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3.8k
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2.37k
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3.5k
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PASSED | 7555f421bc72b542896fcddd12ec5587 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.io.*;
import java.util.*;
public class A {
static class Pair
{
int f;int s; //
Pair(){}
Pair(int f,int s){ this.f=f;this.s=s;}
}
static class Fast {
BufferedReader br;
StringTokenizer st;
public Fast() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readArray1(int n) {
long a[] = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
/* static long noOfDivisor(long a)
{
long count=0;
long t=a;
for(long i=1;i<=(int)Math.sqrt(a);i++)
{
if(a%i==0)
count+=2;
}
if(a==((long)Math.sqrt(a)*(long)Math.sqrt(a)))
{
count--;
}
return count;
}*/
static boolean isPrime(long a) {
for (long i = 2; i <= (long) Math.sqrt(a); i++) {
if (a % i == 0)
return false;
}
return true;
}
static void primeFact(int n) {
int temp = n;
HashMap<Integer, Integer> h = new HashMap<>();
for (int i = 2; i * i <= n; i++) {
if (temp % i == 0) {
int c = 0;
while (temp % i == 0) {
c++;
temp /= i;
}
h.put(i, c);
}
}
if (temp != 1)
h.put(temp, 1);
}
static void reverseArray(int a[]) {
int n = a.length;
for (int i = 0; i < n / 2; i++) {
a[i] = a[i] ^ a[n - i - 1];
a[n - i - 1] = a[i] ^ a[n - i - 1];
a[i] = a[i] ^ a[n - i - 1];
}
}
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);
}
public static void main(String args[]) throws IOException {
Fast sc = new Fast();
PrintWriter out = new PrintWriter(System.out);
int t1 = sc.nextInt();
outer: while (t1-- > 0) {
int n=sc.nextInt();int r=sc.nextInt();int b=sc.nextInt();
StringBuffer ans=new StringBuffer();
if(b==0)
{
for(int i=0;i<r;i++)
ans.append('R');
out.println(ans);continue outer;
}
int max=(int)(Math.ceil(r*1.0/(b+1)));//out.println(max);
while(b<r&&b!=0)
{
for(int i=0;i<max&&r>0;i++) {
ans.append('R');
r--;
}
ans.append('B');
b--;
}
while(r!=0&&b!=0) {
ans.append('R');
ans.append('B');
r--;b--;
}
while(r!=0)
{
ans.append('R');r--;
}
while(b!=0)
{
ans.append('B');b--;
}
out.println(ans);
}
out.close();
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 41d16717602975e2b69368e105afff67 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(System.out);
int n = nextInt();
for (int j = 0; j < n; j++) {
int abc = nextInt();
int a = nextInt();
int b = nextInt();
int k = (a+b)/(b+1);
int l = a%(b+1);
int schet = 0;
for (int i = 0; i < abc; i++) {
if (schet < k && a > b) {
a--;
System.out.print("R");
schet++;
} else {
System.out.print("B");
b--;
schet = 0;
}
}
System.out.println();
}
}
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(System.out);
static StringTokenizer in = new StringTokenizer("");
public static String nextToken() throws IOException {
while (in == null || !in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public static int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public static double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
public static long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | e97bfc41be1ff4c6f779f656d2dd3c41 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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();
int r=in.nextInt();
int b=in.nextInt();
int k=r/(b+1);
if(r%(b+1)!=0) k++;
int c=r%(b+1);
for(int i=1;i<=c;i++){
for(int j=1;j<=k;j++){
out.print("R");
}
out.print("B");
}
b-=c;
r-=c*k;
k=r/(b+1);
for(int i=0;i<b;i++){
for(int j=1;j<=k;j++) out.print("R");
out.print("B");
}
for(int j=1;j<=k;j++) out.print("R");
out.println();
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 74196376b26da2c95afebb199c640d9f | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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 A782 {
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();
}
static class Solver {
void solve(InputReader in, PrintWriter out) {
int k = in.nextInt();
while (k-- > 0) {
int n, r, b;
n = in.nextInt();
r = in.nextInt();
b = in.nextInt();
int x = r / (b + 1);
int ext = r - x * (b + 1);
for (int i = 1; i <= r; i++) {
out.print('R');
if (i % x == 0 && ext > 0) {
out.print('R');
ext--;
r--;
}
if (i % x == 0 && b > 0) {
out.print('B');
b--;
}
}
while (b > 0) {
out.print('B');
b--;
}
out.println();
}
}
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 5e4a64e1be1cb05e43023aa56a3a58b9 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc = new Scanner(System.in);
int t=sc.nextInt();
StringBuilder st=new StringBuilder();
while(t-->0)
{
int n=sc.nextInt();
double r=sc.nextDouble();
double b=sc.nextDouble();
String s= "";
int c=(int)Math.ceil(r/(b+1));
//System.out.println(c);
int k=c;
while(r>0&&b>0)
{
while(c!=0&&r>0)
{
s+="R";
--r;
--c;
}
//c=k;
s+="B";
--b;
c=(int)Math.ceil(r/(b+1));
}
while(b>0)
{
s+="B";
--b;
}
while(r>0)
{
s+="R";
--r;
}
st.append(s);
st.append("\n");
}
System.out.println(st);
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 387550c6b73c7b102ef892f73f887f79 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.lang.reflect.Array;
import java.math.*;
import java.util.*;
public class Main {
private final BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
private final BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
private StringTokenizer st;
private final Long MOD = 1000000007L;
private final String endl = "\n";
private final String space = " ";
boolean isConsonantUpperCase(char c) {
return !isVowelUpperCase(c);
}
// only for upper case
