dae807be760ef8d16ed2 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | /*
"Everything in the universe is balanced. Every disappointment
you face in life will be balanced by something good for you!
Keep going, never give up."
Just have Patience + 1...
*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Solution {
static long INF = (long) 1e18;
public static void main(String[] args) throws java.lang.Exception {
out = new PrintWriter(new BufferedOutputStream(System.out));
sc = new FastReader();
int test = sc.nextInt();
for (int t = 1; t <= test; t++) {
solve();
}
out.close();
}
private static void solve() {
int n = sc.nextInt();
int k = sc.nextInt();
char[] arr = sc.next().toCharArray();
int[] operationsPerformed = new int[n];
int remainingOperations = k;
for (int i = 0; i < n - 1 && remainingOperations > 0; i++) {
// you need to perform even flips on it so that it remains '1', that means on the remaining string we do even operations.
if (arr[i] == '1' && k % 2 == 1) {
operationsPerformed[i] = 1;
remainingOperations--;
}
// to make arr[i] = '1' you need to perform odd flips on it, that means on the remaining string we do odd operations.
if (arr[i] == '0' && k % 2 == 0) {
operationsPerformed[i] = 1;
remainingOperations--;
}
}
operationsPerformed[n - 1] = remainingOperations;
for (int i = 0; i < n; i++) {
if ((k - operationsPerformed[i]) % 2 == 1) {
if (arr[i] == '0') {
arr[i] = '1';
}else {
arr[i] = '0';
}
}
}
out.println(String.valueOf(arr));
for (int i = 0; i < n; i++) {
out.print(operationsPerformed[i] + " ");
}
out.println();
}
public static FastReader sc;
public static PrintWriter out;
static class FastReader
{
BufferedReader br;
StringTokenizer str;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (str == null || !str.hasMoreElements())
{
try
{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
}
return str.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 lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 0329b60a100a69b1499b5ad66e91d468 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | /*
"Everything in the universe is balanced. Every disappointment
you face in life will be balanced by something good for you!
Keep going, never give up."
Just have Patience + 1...
*/
import java.util.*;
import java.lang.*;
import java.io.*;
public class Solution {
public static void main(String[] args) throws java.lang.Exception {
out = new PrintWriter(new BufferedOutputStream(System.out));
sc = new FastReader();
int test = sc.nextInt();
for (int t = 1; t <= test; t++) {
solve();
}
out.close();
}
private static void solve() {
int n = sc.nextInt();
int k = sc.nextInt();
int remaining = k;
String s = sc.next();
char[] from = s.toCharArray();
char[] to = s.toCharArray();
int[] res = new int[n];
int bit = 0;
for (int i = 0; i < n && k > 0; i++) {
from[i] ^= bit;
if (k % 2 != from[i] - '0' || i == n - 1) {
continue;
}
bit ^= 1;
res[i]++;
k--;
}
res[n - 1] += k;
for (int i = 0; i < n; i++) {
if ((remaining - res[i]) % 2 == 1) {
to[i] ^= 1;
}
}
out.println(String.valueOf(to));
for (int freq : res) {
out.print(freq + " ");
}
out.println();
}
public static FastReader sc;
public static PrintWriter out;
static class FastReader
{
BufferedReader br;
StringTokenizer str;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (str == null || !str.hasMoreElements())
{
try
{
str = new StringTokenizer(br.readLine());
}
catch (IOException lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
}
return str.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 lastMonthOfVacation)
{
lastMonthOfVacation.printStackTrace();
}
return str;
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | a527a7e6e7d67976d4bd0ee37811feed | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class B_Bit_Flipping {
public static void main(String[] args) {
FastReader rd = new FastReader();
StringBuilder bd = new StringBuilder();
int t = rd.nextInt();
while(t!=0) {
bd.setLength(0);
int n = rd.nextInt(), k = rd.nextInt();
String s = rd.nextLine();
int c1 = 0, c0 = 0;
for(int i=0;i<n-1;i++) {
if(s.charAt(i) == '0') c0++;
else c1++;
}
int a[] = new int[n];
if(k%2==0) {
if(k<=c0) {
for(int i=0;i<n && k>0;i++) {
if(s.charAt(i) == '0') {
a[i] = 1;
k--;
}
}
for(int i=0;i<n;i++) {
if(a[i] == 0) bd.append(s.charAt(i));
else {
if(s.charAt(i) == '0') bd.append(1);
else bd.append(0);
}
}
bd.append("\n");
for(int i=0;i<n;i++) {
bd.append(a[i] + " ");
}
bd.append("\n");
}
else {
for(int i=0;i<n-1;i++) {
if(s.charAt(i) == '0') a[i] = 1;
}
a[n-1] = k-c0;
for(int i=0;i<n-1;i++) {
if(a[i] == 0) bd.append(s.charAt(i));
else {
if(s.charAt(i) == '0') bd.append(1);
else bd.append(0);
}
}
if(c0%2==0) {
bd.append(s.charAt(n-1));
}
else {
if(s.charAt(n-1) == '0') bd.append(1);
else bd.append(0);
}
bd.append("\n");
for(int i=0;i<n;i++) {
bd.append(a[i] + " ");
}
bd.append("\n");
}
}
else {
if(k<=c1) {
for(int i=0;i<n && k>0;i++) {
if(s.charAt(i) == '1') {
a[i] = 1;
k--;
}
}
for(int i=0;i<n;i++) {
if(a[i] == 0) {
if(s.charAt(i) == '0') bd.append(1);
else bd.append(0);
}
else {
bd.append(s.charAt(i));
}
}
bd.append("\n");
for(int i=0;i<n;i++) {
bd.append(a[i] + " ");
}
bd.append("\n");
}
else {
for(int i=0;i<n-1;i++) {
if(s.charAt(i) == '1') a[i] = 1;
}
a[n-1] = k-c1;
for(int i=0;i<n-1;i++) {
if(a[i] == 0) {
if(s.charAt(i) == '0') bd.append(1);
else bd.append(0);
}
else {
bd.append(s.charAt(i));
}
}
if(c1%2==0) {
bd.append(s.charAt(n-1));
}
else {
if(s.charAt(n-1) == '0') bd.append(1);
else bd.append(0);
}
bd.append("\n");
for(int i=0;i<n;i++) {
bd.append(a[i] + " ");
}
bd.append("\n");
}
}
out(bd.toString());
t--;
}
}
public static void out(Object obj) {
System.out.print(obj);
}
public static void outn(Object obj) {
System.out.println(obj);
}
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;
}
}
}
// class Pair {
// int first;
// int second;
// public Pair(int first, int second) {
// this.first = first;
// this.second = second;
// }
// }
// class Compare {
// static void compare(Pair arr[], int PRIORITY) {
// Arrays.sort(arr, new Comparator<Pair>(){
// @Override
// public int compare(Pair p1, Pair p2) {
// if(PRIORITY == 1) return p1.second - p2.second;
// else return p1.first - p2.first;
// }
// });
// }
// }
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 7ba14c60a6a8310e9569cd9e5ec459b0 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | // Source: https://usaco.guide/general/io