boolean isVowelUpperCase(char c) {
c = (c + "").toUpperCase().charAt(0);
return c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
}
// IO UTILS
void log(Object... args) {
if (!TESTING) return;
for (Object s : args) System.out.print(s.toString() + " ");
System.out.println();
}
String line() throws IOException {
return in.readLine().trim();
}
StringTokenizer tokens() throws IOException {
return new StringTokenizer(line());
}
String nextToken() throws IOException {
if (st == null || !st.hasMoreTokens()) st = new StringTokenizer(line());
return st.nextToken();
}
long l(String s) {
return Long.parseLong(s);
}
long rl() throws IOException {
return l(nextToken());
}
long[] rla(int n) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = rl();
return arr;
}
int i(String s) {
return Integer.parseInt(s);
}
int ri() throws IOException {
return i(nextToken());
}
int[] ria(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = ri();
return arr;
}
String rs() throws IOException {
return nextToken();
}
String[] rsa(int n) throws IOException {
String[] s = new String[n];
for (int i = 0; i < n; i++)
s[i] = nextToken();
return s;
}
double d(String s) {
return Double.parseDouble(s);
}
double rd() throws IOException {
return d(nextToken());
}
double[] rda(int n) throws IOException {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = rd();
}
return arr;
}
void ol(Object... arg) throws IOException {
int n = arg.length;
for (int i = 0; i < n - 1; i += 1) {
os(arg[i].toString());
}
if (n > 0) ol(arg[n - 1].toString());
}
void ol() throws IOException {
out.write(endl);
}
void os() throws IOException {
out.write(space);
}
void ol(String s) throws IOException {
out.write(s + endl);
}
void os(String s) throws IOException {
out.write(s + space);
}
void ol(int n) throws IOException {
out.write(n + endl);
}
void os(int s) throws IOException {
out.write(s + space);
}
void ol(long s) throws IOException {
out.write(s + endl);
}
void os(long s) throws IOException {
out.write(s + space);
}
void o(String s) throws IOException {
out.write(s);
}
void o(int n) throws IOException {
out.write(n + "");
}
void o(long s) throws IOException {
out.write(s + "");
}
// IO UTILS END
// MATH UTILS
long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
long lcm(long a, long b) {
long g = gcd(a, b);
return (a / g) * b; // directly multiply a and b can overflow
}
public long modInv(long a, long MOD) {
return new BigInteger(a + "").modPow(new BigInteger((MOD - 2) + ""), new BigInteger(MOD + "")).longValue();
}
public long modPro(long a, long b, long MOD) {
return ((a % MOD) * (b % MOD)) % MOD;
}
private long factorialMod(int n, long MOD) {
long pro = 1;
for (long i = 1; i <= n; i++) {
pro = pro * i;
pro %= MOD;
}
return pro;
}
static void ruffleSort(int[] a) {
// ruffle
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n), temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
// ruffle
int n = a.length;
Random r = new Random();
for (int i = 0; i < a.length; i++) {
int oi = r.nextInt(n);
long temp = a[i];
a[i] = a[oi];
a[oi] = temp;
}
Arrays.sort(a);
}
boolean[] sieveOfEratosthenes(int n) {
boolean[] prime = new boolean[n + 1];
for (int i = 0; i < n; i++)
prime[i] = true;
prime[0] = false;
prime[1] = false;
for (int i = 4; i <= n; i += 2) {
prime[i] = false;
}
for (int p = 3; p * p <= n; p += 2) {
// If prime[p] is not changed, then it is a prime
if (prime[p]) {
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
// MATH UTILS END
private final boolean TESTING = false;
public static void main(String[] args) throws Exception {
new Main().solve();
}
private void solve() throws Exception {
int testCases = 1;
testCases = Integer.parseInt(in.readLine().trim());
preProcess();
for (int i = 1; i <= testCases; i++) {
solveTestCase(i);
}
out.flush();
out.close();
}
void preProcess() {
}
// make debugging flag based
private void solveTestCase(int testCaseNumber) throws Exception {
int n = ri(), r = ri(), b = ri();
StringBuilder ans = new StringBuilder();
int rem = n;
int g = r / (b+1) ;
int mod = r % (b+1);
for (int i = 0; i < b; i++) {
int temp = g;
while(temp-->0) ans.append("R");
if (mod > 0 ) {
ans.append("R");
rem -= 1;
mod -= 1;
}
ans.append("B");
rem -= g + 1;
}
if (rem > 0 ) {
while(rem-->0) ans.append("R");
}
ol(ans.toString());
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 6ae771b8b8df1eaf69dbb8ffe8b1b1e1 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long mod=(long)1e9+7;
static long[]fac=new long[1002];
static int n, x=0,me,op;
static int[]pe,a,aa, prime=new int[(int)1e7+1];
static int[][]perm;
static long[][]memo;
static Integer[]ps;
static TreeSet<Long>p=new TreeSet<Long>();
public static void main(String[] args) throws Exception{
int t =sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int r =sc.nextInt();
int b=sc.nextInt();
char[]a = new char[n];
int x = r/(b+1);
int y = r%(b+1);
int c =0;
for (int i = 0; i < a.length; i++) {
if(c<x) {
a[i]='R';
r--;
c++;
}else {
if(y!=0) {
a[i++]='R';
r--;
b--;
a[i]='B';
y--;
}else {
b--;
a[i]='B';
}
c=0;
}
}
pw.println(new String(a));
}
pw.close();
}
public static long solFul(int m, int o) {
if(o<m)return 0;
if(o == op)return 1;
if(memo[m][o]!=-1) {
return memo[m][o];
}
return memo[m][o] = (solFul(m+1,o)+solFul(m, o+1))%mod;
}
public static long solFree(int m, int o) {
if(m<=o)return 0;
if(m == me)return 1;
if(o == op&&m>op)return 1;
if(memo[m][o]!=-1)return memo[m][o];
return memo[m][o] = (solFree(m+1,o)+solFree(m, o+1))%mod;
}
public static String rev(String s) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