import java.io.*;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
StringTokenizer st = new StringTokenizer(br.readLine());
int T= Integer.parseInt(st.nextToken());
int n,k,count;
String input= "";
StringBuilder result=new StringBuilder();
StringBuilder secondResult=new StringBuilder();
for (int i=0; i<T; i++){
st= new StringTokenizer(br.readLine());
n= Integer.parseInt(st.nextToken());
k= Integer.parseInt(st.nextToken());
st= new StringTokenizer(br.readLine());
input= st.nextToken();
count=k;
for (int z=0; z<n-1; z++){
if ((k + (int) input.charAt(z))%2 == 1){
result.append("1");
secondResult.append( "0 ");
}else if (count!=0){
result.append("1");
count--;
secondResult.append("1 ");
}else{
result.append("0");
secondResult.append("0 ");
}
}
secondResult.append(count);
result.append(((input.charAt(n-1) + k%2+ count)%2));
pw.println(result);
pw.println(secondResult);
result= new StringBuilder();
secondResult= new StringBuilder();
}
pw.close();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | dad8fe4c3095e637cef410ef9adc8f9e | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Solution {
static boolean cases = true;
// Solution
static void solve(int t) {
int n = sc.nextInt(), k = sc.nextInt();
char a[] = sc.next().toCharArray();
char b[] = a.clone();
int ans[] = new int[n];
int x = 0, temp = k;
for (int i = 0; i < n - 1 && k > 0; i++) {
if (x == 1) { b[i] = b[i] == '1' ? '0' : '1'; }
if (k % 2 != (int) (b[i] - '0')) continue;
ans[i]++; k--; x ^= 1;
}
ans[n - 1] += k;
for (int i = 0; i < n; ++i) { if ((temp - ans[i]) % 2 != 0) { a[i] = a[i] == '1' ? '0' : '1'; } }
System.out.println(new String(a));
for (int i : ans) System.out.print(i + " ");
System.out.println();
}
public static void main(String[] args) {
int c = 1;
for (int t = (cases ? sc.nextInt() : 0); t > 1; t--, c++) solve(c);
solve(c);
}
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[] 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()); }
}
private static final FastScanner sc = new FastScanner();
private static PrintWriter out = new PrintWriter(System.out);
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | ea9bee2a366d4adc05863fdc649fe68c | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.Closeable;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.time.Clock;
import java.time.LocalDateTime;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Objects;
import java.util.Random;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.function.Supplier;
public class Solution {
private void solve() throws IOException {
int n = nextInt();
int k = nextInt();
boolean[] b = new boolean[n];
{
char[] s = nextTokenChars();
for (int i = 0; i < n; i++) {
b[i] = (s[i] == '1') ^ (k % 2 != 0);
}
}
int[] a = new int[n];
for (int i = 0; i < n - 1 && k > 0; i++) {
if (!b[i]) {
b[i] = !b[i];
a[i]++;
k--;
}
}
a[n - 1] = k;
if (k % 2 != 0) {
b[n - 1] = !b[n - 1];
}
for (boolean i : b) {
out.print(i ? 1 : 0);
}
out.println();
for (int i : a) {
out.print(i + " ");
}
out.println();
}
private static final boolean runNTestsInProd = true;
private static final boolean printCaseNumber = false;
private static final boolean assertInProd = false;
private static final boolean logToFile = false;
private static final boolean readFromConsoleInDebug = false;
private static final boolean writeToConsoleInDebug = true;
private static final boolean testTimer = false;
private static Boolean isDebug = null;
private BufferedReader in;
private StringTokenizer line;
private PrintWriter out;
public static void main(String[] args) throws Exception {
isDebug = Arrays.asList(args).contains("DEBUG_MODE");
if (isDebug) {
log = logToFile ? new PrintWriter("logs/j_solution_" + System.currentTimeMillis() + ".log") : new PrintWriter(System.out);
clock = Clock.systemDefaultZone();
}
new Solution().run();
}
private void run() throws Exception {
in = new BufferedReader(new InputStreamReader(!isDebug || readFromConsoleInDebug ? System.in : new FileInputStream("input.txt")));
out = !isDebug || writeToConsoleInDebug ? new PrintWriter(System.out) : new PrintWriter("output.txt");
try (Timer totalTimer = new Timer("total")) {
int t = runNTestsInProd || isDebug ? nextInt() : 1;
for (int i = 0; i < t; i++) {
if (printCaseNumber) {
out.print("Case #" + (i + 1) + ": ");
}
if (testTimer) {
try (Timer testTimer = new Timer("test #" + (i + 1))) {
solve();
}
} else {
solve();
}
if (isDebug) {
out.flush();
}
}
}
in.close();
out.flush();
out.close();
}
private void println(Object... objects) {
boolean isFirst = true;
for (Object o : objects) {
if (!isFirst) {
out.print(" ");
} else {
isFirst = false;
}
out.print(o.toString());
}
out.println();
}
private int[] nextIntArray(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
private long[] nextLongArray(int n) throws IOException {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextLong();
}
return res;
}
private int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
private long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
private double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private char[] nextTokenChars() throws IOException {
return nextToken().toCharArray();
}
private String nextToken() throws IOException {
while (line == null || !line.hasMoreTokens()) {
line = new StringTokenizer(in.readLine());
}
return line.nextToken();
}
private static void assertPredicate(boolean p) {
if ((isDebug || assertInProd) && !p) {
throw new RuntimeException();
}
}
private static void assertPredicate(boolean p, String message) {
if ((isDebug || assertInProd) && !p) {
throw new RuntimeException(message);
}
}
private static <T> void assertNotEqual(T unexpected, T actual) {
if ((isDebug || assertInProd) && Objects.equals(actual, unexpected)) {
throw new RuntimeException("assertNotEqual: " + unexpected + " == " + actual);
}
}
private static <T> void assertEqual(T expected, T actual) {
if ((isDebug || assertInProd) && !Objects.equals(actual, expected)) {
throw new RuntimeException("assertEqual: " + expected + " != " + actual);
}
}
private static PrintWriter log = null;
private static Clock clock = null;
private static void log(Object... objects) {
log(true, objects);
}
private static void logNoDelimiter(Object... objects) {
log(false, objects);
}