sb.append(s.charAt(s.length()-1-i));
}
return sb.toString();
}
public static int divNum(int n) {
int total = 1;
while(n>1) {
int p=prime[n];
int count =1;
while(n%p==0) {
n/=p;
count++;
}
total*=count;
}
return total;
}
public static void sieve() {
for (int i = 2; i < prime.length; i++) {
if(prime[i]==0) {
p.add((long)i);
for (int j = i; j < prime.length; j+=i) {
prime[j]++;
}
}
}
}
public static long[] Extended(long p, long q) {
if (q == 0)
return new long[] { p, 1, 0 };
long[] vals = Extended(q, p % q);
long d = vals[0];
long a = vals[2];
long b = vals[1] - (p / q) * vals[2];
return new long[] { d, a, b };
}
public static int LIS(int[] a) {
int n = a.length;
int[] ser = new int[n];
Arrays.fill(ser, Integer.MAX_VALUE);
int cur = -1;
for (int i = 0; i < n; i++) {
int low = 0;
int high = n - 1;
int mid = (low + high) / 2;
while (low <= high) {
if (ser[mid] < a[i]) {
low = mid + 1;
} else {
high = mid - 1;
}
mid = (low + high) / 2;
}
cur = Math.max(cur, high + 1);
ser[high + 1] = Math.min(ser[high + 1], a[i]);
}
return cur + 1;
}
public static void permutation(int idx,int v) {
if(v==(1<<n)-1) {
perm[x++]=pe.clone();
return ;
}
for (int i = 0; i < n; i++) {
if((v&1<<i)==0) {
pe[idx]=aa[i];
permutation(idx+1, v|1<<i);
}
}
return ;
}
public static void sort(int[]a) {
mergesort(a, 0, a.length-1);
}
public static void sortIdx(long[]a,long[]idx) {
mergesortidx(a, idx, 0, a.length-1);
}
public static long C(int a,int b) {
long x=fac[a];
long y=fac[a-b]*fac[b];
return x*pow(y,mod-2)%mod;
}
public static long pow(long a,long b) {
long ans=1;a%=mod;
for(long i=b;i>0;i/=2) {
if((i&1)!=0)
ans=ans*a%mod;
a=a*a%mod;
}
return ans;
}
public static void pre(){
fac[0]=1;
fac[1]=1;
fac[2]=2;
for (int i = 3; i < fac.length; i++) {
fac[i]=fac[i-1]*i%mod;
}
}
public static long eval(String s) {
long p=1;
long res=0;
for (int i = 0; i < s.length(); i++) {
res+=p*(s.charAt(s.length()-1-i)=='1'?1:0);
p*=2;
}
return res;
}
public static String binary(long x) {
String s="";
while(x!=0) {
s=(x%2)+s;
x/=2;
}
return s;
}
public static boolean allSame(String s) {
char x=s.charAt(0);
for (int i = 0; i < s.length(); i++) {
if(s.charAt(i)!=x)return false;
}
return true;
}
public static boolean isPalindrom(String s) {
int l=0;
int r=s.length()-1;
while(l<r) {
if(s.charAt(r--)!=s.charAt(l++))return false;
}
return true;
}
public static boolean isSubString(String s,String t) {
int ls=s.length();
int lt=t.length();
boolean res=false;
for (int i = 0; i <=lt-ls; i++) {
if(t.substring(i, i+ls).equals(s)) {
res=true;
break;
}
}
return res;
}
public static boolean isSorted(long[]a) {
for (int i = 0; i < a.length-1; i++) {
if(a[i]>a[i+1])return false;
}
return true;
}
public static long evaln(String x,int n) {
long res=0;
for (int i = 0; i < x.length(); i++) {
res+=Long.parseLong(x.charAt(x.length()-1-i)+"")*Math.pow(n, i);
}
return res;
}
static void mergesort(int[] arr,int b,int e) {
if(b<e) {
int m=b+(e-b)/2;
mergesort(arr,b,m);
mergesort(arr,m+1,e);
merge(arr,b,m,e);
}
return;
}
static void merge(int[] arr,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
int[] l=new int[len1];
int[] r=new int[len2];
for(int i=0;i<len1;i++)l[i]=arr[b+i];
for(int i=0;i<len2;i++)r[i]=arr[m+1+i];
int i=0,j=0,k=b;
while(i<len1 && j<len2) {
if(l[i]<r[j])arr[k++]=l[i++];
else arr[k++]=r[j++];
}
while(i<len1)arr[k++]=l[i++];
while(j<len2)arr[k++]=r[j++];
return;
}
static void mergesortidx(long[] arr,long[]idx,int b,int e) {
if(b<e) {
int m=b+(e-b)/2;
mergesortidx(arr,idx,b,m);
mergesortidx(arr,idx,m+1,e);
mergeidx(arr,idx,b,m,e);
}
return;
}
static void mergeidx(long[] arr,long[]idx,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
long[] l=new long[len1];
long[] lidx=new long[len1];
long[] r=new long[len2];
long[] ridx=new long[len2];
for(int i=0;i<len1;i++) {
l[i]=arr[b+i];
lidx[i]=idx[b+i];
}
for(int i=0;i<len2;i++) {
r[i]=arr[m+1+i];
ridx[i]=idx[m+1+i];
}
int i=0,j=0,k=b;
while(i<len1 && j<len2) {
if(l[i]<=r[j]) {
arr[k++]=l[i++];
idx[k-1]=lidx[i-1];
}
else {
arr[k++]=r[j++];
idx[k-1]=ridx[j-1];
}
}
while(i<len1) {
idx[k]=lidx[i];
arr[k++]=l[i++];
}
while(j<len2) {
idx[k]=ridx[j];
arr[k++]=r[j++];
}
return;
}
static long mergen(int[] arr,int b,int m,int e) {
int len1=m-b+1,len2=e-m;
int[] l=new int[len1];
int[] r=new int[len2];
for(int i=0;i<len1;i++)l[i]=arr[b+i];
for(int i=0;i<len2;i++)r[i]=arr[m+1+i];
int i=0,j=0,k=b;
long c=0;
while(i<len1 && j<len2) {
if(l[i]<r[j])arr[k++]=l[i++];
else {
arr[k++]=r[j++];
c=c+(long)(len1-i);
}
}
while(i<len1)arr[k++]=l[i++];
while(j<len2)arr[k++]=r[j++];
return c;
}
static long mergesortn(int[] arr,int b,int e) {
long c=0;
if(b<e) {
int m=b+(e-b)/2;
c=c+(long)mergesortn(arr,b,m);
c=c+(long)mergesortn(arr,m+1,e);
c=c+(long)mergen(arr,b,m,e);
}
return c;
}
public static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static long sumd(long x) {
long sum=0;
while(x!=0) {
sum+=x%10;
x=x/10;
}
return sum;
}
public static ArrayList<Integer> findDivisors(int n){
ArrayList<Integer>res=new ArrayList<Integer>();
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
// If divisors are equal, print only one
if (n/i == i)
res.add(i);
else {
res.add(i);
res.add(n/i);
}
}
}
return res;
}
public static void sort2darray(Integer[][]a){
Arrays.sort(a,Comparator.<Integer[]>comparingInt(x -> x[0]).thenComparingInt(x -> x[1]));
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(String file) throws FileNotFoundException {
br = new BufferedReader(new FileReader(file));
}
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 {
return Double.parseDouble(next());
}
public boolean ready() throws IOException {
return br.ready();
}
public int[] nextArrint(int size) throws IOException {
int[] a=new int[size];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextInt();
}
return a;
}
public long[] nextArrlong(int size) throws IOException {
long[] a=new long[size];
for (int i = 0; i < a.length; i++) {
a[i]=sc.nextLong();
}
return a;
}
public int[][] next2dArrint(int rows,int columns) throws IOException{
int[][]a=new int[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j]=sc.nextInt();
}
}
return a;
}
public long[][] next2dArrlong(int rows,int columns) throws IOException{
long[][]a=new long[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j]=sc.nextLong();
}
}
return a;
}
}
static class Pair{
long x;
long y;
public Pair(long x,long y) {
this.x=x;
this.y=y;
}
}
static Scanner sc=new Scanner(System.in);
static PrintWriter pw=new PrintWriter(System.out);
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 0482f4a58b87eb8fd5ad1ba28abc2926 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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) {
long n = r.nl();
long rr = r.nl();
long b = r.nl();
long can = rr / (b + 1);
long add = rr % (b + 1);
if (rr != 0) {
do {
long need = can;
if (add > 0) need++;
add = Math.max(0, add - 1);
rr -= need;
while (need-- > 0) out.write(("R" + "").getBytes());
if (b > 0) {
out.write(("B" + "").getBytes());
b--;
}
} while (rr != 0);
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 2d7dc8e21f5f420caca92a4703936fc0 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
public static void main(String[] args) {
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int T = in.nextInt();
for(int ttt = 1; ttt <= T; ttt++) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
int max = r/(b+1);
int left = r-max*(b+1);
String ans = "";
for(int i = 0; i < left; i++) {
for(int j = 0; j <= max; j++) ans += "R";
ans += "B";
}
// String res = "";
// res += ans;
for(int i = 0; i < b-left; i++) {
for(int j = 0; j < max; j++) ans += "R";
ans += "B";
}
// res += ans;
int cnt = (max+1)*left + (b-left)*max;
while(cnt < r) {
ans += "R";
cnt++;
}
out.println(ans);
}
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;
}
int[] readInt(int size) {
int[] arr = new int[size];
for(int i = 0; i < size; i++)
arr[i] = Integer.parseInt(next());
return arr;
}
long[] readLong(int size) {
long[] arr = new long[size];
for(int i = 0; i < size; i++)
arr[i] = Long.parseLong(next());
return arr;
}
int[][] read2dArray(int rows, int cols) {
int[][] arr = new int[rows][cols];
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++)
arr[i][j] = Integer.parseInt(next());
}
return arr;
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 915ddbc2682d884b5f6e750e0af9b634 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Roy
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
ARedVersusBlue solver = new ARedVersusBlue();
solver.solve(1, in, out);
out.close();
}
static class ARedVersusBlue {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int tCases = in.readInteger();
for (int cs = 1; cs <= tCases; ++cs) {
int n = in.readInteger();
int r = in.readInteger();
int b = in.readInteger();
int min = Math.min(r, b);
List<StringBuilder> red = new ArrayList<>();
List<StringBuilder> blue = new ArrayList<>();
for (int i = 0; i < min; i++) {
red.add(new StringBuilder("R"));
blue.add(new StringBuilder("B"));
}
int redRemains = r - min;
int blueRemains = b - min;
if (redRemains == 0 && blueRemains > 0) {
blue.add(new StringBuilder("B"));
blueRemains--;
int index = 0, blueSz = blue.size();
while (blueRemains != 0) {
blue.get(index % blueSz).append("B");
index++;
blueRemains--;
}
} else if (blueRemains == 0 && redRemains > 0) {
red.add(new StringBuilder("R"));
redRemains--;
int index = 0, redSz = red.size();
while (redRemains != 0) {
red.get(index % redSz).append("R");
index++;
redRemains--;
}
}
StringBuilder ans = new StringBuilder();
int redSz = red.size(), blueSz = blue.size();
int redIndex = 0, blueIndex = 0;
if (redSz >= blueSz) {
while (redIndex < redSz || blueIndex < blueSz) {
if (redIndex < redSz) {
ans.append(red.get(redIndex++));
}
if (blueIndex < blueSz) {
ans.append(blue.get(blueIndex++));
}
}
} else {
while (redIndex < redSz || blueIndex < blueSz) {
if (blueIndex < blueSz) {
ans.append(blue.get(blueIndex++));
}
if (redIndex < redSz) {
ans.append(red.get(redIndex++));
}
}
}
out.printLine(ans);
}
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void 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) {
this.print(objects);
writer.println();
}
public void close() {
writer.flush();
writer.close();
}
}
static class InputReader {
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
private final InputStream stream;
private final byte[] buf = new byte[1024];
public InputReader(InputStream stream) {
this.stream = stream;
}
private long readWholeNumber(int c) {
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res;
}
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 readInteger() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = (int) readWholeNumber(c);
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
boolean isSpaceChar(int ch);
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 94694378076a4ce13897d301783b4b46 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Roy
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
ARedVersusBlue solver = new ARedVersusBlue();