private static void log(boolean printDelimiter, Object[] objects) {
if (isDebug) {
StringBuilder sb = new StringBuilder();
sb.append(LocalDateTime.now(clock)).append(" - ");
boolean isFirst = true;
for (Object o : objects) {
if (!isFirst && printDelimiter) {
sb.append(" ");
} else {
isFirst = false;
}
sb.append(o.toString());
}
log.println(sb);
log.flush();
}
}
private static class Timer implements Closeable {
private final String label;
private final long startTime = isDebug ? System.nanoTime() : 0;
public Timer(String label) {
this.label = label;
}
@Override
public void close() throws IOException {
if (isDebug) {
long executionTime = System.nanoTime() - startTime;
String fraction = Long.toString(executionTime / 1000 % 1_000_000);
logNoDelimiter("Timer[", label, "]: ", executionTime / 1_000_000_000, '.', "00000".substring(0, 6 - fraction.length()), fraction, 's');
}
}
}
private static <T> T timer(String label, Supplier<T> f) throws Exception {
if (isDebug) {
try (Timer timer = new Timer(label)) {
return f.get();
}
} else {
return f.get();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 8412e6908e547f57f0c1751702da3616 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Codeforces{
static long mod = 1000000007L;
// map.put(a[i],map.getOrDefault(a[i],0)+1);
// map.putIfAbsent;
static ArrayList<Integer> li = new ArrayList<>();
static long[] a = new long[40005];
static MyScanner sc = new MyScanner();
//<----------------------------------------------WRITE HERE------------------------------------------->
static void solve(){
int n = sc.nextInt();
int k = sc.nextInt();
char[] c = sc.next().toCharArray();
int[] a = new int[n];
int tmpk = k;
String ans = "";
for(int i=0;i<n&& tmpk>0;i++) {
if(k%2 == c[i]-'0') {
a[i] = 1;
tmpk--;
}
}
a[n-1] += tmpk;
for(int i=0;i<n;i++) {
if((k-a[i])%2 == 1) {
c[i] = (char)('1' - (c[i]-'0'));
// out.println(('1' - (c[i]-'0'));
}
}
out.println(c);
priArr(a);
}
//<----------------------------------------------WRITE HERE------------------------------------------->
static void swap(int[] a,int i,int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
static void priArr(int[] a) {
for(int i=0;i<a.length;i++) {
out.print(a[i] + " ");
}
out.println();
}
static long gcd(long a,long b){
if(b==0) return a;
return gcd(b,a%b);
}
static int gcd(int a,int b){
if(b==0) return a;
return gcd(b,a%b);
}
static class Pair implements Comparable<Pair>{
Integer val;
Integer ind;
Pair(int v,int f){
val = v;
ind = f;
}
public int compareTo(Pair p){
int s1 = this.ind-this.val;
int s2 = p.ind-p.val;
if(s1 != s2) return s1-s2;
return this.val - p.val;
}
}
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
// int t= 1;
while(t-- >0){
solve();
}
// Stop writing your solution here. -------------------------------------
out.close();
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------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();
}
public int[] readIntArray(int n){
int arr[] = new int[n];
for(int i = 0;i<n;i++){
arr[i] = Integer.parseInt(next());
}
return arr;
}
int[] reverse(int arr[]){
int n= arr.length;
int i = 0;
int j = n-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
j--;i++;
}
return arr;
}
long[] readLongArray(int n){
long arr[] = new long[n];
for(int i = 0;i<n;i++){
arr[i] = Long.parseLong(next());
}
return arr;
}
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;
}
}
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);
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 5ac5466ed3e7209b4917f6ee8daaf3d6 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | //package com.rajan.codeforces.level_1300_1400;
import java.io.*;
public class BitFlipping {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
int tt = Integer.parseInt(reader.readLine());
while (tt-- > 0) {
String[] temp = reader.readLine().split("\\s+");
int n = Integer.parseInt(temp[0]), k = Integer.parseInt(temp[1]);
String s = reader.readLine();
int[] opFreq = new int[n];
int p = k;
for (int i = 0; i < n && p > 0; i++) {
char c = s.charAt(i);
if (c == '1') {
if (((k - 1) & 1) == 0) {
opFreq[i] = 1;
p--;
}
} else {
if (((k - 1) & 1) != 0) {
opFreq[i] = 1;
p--;
}
}
}
if ((p & 1) != 0) {
opFreq[n - 1] += 1;
p--;
}
opFreq[0] += p;
for (int i = 0; i < n; i++)
if (((k - opFreq[i]) & 1) == 0)
writer.write(s.charAt(i) + "");
else
writer.write((s.charAt(i) == '1' ? "0" : "1"));
writer.write("\n");
for (int i = 0; i < n; i++)
writer.write((i == 0 ? "" : " ") + opFreq[i]);
writer.write("\n");
}
writer.flush();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f2405c95c2197c34de8aff8b313cef4f | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.util.*;
import java.io.OutputStream;
import java.io.PrintWriter;
public class Solver {
public static void main(String[] args) {
InputStream inputStream = System.in;
InputReader in = new InputReader(inputStream);
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
Task task = new Task();
int t = in.nextInt();
while (t-- > 0) {
task.solve(in, out);
}
out.close();
}
static class Task {
public void solve(InputReader in, PrintWriter out) {
int n = in.nextInt();
int k = in.nextInt();
String s = in.next();
int[] f = new int[n];
Arrays.fill(f, 0);
int cnt = 0;
for (int i = 0; i < n && cnt < k; i++) {
if ((s.charAt(i) == '0' && k % 2 == 0) || (s.charAt(i) == '1' && k % 2 == 1)) {
f[i] = 1;
cnt += 1;
}
}
f[n - 1] += Math.max(k - cnt, 0);
for (int i = 0; i < n; i++) {
int x = (int) s.charAt(i) - '0';
if ((k - f[i]) % 2 > 0) {
x ^= 1;
}
if (i == n - 1) {
out.println(x);
} else {
out.print(x);
}
}
for (int i = 0; i < n; i++) {
out.print(f[i] + (i == n - 1 ? "\n" : " "));
}
}
}
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());
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 9ce646a3d3cf96e99c281e6e8196c4ed | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.math.*;
import java.security.KeyStore.Entry;
import java.util.*;
public class code {
private static boolean[] vis;
private static long[] dist;
private static long mod = 1000000000 + 7;
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 {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void merge(int arr[], int l, int m, int r)
{
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
int L[] = new int [n1];
int R[] = new int [n2];
/*Copy data to temp arrays*/