solver.solve(1, in, out);
out.close();
}
static class ARedVersusBlue {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int tCases = in.readInteger();
for (int cs = 1; cs <= tCases; ++cs) {
int n = in.readInteger();
int r = in.readInteger();
int b = in.readInteger();
int min = Math.min(r, b);
List<StringBuilder> red = new ArrayList<>();
List<StringBuilder> blue = new ArrayList<>();
for (int i = 0; i < min; i++) {
red.add(new StringBuilder("R"));
blue.add(new StringBuilder("B"));
}
int redRemains = r - min;
int blueRemains = b - min;
if (redRemains == 0 && blueRemains > 0) {
blue.add(new StringBuilder("B"));
blueRemains--;
int index = 0, blueSz = blue.size();
while (blueRemains != 0) {
blue.get(index % blueSz).append("B");
index++;
blueRemains--;
}
} else if (blueRemains == 0 && redRemains > 0) {
red.add(new StringBuilder("R"));
redRemains--;
int index = 0, redSz = red.size();
while (redRemains != 0) {
red.get(index % redSz).append("R");
index++;
redRemains--;
}
}
StringBuilder ans = new StringBuilder();
int redSz = red.size(), blueSz = blue.size();
int redIndex = 0, blueIndex = 0;
if (redSz >= blueSz) {
while (redIndex < redSz || blueIndex < blueSz) {
if (redIndex < redSz) {
ans.append(red.get(redIndex));
redIndex++;
}
if (blueIndex < blueSz) {
ans.append(blue.get(blueIndex));
blueIndex++;
}
}
} else {
while (redIndex < redSz || blueIndex < blueSz) {
if (blueIndex < blueSz) {
ans.append(blue.get(blueIndex));
blueIndex++;
}
if (redIndex < redSz) {
ans.append(red.get(redIndex));
redIndex++;
}
}
}
out.printLine(ans);
}
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void 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) {
this.print(objects);
writer.println();
}
public void close() {
writer.flush();
writer.close();
}
}
static class InputReader {
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
private final InputStream stream;
private final byte[] buf = new byte[1024];
public InputReader(InputStream stream) {
this.stream = stream;
}
private long readWholeNumber(int c) {
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res;
}
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 readInteger() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = (int) readWholeNumber(c);
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
boolean isSpaceChar(int ch);
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9e5d74f734f74e9c67683758a1295339 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class Solution {
public static void main(String... args){
Solution sol = new Solution();
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-->0){
sol.solve(in);
}
}
private String giveString(int x){
String ans = "";
for(int i=0;i<x;i++) ans+="R";
return ans;
}
private void solve(Scanner in){
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
b++;
String ans = "";
int i = 0;
while(i<n){
if(b==1){
ans+="R";
r--;
i++;
continue;
}
int x = (int)Math.ceil(r/(float)b);
ans+=giveString(x);
ans+="B";
r-=x;
b--;
i+=x+1;
}
// System.out.println("R:"+countit(ans, 'R') + " B:"+countit(ans, 'B'));
System.out.println(ans);
}
int countit(String ans, char s){
int n = ans.length();
int count = 0;
for(int i=0;i<n;i++){
if(ans.charAt(i) == s){
count++;
}
}
return count;
}
/**
0 1 2 3 4 5 6 7 8 9
R R B R R B R R B
19 13 6
6 5 1
R R B R R R
6 5 2
R R B R R B R
7 4 3
R R R R
R B R B R B R
19 13 6
R R R R R R R R R R R R R
R R B R R B R R B R R B R R B R R B R
11 9 2
R R R R R R R R R
9/X=2+1
9/(2+1)
R R R R B R R R R B R
B
10 9 1
R R R R R R R R R
9/2
10 8 2
R R R R R R R R
R R R R R R R R
10 6 4
6/5=1.2=2
0 1 2 3 4 5 6 7 8
R R B R
R R R R R R
R R B R R B R R B B
*/
}
/*
|
10110101
|
40
0001
*/ | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 6c908c4536404615faf82c044f0ad5d1 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class Largest
{
public static void solve() throws Exception
{
FastReader fr=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int T=fr.nextInt();
while(T-->0)
{
// out.println("////////////");
int N=fr.nextInt();
int R=fr.nextInt();
int B=fr.nextInt();
int p=R/(B+1);
int x=R%(B+1);
int i=0;
while(i<x)
{
int j=0;
while(j<=p)
{
out.print("R");
++j;
}
out.print("B");
++i;
}
while(i<=B)
{
int j=0;
while(j<p)
{
out.print("R");
++j;
}
if(i!=B) out.print("B");
++i;
}
out.println();
}
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;
}
}
public static int find_max(int[] arr)
{
int max=0;
int i=0;
while(i<arr.length)
{
if(arr[i]>arr[max]) max=i;
++i;
}
return max;
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int countSetBits(int n)
{
int count = 0;
while (n > 0) {
count += n & 1;
n >>= 1;
}
return count;
}
static int countBits(int n)
{
int count = 0;
while (n != 0)
{
count++;
n >>= 1;
}
return count;
}
public static String reverseStr(String str)
{
StringBuilder sb=new StringBuilder();
sb.append(str);
sb.reverse();
return sb.toString();
}
public static long getSum(long N)
{
long res=0;
while(N!=0)
{
res+=N%10;
N/=10;
}
return res;
}
public static int __(int x, int y) {return x;}
public static void reverseArray(int []arr)
{
int n=arr.length;
for (int i = 0; i < n / 2; i++)
arr[i] = __(arr[n - i - 1],
arr[n - i - 1] = arr[i]);
}
public static void main(String args[])
{
try
{
solve();
}
catch(Exception e)
{
return;
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 85f7f64487cb7e2a730ce5b1fc69c7d1 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
/*
!!!!Hello World,Prakhar here!!!!
codechef handle prakhar_3011
codeforces handle prakhar_30
trying to get good at CP
PEACE OUT.........