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
public static void sort(int arr[], int l, int r)
{
if (l < r)
{
// Find the middle point
int m = (l+r)/2;
// Sort first and second halves
sort(arr, l, m);
sort(arr , m+1, r);
// Merge the sorted halves
merge(arr, l, m, r);
}
}
private static long gcd(long n, long l) {
if (l == 0)
return n;
return gcd(l, n % l);
}
private static int gcd(int n, int l) {
if (l == 0)
return n;
return gcd(l, n % l);
}
public static void dfs(ArrayList<Integer>[] gr,int v)
{
vis[v]=true;
ArrayList<Integer> arr=gr[v];
for(int i=0;i<arr.size();i++)
{
if(!vis[arr.get(i)])
{
dfs(gr,arr.get(i));
}
}
}
private static class Pairs implements Comparable<Pairs>{
String v;int i;int j;
public Pairs(String a,int b,int c){
v=a;
i=b;
j=c;
}
@Override
public int compareTo(Pairs arg0) {
if(!this.v.equals(arg0.v))
return this.v.compareTo(arg0.v);
else
return arg0.i-this.i;
}
}
private static class Pair implements Comparable<Pair>{
int v,i;Pair j;
public Pair(int a,int b){
v=a;
i=b;
}
public Pair(int a,Pair b)
{
v=a;
j=b;
}
@Override
public int compareTo(Pair arg0) {
{
if(this.v!=arg0.v)
return this.v-arg0.v;
else
return this.i-arg0.i;
}
}
}
public static long mmi(long a, long m)
{
long m0 = m;
long y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1)
{
// q is quotient
long q = a / m;
long t = m;
// m is remainder now, process same as
// Euclid's algo
m = a % m; a = t;
t = y;
// Update y and x
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
private static long pow(long a, long b, long c) {
if (b == 0)
return 1;
long p = pow(a, b / 2, c);
p = (p * p) % c;
return (b % 2 == 0) ? p : (a * p) % c;
}
public static int ub(ArrayList<Integer> array, int low,int high, int value) {
while (low < high) {
final int mid = (low + high) / 2;
if (value >= array.get(mid)) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
public static void main(String[] args) throws Exception
{
FastReader sc = new FastReader();
BufferedWriter pw = new BufferedWriter(
new OutputStreamWriter(System.out));
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int k = sc.nextInt();
int tot = k;
String s = sc.nextLine();
int[] arr = new int[n];
int[] ans = new int[n];
for(int i=0;i<n-1;i++) {
if(s.charAt(i)=='0' && k%2==0 && tot>0) {
arr[i]=1;
ans[i]=1;
tot--;
} else if(s.charAt(i)=='1' && k%2==1 && tot>0) {
arr[i] = 1;
ans[i]=1;
tot--;
} else {
if(k%2==0) {
ans[i] = (s.charAt(i)=='1') ? 1 : 0;
} else {
ans[i] = (s.charAt(i)=='1') ? 0 : 1;
}
}
}
k-=tot;
if(k%2==0) {
ans[n-1] = (s.charAt(n-1)=='1') ? 1 : 0;
} else {
ans[n-1] = (s.charAt(n-1)=='1') ? 0 : 1;
}
arr[n-1] += tot;
for(int i=0;i<n;i++) pw.append(ans[i]+"");
pw.append("\n");
for(int i=0;i<n;i++) pw.append(arr[i]+" ");
pw.append("\n");
}
pw.flush();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | c2e8380f19a56d4b3cc5accd0c88327d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Q1659B {
static int mod = (int) (1e9 + 7);
static StringBuilder sb=new StringBuilder();
static void solve() {
int n = i();
int k = i();
char a[] = s().toCharArray();
int b[] = new int[n];
int tempk = k;
for (int i = 0; i < n && tempk > 0; i++) {
if (a[i] == '1' && k % 2 != 0) {
b[i]++;
tempk--;
}
if (a[i] == '0' && k % 2 == 0) {
b[i]++;
tempk--;
}
}
b[n-1]+=tempk;
for (int i = 0; i < n; i++) {
if(((k-b[i])%2)==1){
if(a[i]=='1'){
a[i]='0';
}else{
a[i]='1';
}
}
}
for(char val:a){
sb.append(val);
// System.out.print(val);
}
// System.out.println();
sb.append("\n");
for(int val:b){
sb.append(val+" ");
// System.out.print(val+" ");
}
sb.append("\n");
// System.out.println();
}
public static void main(String[] args) {
int test = i();
while (test-- > 0) {
solve();
}
System.out.println(sb);
}
// -----> POWER ---> long power(long x, long y) <----
// -----> LCM ---> long lcm(long x, long y) <----
// -----> GCD ---> long gcd(long x, long y) <----
// -----> SIEVE --> ArrayList<Integer> sieve(int N) <-----
// -----> NCR ---> long ncr(int n, int r) <----
// -----> BINARY_SEARCH ---> int binary_Search(long[]arr,long x) <----
// -----> (SORTING OF LONG, CHAR,INT) -->long[] sortLong(long[] a2)<--
// ----> DFS ---> void dfs(ArrayList<ArrayList<Integer>>edges,int child,int
// parent)<---
// ---> NODETOROOT --> ArrayList<Integer>
// node2Root(ArrayList<ArrayList<Integer>>edges,int child,int parent,int tofind)
// <--
// ---> LEVELS_TREE -->levels_Trees(ArrayList<HashSet<Integer>> edges, int
// child, int parent,int[]level,int currLevel) <--
// ---> K_THPARENT --> int k_thparent(int node,int[][]parent,int k) <---
// ---> TWOPOWERITHPARENTFILL --> void twoPowerIthParentFill(int[][]parent) <---
// -----> (INPUT OF INT,LONG ,STRING) -> int i() long l() String s()<-
// tempstart
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 int Int() {
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 String String() {
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 String();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static InputReader in = new InputReader(System.in);
public static int i() {
return in.Int();
}
public static long l() {
String s = in.String();
return Long.parseLong(s);
}
public static String s() {
return in.String();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 2cd469e8418433f691d2952fbc8121f4 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 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.InputStreamReader;
import java.io.InputStream;
import java.util.*;
/**
* 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);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
int testNum = in.nextInt();
solver.solve(testNum, in, out);
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
for (int z = 0; z < testNumber; z++) {
int n = in.nextInt();
int k = in.nextInt();
int[] numSelected = new int[n];
String s = in.next();
char[] sChar = new char[n];
for (int i = 0; i < n; i++) {
sChar[i] = s.charAt(i);
}
if (k%2 == 1) {
for (int i = 0; i < s.length(); i++) {
if (sChar[i] == '1') {
sChar[i] = '0';
} else {
sChar[i] = '1';
}
}
}
for (int i = 0; i < sChar.length; i++) {
if (sChar[i] == '0' && k > 0) {
sChar[i] = '1';
numSelected[i]++;
k--;
}
}
if (k > 0) {
numSelected[n-1] += k;
if (k % 2 == 1) {
sChar[n-1] = '0';
}
}
for (int i =0; i < n; i++) {
out.print(sChar[i]);
}
out.println();
for (int i = 0; i < n; i++) {
out.print(numSelected[i] + " ");
}
out.println();
}
}
}
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());
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 760e7ebdc31832ff3f56c2d3c9c22310 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.lang.*;
import java.util.*;
public class BitFlipping {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-- > 0) {
int n = s.nextInt();
int k = s.nextInt();
String st = s.next();
StringBuilder ans = new StringBuilder();
int x = k;
int[] flips = new int[n];
for(int i = 0 ; i < n ; i++) {
if(x <= 0)
break;
char ch = st.charAt(i);
if(i == n - 1) {
flips[i] = x;
}
else {
if(ch == '1') {
if(k % 2 != 0) {
flips[i] = 1;
x--;
}
}
else {
if(k % 2 == 0) {
flips[i] = 1;
x--;
}
}
}
}
for(int i = 0 ; i < n ; i++) {
char ch = st.charAt(i);
int rem = k - flips[i];
if(ch == '1') {
if(rem % 2 == 0)
ans.append("1");
else
ans.append("0");
}
else {
if(rem % 2 != 0)
ans.append("1");
else
ans.append("0");
}
}
System.out.println(ans);
StringBuilder temp = new StringBuilder();
for(int i : flips) {
temp.append(i + " ");
}
System.out.println(temp);
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 2efe1724807a20bcb954b4e147d34b62 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.util.*;
import java.io.*;
public class Main {
static StringBuilder sb;
static dsu dsu;
static long fact[];
static int mod = (int) (1e9 + 7);
static void solve() {
int n=i();
int k=i();
String s=s();
int curr=0;
int[]arr=new int[n];
int []ans=new int[n];
for(int i=0;i<n;i++){
int bit=s.charAt(i)-'0';
if(curr%2==1){
if(bit==0)bit++;
else bit--;
}
int cans=0;
if(bit==0){
int rem=k-curr;
if(rem%2==0){
if(rem!=0){
cans++;
ans[i]=1;
}
}
else ans[i]=1;
}
else{
int rem=k-curr;
if(rem%2==1){
cans=1;
ans[i]=1;
}
else{
ans[i]=1;
}
}
curr+=cans;
arr[i]=cans;
}
int rem=k-curr;
if(rem%2==0){
for(int i=0;i<n;i++){
sb.append(ans[i]+"");
}
sb.append("\n");
sb.append((arr[0]+rem)+" ");
for(int i=1;i<n;i++){
sb.append(arr[i]+" ");
}
}
else{
for(int i=0;i<n;i++){
if(i==n-1)ans[i]=0;
sb.append(ans[i]+"");
}
sb.append("\n");
for(int i=0;i<n-1;i++){
sb.append(arr[i]+" ");
}
sb.append((arr[n-1]+rem)+" ");
}
sb.append("\n");
}
public static void main(String[] args) {
sb = new StringBuilder();
int test = i();
fact=new long[(int)1e6+10];
fact[0]=fact[1]=1;
for(int i=2;i<fact.length;i++)
{ fact[i]=((long)(i%mod)*(long)(fact[i-1]%mod))%mod; }
while (test-- > 0) {
solve();
}
System.out.println(sb);
}
//**************NCR%P******************
static long ncr(int n, int r) {
if (r > n)
return (long) 0;
long res = fact[n] % mod;
// System.out.println(res);
res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod;
res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod;
// System.out.println(res);
return res;
}
static long p(long x, long y)// POWER FXN //
{
if (y == 0)
return 1;
long res = 1;
while (y > 0) {
if (y % 2 == 1) {
res = (res * x) % mod;
y--;
}
x = (x * x) % mod;
y = y / 2;
}
return res;
}
//**************END******************
// *************Disjoint set
// union*********//
static class dsu {
int parent[];
dsu(int n) {
parent = new int[n];
for (int i = 0; i < n; i++)
parent[i] = i;
}
int find(int a) {
if (parent[a] ==a)
return a;
else {
int x = find(parent[a]);
parent[a] = x;
return x;
}
}
void merge(int a, int b) {
a = find(a);
b = find(b);
if (a == b)
return;
parent[b] = a;
}
}
//**************PRIME FACTORIZE **********************************//
static TreeMap<Integer, Integer> prime(long n) {
TreeMap<Integer, Integer> h = new TreeMap<>();
long num = n;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (n % i == 0) {
int nt = 0;
while (n % i == 0) {
n = n / i;
nt++;
}
h.put(i, nt);
}
}
if (n != 1)
h.put((int) n, 1);
return h;
}
//****CLASS PAIR ************************************************
static class Pair implements Comparable<Pair> {
int x;
long y;
Pair(int x, long y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o) {
return (int) (this.y - o.y);
}
}
//****CLASS PAIR **************************************************
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 int Int() {
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 String String() {
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 String();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
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 printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
static InputReader in = new InputReader(System.in);
static OutputWriter out = new OutputWriter(System.out);
public static long[] sort(long[] a2) {
int n = a2.length;
ArrayList<Long> l = new ArrayList<>();
for (long i : a2)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a2[i] = l.get(i);
return a2;
}
public static char[] sort(char[] a2) {
int n = a2.length;
ArrayList<Character> l = new ArrayList<>();
for (char i : a2)
l.add(i);
Collections.sort(l);
for (int i = 0; i < l.size(); i++)
a2[i] = l.get(i);
return a2;
}
public static long pow(long x, long y) {
long res = 1;
while (y > 0) {
if (y % 2 != 0) {
res = (res * x);// % modulus;
y--;
}
x = (x * x);// % modulus;
y = y / 2;
}
return res;
}
//GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
public static long gcd(long x, long y) {
if (x == 0)
return y;
else
return gcd(y % x, x);
}
// ******LOWEST COMMON MULTIPLE
// *********************************************
public static long lcm(long x, long y) {
return (x * (y / gcd(x, y)));
}
//INPUT PATTERN********************************************************
public static int i() {
return in.Int();
}
public static long l() {
String s = in.String();
return Long.parseLong(s);
}
public static String s() {
return in.String();
}
public static int[] readArrayi(int n) {
int A[] = new int[n];
for (int i = 0; i < n; i++) {
A[i] = i();
}
return A;
}
public static long[] readArray(long n) {
long A[] = new long[(int) n];
for (int i = 0; i < n; i++) {
A[i] = l();
}
return A;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 93f8ff2f035cef87a9186c239a3ab2c3 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class test {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
StringTokenizer tokenizer = new StringTokenizer(br.readLine());
int n = Integer.parseInt(tokenizer.nextToken());
int k = Integer.parseInt(tokenizer.nextToken());
String bits = br.readLine();
pw.println(solve(bits, n, k));
}
pw.flush();
pw.close();
br.close();
}