*/
/*
n r b
10 9 1 rrrrbrrrr
10 8 2 rrrbrrrbrr
11 9 2 rrrbrrrbrrr
6 5 1
*/
import java.io.*;
import java.util.*;
public class redvsblue {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n=sc.nextInt();
int r=sc.nextInt();
int b=sc.nextInt();
int div=r/(b+1);int count=0;
int exr=r-((b+1)*div);
int x=b+1;
while(x>0){
System.out.print("R");
count++;
if(count==div){
if(exr>0) {System.out.print("R");exr--;}
if(b>0){System.out.print("B");b--;}
count=0;x--;
}
}
System.out.println();
}
}
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 revsort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l, Collections.reverseOrder());
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
/* ......FAST SCANNER template taken from secondthread...... */
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());
}
double nextdouble() {
return Double.parseDouble(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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | aa422f314cc5638739987083693a428d | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | /**
* @Jai_Bajrang_Bali
* @Har_Har_Mahadev
*/
import java.util.*;
public class practice2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tt = sc.nextInt();
while (tt-- > 0) {
int n=sc.nextInt();
int r=sc.nextInt();
int b=sc.nextInt();
String ans="";
while(b>0)
{
int max=(int)Math.ceil((float)r/(b+1));
for(int i=0;i<max;i++){
ans+='R';
}
ans+='B';
b-=1;
r-=max;
}
for(int i=0;i<r;i++){
ans+='R';
}
System.out.println(ans);
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 7dc25d2a240d1c32103cf949f9b6611e | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class RedVsBlue {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int parts = r/(b+1);
int remain = r % (b+1);
while(r > 0) {
for(int i = 1; i <= parts; i++) {
System.out.print("R");
}
r -= parts;
if(remain > 0) {
System.out.print("R");
remain--;
r--;
}
if(b > 0) {
System.out.print("B");
b--;
}
}
System.out.println();
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 77e6c83c040dfafd4c3b3a1e052b7eca | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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 r=sc.nextInt();
int b=sc.nextInt();
while(r>0 && b>0) {
int cnt=r/(b+1);
if(r%(b+1)!=0)
cnt++;
for(int i=0;i<cnt;i++)
System.out.print("R");
System.out.print("B");
r-=cnt;
b-=1;
}
while(r>0) {
System.out.print("R");
r--;
}
while(b>0) {
System.out.print("B");
b--;
}
System.out.println();
}
sc.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 1a72e61ab2a0660d6457a580085bb517 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) throws IOException{
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(br.readLine());
while(t--> 0){
String s[] = br.readLine().split("\\s+");
int n = Integer.parseInt(s[0]);
int r = Integer.parseInt(s[1]);
int b = Integer.parseInt(s[2]);
int a = r/(b+1);
int c = a;
r = r-a*(b+1);
int d = r;
for(int i=0;i<n;i++){
if(a>0){
bw.write("R");
a--;
}else{
if(r>0){
bw.write("R");
r=0;
}else{
bw.write("B");
a = c;
r=d--;
}
}
}
bw.write("\n");
}
bw.flush();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 7f09c0716bb6b6624dbcaae22c498dca | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution{
public static void main(String args[]) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(br.readLine());
while(t--> 0){
String s[] = br.readLine().split("\\s+");
int n = Integer.parseInt(s[0]);
int r = Integer.parseInt(s[1]);
int b = Integer.parseInt(s[2]);
int a = r/(b+1);
int c = a;
r = r-a*(b+1);
int d = r;
for(int i=0;i<n;i++){
if(a>0){
bw.write("R");
a--;
}else{
if(r>0){
bw.write("R");
r = 0;
}else{
bw.write("B");
a = c;
r = --d;
}
}
}
bw.write("\n");
}
bw.flush();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | a500ce6179a2fadedb48503275dd34b9 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // package Round782DIV2;
import java.io.*;
import java.util.*;
public class A { // GOGIGO!!!
static long[] fac;
static int mod = 998244353;
public static void main(String[] args) throws IOException {
// Scanner sc = new Scanner(new FileReader("input.in"));
// PrintWriter pw = new PrintWriter(new FileWriter(""));
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
// int t = 1;
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
StringBuilder sb = new StringBuilder();
if(b==0){
for(int i =0;i<n;i++){
sb.append("R");
}
pw.println(sb.toString());
continue;
}
if(b==1){
int max= r/2;
for(int i =0;i<max;i++){
sb.append("R");
}
sb.append("B");
while(sb.length()<n){
sb.append("R");
}
}else{
int max = r/(b); // kam mara ha7ot el r
int rem= r%(b);
int tmpM=max;
int tmpR=rem;
int count=0;
int max2=max+1;
int b1=0;
int b2=b;
if(rem-1>max){
int diff= rem-max;
b1=diff;
b2-=diff;
rem=max;
}
if(tmpM>tmpR)
while(true){
if(tmpM<tmpR){
max-=(count-1);
rem+=((count-1)*b);
break;
}
tmpM--;
tmpR+=b;
count++;
}
for(int i=0;i<b1;i++){
for (int j=0;j<max2;j++){
sb.append("R");
}
sb.append("B");
}
// pw.println(max+" "+rem);
for(int i =0;i<b2;i++){
for(int j=0;j<max;j++){
sb.append("R");
}
sb.append("B");
}
for(int i =0;i<rem;i++){
sb.append("R");
}
}
pw.println(sb.toString());
}
pw.close();
}
// -------------------------------------------------------Basics----------------------------------------------------
public static void fac(int MAXN) {
fac = new long[MAXN + 1];
fac[0] = 1;
for (int i = 1; i < MAXN; i++) {
fac[i] = (fac[i - 1] * i) % mod;
}
}
//------------------------------------------------------ BINARYSEARCH ------------------------------------------------
public static int binarySearch(long x, Long [] a){
int i =0;
int j = a.length-1;
int mid ;
while(i<=j){
mid = (i+j)/2;
if(a[mid]<=x){
i=mid+1;
}else{
j=mid-1;
}
}
return i;
}
// ------------------------------------------------------- MATH ----------------------------------------------------
private static int gcd(int a, int b) {
return (b == 0)? a : gcd(b, a % b);
}
private static long gcd(long a, long b) {
return (b == 0)? a : gcd(b, a % b);
}
private static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
private static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
// ------------------------------------------------------ Objects --------------------------------------------------
static class Pair implements Comparable<Pair>{
long x ;
long y ;
Pair(long x , long y){
this.x=x;
this.y=y;
}
@Override
public int compareTo(Pair o) {
if(this.x==o.x)return 0;
if(this.x>o.x)return 1;
return -1;
}
@Override
public String toString() {
return x +" " + y ;
}
}
static class Tuple implements Comparable<Tuple>{
int x;
int y;
int z;
Tuple(int x, int y , int z){
this.x=x;
this.y=y;
this.z=z;
}
@Override
public int compareTo(Tuple o) {
if(this.x==o.x){
if(this.y==o.y)return this.z-o.z;
return this.y-o.y;
}
return this.x-o.x;
}
@Override
public String toString() {
return x +" " + y + " " + z ;
}
}
// -------------------------------------------------------Scanner---------------------------------------------------
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();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9f5acc5ad70cbf2e37c7cfa8987f197a | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.*;
public class Codeforces {
static int mod= 1000000007 ;
public static void main(String[] args) throws Exception {
PrintWriter out=new PrintWriter(System.out);
FastScanner fs=new FastScanner();
DecimalFormat formatter= new DecimalFormat("#0.000000");