public static String solve(String bits, int n, int k) {
int[] b = new int[n];
for (int i = 0; i < n; i++) {
b[i] = k;
}
int c = k;
for (int i = 0; i < n; i++) {
if (c == 0)
break;
if (k % 2 == 0 && bits.charAt(i) == '0') {
b[i]--;
c--;
} else if (k % 2 == 1 && bits.charAt(i) == '1') {
b[i]--;
c--;
}
}
b[n - 1] -= c;
StringBuilder result = new StringBuilder();
for (int i = 0; i < n; i++) {
if (b[i] % 2 != 0) {
result.append(bits.charAt(i) == '0' ? '1' : '0');
} else {
result.append(bits.charAt(i));
}
}
result.append("\n");
for (int i = 0; i < n; i++) {
if (i > 0) {
result.append(" ");
}
result.append(k - b[i]);
}
return result.toString();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | c09a54600f7b48d6dba2cde94cc21a50 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class R782B {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(System.out);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0)
{
StringTokenizer st= new StringTokenizer(br.readLine());
int n= Integer.parseInt(st.nextToken());
int k= Integer.parseInt(st.nextToken());
String s= br.readLine();
int[] count= new int[n];
StringBuilder res= new StringBuilder("");
char search='1';
if(k%2==0) {
search='0';
}
int total=k;
for(int i=0;i<n; i++)
{
if(s.charAt(i)==search&& k>0)
{
count[i]++;
res.append("1");
k--;
}else{
if(total%2==0)
res.append(s.charAt(i));
else
res.append(s.charAt(i)=='0'?'1':'0');
}
}
if(k>0)
{
if(k%2==0)
{
res.setCharAt(n-1,res.charAt(n-1));
}else
res.setCharAt(n-1,res.charAt(n-1)=='0'?'1':'0');
count[n-1]+=k;
}
pw.println(res.toString());
for(int i=0; i<n;i++)
pw.print(count[i]+" ");
pw.println();
pw.flush();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 99f12f6c6f5aaf938226a34d4c2add88 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static long mod = (int)1e9+7;
static PrintWriter out=new PrintWriter(new BufferedOutputStream(System.out));
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
// int t=1;
O : while(t-->0)
{
int n=sc.nextInt();
int k=sc.nextInt();
String s=sc.next();
int a[]=new int[n];
for(int i=0;i<n;i++)
a[i]=(s.charAt(i)=='1')?1:0;
int ans[]=new int[n];
for(int i=0;i<n;i++){
if(a[i]==1){
if(k%2!=0){
ans[i]=1;
}
}
else{
if(k%2==0){
ans[i]=1;
}
}
}
int count=0;
for(int i=0;i<n;i++){
if(ans[i]==1){
if (count<k) count++;
else ans[i]=0;
}
}
if(count<k){
ans[n-1]+=(k-count);
}
for(int i=0;i<n;i++){
if(a[i]==1){
if(k%2!=0){
if(ans[i]%2==1) out.print(1);
else out.print(0);
}
else{
if(ans[i]%2==1) out.print(0);
else out.print(1);
}
}
else{
if(k%2==0){
if(ans[i]%2==1) out.print(1);
else out.print(0);
}
else{
if(ans[i]%2==1) out.print(0);
else out.print(1);
}
}
}
out.println();
for(int i=0;i<n;i++)
out.print(ans[i]+" ");
out.println();
}
out.flush();
}
static void printN()
{
System.out.println("NO");
}
static void printY()
{
System.out.println("YES");
}
static int findfrequencies(int a[],int n)
{
int count=0;
for(int i=0;i<a.length;i++)
{
if(a[i]==n)
{
count++;
}
}
return count;
}
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());
}
float nextFloat()
{
return Float.parseFloat(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;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
public static int[] radixSort2(int[] a)
{
int n = a.length;
int[] c0 = new int[0x101];
int[] c1 = new int[0x101];
int[] c2 = new int[0x101];
int[] c3 = new int[0x101];
for(int v : a) {
c0[(v&0xff)+1]++;
c1[(v>>>8&0xff)+1]++;
c2[(v>>>16&0xff)+1]++;
c3[(v>>>24^0x80)+1]++;
}
for(int i = 0;i < 0xff;i++) {
c0[i+1] += c0[i];
c1[i+1] += c1[i];
c2[i+1] += c2[i];
c3[i+1] += c3[i];
}
int[] t = new int[n];
for(int v : a)t[c0[v&0xff]++] = v;
for(int v : t)a[c1[v>>>8&0xff]++] = v;
for(int v : a)t[c2[v>>>16&0xff]++] = v;
for(int v : t)a[c3[v>>>24^0x80]++] = v;
return a;
}
static int[] EvenOddArragement(int a[])
{
ArrayList<Integer> list=new ArrayList<>();
for(int i=0;i<a.length;i++)
{
if(a[i]%2==0)
{
list.add(a[i]);
}
}
for(int i=0;i<a.length;i++)
{
if(a[i]%2!=0)
{
list.add(a[i]);
}
}
for(int i=0;i<a.length;i++)
{
a[i]=list.get(i);
}
return a;
}
static int gcd(int a, int b) {
while (b != 0) {
int t = a;
a = b;
b = t % b;
}
return a;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
public static HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm)
{
// Create a list from elements of HashMap
List<Map.Entry<Integer, Integer> > list =
new LinkedList<Map.Entry<Integer, Integer> >(hm.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1,
Map.Entry<Integer, Integer> o2)
{
return (o1.getValue()).compareTo(o2.getValue());
}
});
// put data from sorted list to hashmap
HashMap<Integer, Integer> temp = new LinkedHashMap<Integer, Integer>();
for (Map.Entry<Integer, Integer> aa : list) {
temp.put(aa.getKey(), aa.getValue());
}
return temp;
}
static int DigitSum(int n)
{
int r=0,sum=0;
while(n>=0)
{
r=n%10;
sum=sum+r;
n=n/10;
}
return sum;
}
static boolean checkPerfectSquare(int number)
{
double sqrt=Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static boolean isPowerOfTwo(int n)
{
if(n==0)
return false;
return (int)(Math.ceil((Math.log(n) / Math.log(2)))) == (int)(Math.floor(((Math.log(n) / Math.log(2)))));
}
static boolean isPrime2(int n)
{
if (n <= 1)
{
return false;
}
if (n == 2)
{
return true;
}
if (n % 2 == 0)
{
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
static String minLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for(int i=0;i<n;i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[0];
}
static String maxLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for (int i = 0; i < n; i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[arr.length-1];
}
static class P implements Comparable<P> {
int i, j;
public P(int i, int j) {
this.i=i;
this.j=j;
}
public int compareTo(P o) {
return Integer.compare(i, o.i);
}
}
static class pair{
int i,j;
pair(int x,int y){
i=x;
j=y;
}
}
static int binary_search(int a[],int value)
{
int start=0;
int end=a.length-1;
int mid=start+(end-start)/2;
while(start<=end)
{
if(a[mid]==value)
{
return mid;
}
if(a[mid]>value)
{
end=mid-1;
}
else
{
start=mid+1;
}
mid=start+(end-start)/2;
}
return -1;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 004bbec9ec7e14269120c31a1b360e4a | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.time.*;
import static java.lang.Math.*;
@SuppressWarnings("unused")
public class A {
static boolean DEBUG = false;
static Reader fs;
static PrintWriter pw;
static void solve() {
int n = fs.nextInt(), k = fs.nextInt();
char a[] = fs.next().toCharArray();
int ans[] = new int[n];
if(k%2==1) {
for(int i = 0 ;i < n;i++) {
a[i] = a[i] == '0'?'1':'0';
}
}
for(int i = 0 ; i < n; i++) {
if(a[i] == '0' && k > 0) {
ans[i]++;
a[i] = '1';
k--;
}
}
ans[n-1] += k;
if(k%2 == 1) {
a[n-1] = '0';
}
for(int i = 0 ; i < n ; i++) {
pw.print(a[i]);
}
pw.println();
for(int i = 0 ; i < n ; i++) {
pw.print(ans[i] + " ");
}
pw.println();
}
public static void main(String[] args) throws IOException {
if (args.length == 2) {
System.setIn(new FileInputStream("D:\\program\\javaCPEclipse\\CodeForces\\src\\input.txt"));
System.setErr(new PrintStream("D:\\program\\javaCPEclipse\\CodeForces\\src\\error.txt"));
DEBUG = true;
}
Instant start = Instant.now();
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\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 3b3873b9994b8c6101d21d34fcafebe6 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Scanner;
import javax.security.auth.kerberos.KerberosCredMessage;
public class B {
public static void main(String[] args) {
/**
* before copy-pasting in to the challenge:
* comment out the "System.setIn(new FileInputStream("input.txt"));" in line 24
* rename the class to Solution
*/
//redirects the input to the file "input.txt"
try {
System.setIn(new FileInputStream("input.txt"));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
//creates a scanner and reads the first line of the input
Scanner sc = new Scanner(System.in);
int testCount = Integer.parseInt(sc.nextLine());
for (int i = 0; i < testCount; i++) {
//seperates the input by " "
String[] inputs = sc.nextLine().split(" ");
int n = Integer.parseInt(inputs[0]);
int k = Integer.parseInt(inputs[1]);
String input = sc.nextLine();
int[] c = new int[input.length()];
char[] chars = input.toCharArray();
if(k%2 == 0) {
for (int j = 0; j < chars.length; j++) {
if(input.charAt(j) == '1') {
chars[j] = '1';
}else if (k > 0) {
chars[j] = '1';
c[j]++;
k--;
}else {
chars[j] = '0';
}
}
}else {
for (int j = 0; j < chars.length; j++) {
if(input.charAt(j) == '0') {
chars[j] = '1';
}else if (k > 0) {
chars[j] = '1';
c[j]++;
k--;
}else {
chars[j] = '0';
}
}
}
if(k%2 == 1) {
chars[chars.length - 1] = chars[chars.length - 1] == '1' ? '0' : '1';
}
c[chars.length - 1]+=k;
System.out.println(String.valueOf(chars));
for (int j = 0; j < c.length; j++) {
System.out.print(c[j] + " ");
}
System.out.println();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 2f7c9d7bb8373ea8270158fd17e5fdda | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.DataInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
public class B {
static final Reader in = new Reader();
static final PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
int t = in.nextInt();
for (int tc = 1; tc <= t; tc++) {
solve();
out.println();
}
out.close();
}
static void solve() {
int n = in.nextInt(), k = in.nextInt();
char[] s = in.next().toCharArray();
int par = k % 2;
int[] freq = new int[n];
int total = 0;
for (int i = 0; i < n && total < k; i++) {
if (s[i] == '0') {
if (par == 0) {
freq[i]++;
total++;
}
}
else {
if (par == 1) {
freq[i]++;
total++;
}
}
}
if (total < k) {
freq[n - 1] += k - total;
}
for (int i = 0; i < n; i++) {
if ((k - freq[i]) % 2 == 1) {
out.print('1' - s[i]);
}
else {
out.print(s[i]);
}
}
out.println();
for (int i = 0; i < n; i++) {
out.print(freq[i] + " ");
}
}
static class Reader {
private final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
Reader(InputStream in) {
try {
din = new DataInputStream(in);
} catch (Exception e) {
throw new RuntimeException();
}
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
String next() {
int c;
while ((c = read()) != -1 && (c == ' ' || c == '\n' || c == '\r')) ;
StringBuilder s = new StringBuilder();
while (c != -1) {
if (c == ' ' || c == '\n' || c == '\r')
break;
s.append((char) c);
c = read();
}
return s.toString();
}
String nextLine() {
int c;
while ((c = read()) != -1 && (c == ' ' || c == '\n' || c == '\r')) ;
StringBuilder s = new StringBuilder();
while (c != -1) {
if (c == '\n' || c == '\r')
break;
s.append((char) c);
c = read();
}
return s.toString();
}
int nextInt() {
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;
}
int[] nextInts(int n) {
int[] ar = new int[n];
for (int i = 0; i < n; ++i)
ar[i] = nextInt();
return ar;
}
long nextLong() {
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;
}
long[] nextLongs(int n) {
long[] ar = new long[n];
for (int i = 0; i < n; ++i)
ar[i] = nextLong();
return ar;
}
double nextDouble() {
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;
}
void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
byte read() {
try {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
} catch (IOException e) {
throw new RuntimeException();
}
}
void close() throws IOException {
din.close();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | bfc161e2dad8fcf24e2f9f99077c0e2e | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | /*LoudSilence*/
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Solution {
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static FastScanner s = new FastScanner();
static FastWriter out = new FastWriter();
final static int mod = (int)1e9 + 7;
final static int INT_MAX = Integer.MAX_VALUE;
final static int INT_MIN = Integer.MIN_VALUE;
final static long LONG_MAX = Long.MAX_VALUE;
final static long LONG_MIN = Long.MIN_VALUE;
final static double DOUBLE_MAX = Double.MAX_VALUE;
final static double DOUBLE_MIN = Double.MIN_VALUE;
final static float FLOAT_MAX = Float.MAX_VALUE;
final static float FLOAT_MIN = Float.MIN_VALUE;
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
static class FastScanner{BufferedReader br;StringTokenizer st;
public FastScanner() {if(System.getProperty("ONLINE_JUDGE") == null){try {br = new BufferedReader(new FileReader("E:\\Competitive Coding\\input.txt"));}
catch (FileNotFoundException e) {br = new BufferedReader(new InputStreamReader(System.in));}}else{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());}
List<Integer> readIntList(int n){List<Integer> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextInt()); return arr;}
List<Long> readLongList(int n){List<Long> arr = new ArrayList<>(); for(int i = 0; i < n; i++) arr.add(s.nextLong()); return arr;}