// int t=1;
int t=fs.nextInt();
outer:for(int time=1;time<=t;time++) {
int n =fs.nextInt(), r=fs.nextInt(), b=fs.nextInt();
StringBuilder ans=new StringBuilder();
int toadd= b+1;
int max= (r+toadd-1)/toadd;
int min = r/toadd;
int x= r-min*toadd;
for(int i=0;i<max;i++) ans.append('R');
r-=max;
x--;
while(b-->0) {
ans.append('B');
if(x>0) {
for(int i=0;i<max;i++) {
ans.append('R');
}
x--;
}
else {
for(int i=0;i<min;i++) ans.append('R');
}
}
out.println(ans);
}
out.close();
}
static long pow(long a,long b) {
if(b<0) return 1;
long res=1;
while(b!=0) {
if((b&1)!=0) {
res*=a;
res%=mod;
}
a*=a;
a%=mod;
b=b>>1;
}
return res;
}
static long gcd(long a,long b) {
if(b==0) return a;
return gcd(b,a%b);
}
static long nck(int n,int k) {
if(k>n) return 0;
long res=1;
res*=fact(n);
res%=mod;
res*=modInv(fact(k));
res%=mod;
res*=modInv(fact(n-k));
res%=mod;
return res;
}
static long fact(long n) {
// return fact[(int)n];
long res=1;
for(int i=2;i<=n;i++) {
res*=i;
res%=mod;
}
return res;
}
static long modInv(long n) {
return pow(n,mod-2);
}
static void sort(int[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
int temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
static void sort(long[] a) {
//suffle
int n=a.length;
Random r=new Random();
for (int i=0; i<a.length; i++) {
int oi=r.nextInt(n);
long temp=a[i];
a[i]=a[oi];
a[oi]=temp;
}
//then sort
Arrays.sort(a);
}
// Use this to input code since it is faster than a Scanner
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
String nextLine() {
String str="";
try {
str= (br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
long[] readArrayL(int n) {
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=nextLong();
return a;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | cef6a7ac01347aabc99e019759fa8e7e | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args) throws IOException
{
try{
Scanner sc =new Scanner(System.in);
int t =sc.nextInt();
for(int o=0;o<t;o++)
{
int n = sc.nextInt();
int r = sc.nextInt();
int b=sc.nextInt();
int[] arr=new int[b+1];
int i =0;
while(r>0)
{
r--;
arr[i]++;
i++;
if(i==b+1)
i=0;
}
char arr1[]=new char[n];
for (int j = 0; j <n ; j++) {
arr1[j]='R';
}int j =0;
for( i =1;i<=b;i++)
{j=j+arr[i-1];
arr1[j]='B';
j++;
}
String string1 = new String(arr1);
System.out.println(string1);
}
}catch(Exception e){
return;
}
}}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c47b2dd66644abe83d7c5ba7c3c572ad | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException,
InterruptedException {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
StringBuilder sb = new StringBuilder();
while (b > 0) {
int x = (int) Math.ceil(r * 1.0 / (b+1));
while (x-- > 0 && r > 0) {
sb.append("R");
r--;
}
sb.append("B");
b--;
}
while (r-- > 0)
sb.append("R");
pw.println(sb.toString());
}
pw.flush();
}
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 readAllLines(BufferedReader reader) throws IOException {
StringBuilder content = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
content.append(line);
content.append(System.lineSeparator());
}
return content.toString();
}
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();
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 8d0b9523d0252124c51d6734b74b9c4c | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.*;
import java.lang.*;
public class red_versus_blue {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t!=0) {
String []st=br.readLine().split(" ");
int n = Integer.parseInt(st[0]);
int r = Integer.parseInt(st[1]);
int b = Integer.parseInt(st[2]);
String s="";
int temp=r/(b+1);
int mod=r%(b+1);
/*boolean flag=true;
if(mod>1) {
temp++;
}
int z=temp;
for(int i=0; i<n; i++) {
//int z=temp;
if(flag==true && z!=0) {
s+='R';
z--;
r--;
if(z==0) {
flag=false;
}
}else if(flag==false && b!=0){
s+='B';
z=temp;
b--;
flag=true;
}else{
s+='R';
}
}*/
int x=0;
int y=0;
while(x<r) {
for(int i=0; i<temp; i++) {
s+='R';
}
x+=temp;
if(mod!=0) {
s+='R';
x+=1;
mod-=1;
}
if(y<b) {
s+='B';
y+=1;
}
}
System.out.println(s);
t--;
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | dde1366f2b231a4e249a3eb5adb67914 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // package CodeForces.contests.contest782div2;
import java.util.*;
/**
* @author SyedAli
* @createdAt 17-04-2022, Sunday, 20:03
*/
public class RedVersusBlue {
private static Scanner s = new Scanner(System.in);
private static void solve() {
int n = s.nextInt(), r = s.nextInt(), b = s.nextInt();
List<StringBuilder> sbs = new ArrayList<>();
for (int i = 0; i < b + 1; i++) {
StringBuilder sb = new StringBuilder();
sbs.add(sb);
}
int start = 0;
for (int i = 0; i < r; i++) {
if (start < sbs.size()) {
sbs.get(start).append("R");
start++;
} else {
start = 0;
sbs.get(start).append("R");
start++;
}
}
for (int i = 0; i < b; i++) {
sbs.get(i).append("B");
}
StringBuilder res = new StringBuilder();
for (int i = 0; i < sbs.size(); i++) res.append(sbs.get(i));
System.out.println(res);
}
public static void main(String[] args) {
int t = s.nextInt();
while (t-- > 0) {
solve();
}
s.close();
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9b4229eda7c697a1a1b149bf8c1f8680 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class A_Red_Versus_Blue {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int testCases, n, r, b;
static void solve(int t) {
char ch[] = {'R', 'B'};
if (r - b == 1) {
int j = 0;
for (int i = 0; i < n; ++i) {
ans.append(ch[(j++) % 2]);
}
} else {
int k = r / (b + 1), need_add = r % (b + 1);
while (r > 0) {
int need = k + ((need_add > 0) ? 1 : 0);
need_add--;
if (need_add < 0) {
need_add = 0;
}
r -= need;
while (need > 0) {
ans.append(ch[0]);
--need;
}
if (b > 0) {
ans.append(ch[1]);
--b;
}
}
}
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();
r = in.nextInt();
b = in.nextInt();
solve(t + 1);
}
out.print(ans.toString());
out.flush();
in.close();
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c02f527f1fac8172ba7fc66691a7fa82 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.*;
import java.util.Random;
import java.util.StringTokenizer;
import static java.lang.Long.compare;
import static java.lang.Long.max;
public class RandomClass {
static final Random random = new Random();
public static class Node implements Comparable<Node>{
int x;
int y;
public Node(int a, int b) {
x = a;
y = b;
}
@Override
public int compareTo(Node o) {
if(this.x > o.x) {
return +1;
} else if(this.x == o.x){
return 0;
} else {
return -1;
}
}
}
public static class Node1 {
int number;
int count;
public Node1(int num, int c){
number = num;
count = c;
}
}
public static void main(String args[]) throws Exception {
FastReader fs = new FastReader();
StringBuilder sb = new StringBuilder();
int t = fs.nextInt();
while(t-->0) {
int n = fs.nextInt();
int r = fs.nextInt();
int b = fs.nextInt();
b = b+1;
int count = r/b;
int remainder = r%b;
StringBuilder sbTemp = new StringBuilder();
for(int i=0 ;i<count; i++) {
sbTemp.append('R');
}
StringBuilder sbAns = new StringBuilder();
for(int i=b; i>0; i--) {
sbAns.append(sbTemp);
sbAns.append('B');
}
sbAns.deleteCharAt(sbAns.length()-1);
/*if(r%b != 0) {
for(int i= r%b ; i>0; i--) {
sbAns.append('R');
}
}*/
int z = sbAns.length();
while(remainder-->0) {
sbAns.insert(z, 'R');
z -=(count+1);
}
sb.append(sbAns + "\n");
}
System.out.println(sb);
}
static int gcd(int a, int b)
{
return (a % b == 0) ?