int[] readIntArr(int n){int[] arr = new int[n]; for (int i = 0; i < n; i++) arr[i] = s.nextInt(); return arr;}
long[] readLongArr(int n){long[] arr = new long[n]; for (int i = 0; i < n; i++) arr[i] = s.nextLong(); return arr;}
String nextLine(){String str = "";try{str = br.readLine();}catch (IOException e){e.printStackTrace();}return str;}}
static class FastWriter{private BufferedWriter bw;public FastWriter(){if(System.getProperty("ONLINE_JUDGE") == null){try {this.bw = new BufferedWriter(new FileWriter("E:\\Competitive Coding\\output.txt"));}
catch (IOException e) {this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}else{this.bw = new BufferedWriter(new OutputStreamWriter(System.out));}}
public void print(Object object) throws IOException{bw.append(""+ object);}
public void println(Object object) throws IOException{print(object);bw.append("\n");}
public void debug(int object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void debug(long object[]) throws IOException{bw.append("["); for(int i = 0; i < object.length; i++){if(i != object.length-1){print(object[i]+", ");}else{print(object[i]);}}bw.append("]\n");}
public void close() throws IOException{bw.close();}}
public static void println(Object str) throws IOException{out.println(""+str);}
public static void println(Object str, int nextLine) throws IOException{out.print(""+str);}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static ArrayList<Integer> seive(int n){ArrayList<Integer> list = new ArrayList<>();int arr[] = new int[n+1];for(long i = 2; i <= n; i++) {if(arr[(int)i] == 1) {continue;}else {list.add((int)i);for(long j = i*i; j <= n; j = j + i) {arr[(int)j] = 1;}}}return list;}
public static long gcd(long a, long b){if(a > b) {a = (a+b)-(b=a);}if(a == 0L){return b;}return gcd(b%a, a);}
public static void swap(int[] arr, int i, int j) {arr[i] = arr[i] ^ arr[j]; arr[j] = arr[j] ^ arr[i]; arr[i] = arr[i] ^ arr[j];}
public static boolean isPrime(long n){if(n < 2){return false;}if(n == 2 || n == 3){return true;}if(n%2 == 0 || n%3 == 0){return false;}long sqrtN = (long)Math.sqrt(n)+1;for(long i = 6L; i <= sqrtN; i += 6) {if(n%(i-1) == 0 || n%(i+1) == 0) return false;}return true;}
public static long mod_add(long a, long b){ return (a%mod + b%mod)%mod;}
public static long mod_sub(long a, long b){ return (a%mod - b%mod + mod)%mod;}
public static long mod_mul(long a, long b){ return (a%mod * b%mod)%mod;}
public static long modInv(long a, long b){ return expo(a, b-2)%b;}
public static long mod_div(long a, long b){return mod_mul(a, modInv(b, mod));}
public static long expo (long a, long n){if(n == 0){return 1;}long recAns = expo(mod_mul(a,a), n/2);if(n % 2 == 0){return recAns;}else{return mod_mul(a, recAns);}}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Pair class
public static class Pair<X extends Comparable<X>,Y extends Comparable<Y>> implements Comparable<Pair<X, Y>>{
X first;
Y second;
public Pair(X first, Y second){
this.first = first;
this.second = second;
}
public String toString(){
return "( " + first+" , "+second+" )";
}
@Override
public int compareTo(Pair<X, Y> o) {
int t = first.compareTo(o.first);
if(t == 0) return second.compareTo(o.second);
return t;
}
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
// Code begins
public static void solve() throws IOException {
int n = s.nextInt();
int k = s.nextInt();
String str = s.nextLine();
int arr[] = new int[n];
int first_one = -1;
for(int i = 0; i < n; i++){
arr[i] = Integer.parseInt(""+str.charAt(i));
if(arr[i] == 1 && first_one == -1){
first_one = i;
}
}
if(first_one == -1){
first_one = n-1;
}
int ans[] = new int[n];
if(k%2 == 1){
ans[first_one] = 1;
for(int i = 0; i < n; i++){
if(i != first_one){
arr[i] = arr[i] ^ 1;
}
}
k--;
}
for(int i = 0; i < n-1; i++){
if(arr[i] == 0 && k > 0){
k--;
arr[i] = 1;
ans[i] += 1;
}
}
if(k%2 == 0){
ans[n-1] += k;
}else{
ans[n-1] += k;
arr[n-1] = arr[n-1]^1;
}
for(int i = 0; i < n; i++){
println(arr[i],1);
}
println("");
for(int i = 0; i < n; i++){
println(ans[i]+" ",1);
}
println("");
}
/*----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------*/
public static void main(String[] args) throws IOException {
int test = s.nextInt();
for(int t = 1; t <= test; t++) {
solve();
}
out.close();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 75ba40d1dc2f636847b20aba7a912e48 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | // Jai Kalash Shah
// Jai shree Ram
// Jai Bajrang Bali
// Jai Saraswati maa
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B {
static FastScanner sn = new FastScanner();
public static void main(String[] args) {
// your code here
int T = sn.nextInt();
while (T-- > 0)
solve();
}
public static void solve() {
int n = sn.nextInt();
int k = sn.nextInt();
String s = sn.next();
char[] arr = s.toCharArray();
if (k % 2 == 1) {
for (int i = 0; i < n; i++) {
if (arr[i] == '0') {
arr[i] = '1';
} else {
arr[i] = '0';
}
}
}
int[] ans = new int[n];
for (int i = 0; i < n; i++) {
ans[i] = 0;
}
for (int i = 0; i < n; i++) {
if (arr[i] == '0' && k > 0) {
arr[i] = '1';
k--;
ans[i]++;
}
}
ans[n - 1] += k;
if (k % 2 == 1) {
arr[n - 1] = '0';
}
s = String.copyValueOf(arr);
System.out.println(s);
for (int i = 0; i < n; i++) {
System.out.print(ans[i] + " ");
}
System.out.println();
}
//----------------------------------------------------------------------------------//
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());
}
char nextChar() {
return next().charAt(0);
}
long[] readArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
static class Debug {
public static void debug(long a) {
System.out.println("--> " + a);
}
public static void debug(long a, long b) {
System.out.println("--> " + a + " + " + b);
}
public static void debug(char a, char b) {
System.out.println("--> " + a + " + " + b);
}
public static void debug(int[] array) {
System.out.print("Array--> ");
System.out.println(Arrays.toString(array));
}
public static void debug(char[] array) {
System.out.print("Array--> ");
System.out.println(Arrays.toString(array));
}
public static void debug(HashMap<Integer, Integer> map) {
System.out.print("Map--> " + map.toString());
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 11 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output |
Subsets and Splits