Math.abs(b) : gcd(b,a%b);
}
static boolean isPossible(int a,
int b, int c)
{
return (c % gcd(a, b) == 0);
}
//Fast Reader
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static void ruffleSort(int[] a) {
int n = a.length;
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);
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9efd352a147db6c420ff2a5efa9ab351 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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 r=sc.nextInt();
int b=sc.nextInt();
int s=b+1;
int rr=r/s;
r%=s;
String temp="";
String ans="";
for(int i=0;i<rr;i++){
temp+="R";
} String temp3= temp;temp+="B";
String temp2="R"+temp;
s-=r;
while(r>0){ans+=temp2;r--;}
while(s>1){ans+=temp;s--;}
ans+=temp3; System.out.println(ans);
t--;
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 57710e65a689503a6ea30b37014b1c8f | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class Solution
{
public static void main(String[] args)
{
Scanner in=new Scanner(System.in);
int t=in.nextInt();
while(--t>=0)
{
int n=in.nextInt(), r=in.nextInt(), b=in.nextInt();
String s="";
int m=r/(b+1), cr=0, cb=0, rmdr=r%(b+1);
while(cr<r)
{
for(int i=0;i<m;i++)
{
s+="R"; ++cr;
}
if(rmdr!=0)
{
s+="R"; ++cr; --rmdr;
}
if(cb<b)
{
s+="B"; ++cb;
}
}
System.out.println(s);
}
in.close();
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 0ebc7d6a4c7d46cc9f3028060cc44910 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class MyClass {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t>0) {
int n,r,b;
n= sc.nextInt();
r=sc.nextInt();
b=sc.nextInt();
int[] arr = new int[b+1];
int j=0;
while(r>0)
{
arr[j] +=1;
r--;
j++;
j=j%(b+1);
}
for(int i=0;i<b;i++) {
for(j=0;j<arr[i];j++)
System.out.print("R");
System.out.print("B");
}
for(int i=0;i<arr[b];i++)
System.out.print("R");
System.out.println();
t--;
}
sc.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 02daa05916c9ca10239ebaca531f66ff | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.lang.*;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.File;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
/* 10^(7) = 1s.
* ceilVal = (a+b-1) / b */
static final int mod = 1000000007;
static final long temp = 998244353;
static final long MOD = 1000000007;
static final long M = (long)1e9+7;
static class Pair implements Comparable<Pair>
{
int first, second;
public Pair(int first, int second)
{
this.first = first;
this.second = second;
}
public int compareTo(Pair ob)
{
return (int)(first - ob.first);
}
}
static class Tuple implements Comparable<Tuple>
{
long first, second,third;
public Tuple(long first, long second, long third)
{
this.first = first;
this.second = second;
this.third = third;
}
public int compareTo(Tuple o)
{
return (int)(o.third - this.third);
}
}
public static class DSU
{
int count = 0;
int[] parent;
int[] rank;
public DSU(int n)
{
count = n;
parent = new int[n];
rank = new int[n];
Arrays.fill(parent, -1);
Arrays.fill(rank, 1);
}
public int find(int i)
{
return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
}
public void union(int a, int b) //Union Find by Rank
{
a = find(a);
b = find(b);
if(a == b) return;
if(rank[a] < rank[b])
{
parent[a] = b;
}
else if(rank[a] > rank[b])
{
parent[b] = a;
}
else
{
parent[b] = a;
rank[a] = 1 + rank[a];
}
count--;
}
public int countConnected()
{
return count;
}
}
static class Reader
{
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) throws IOException {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] longReadArray(int n) throws IOException {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
public static int gcd(int a, int b)
{
if(b == 0)
return a;
else
return gcd(b,a%b);
}
public static long lcm(long a, long b)
{
return (a / LongGCD(a, b)) * b;
}
public static long LongGCD(long a, long b)
{
if(b == 0)
return a;
else
return LongGCD(b,a%b);
}
public static long LongLCM(long a, long b)
{
return (a / LongGCD(a, b)) * b;
}
//Count the number of coprime's upto N
public static long phi(long n) //euler totient/phi function
{
long ans = n;
// for(long i = 2;i*i<=n;i++)
// {
// if(n%i == 0)
// {
// while(n%i == 0) n/=i;
// ans -= (ans/i);
// }
// }
//
// if(n > 1)
// {
// ans -= (ans/n);
// }
for(long i = 2;i<=n;i++)
{
if(isPrime(i))
{
ans -= (ans/i);
}
}
return ans;
}
public static long fastPow(long x, long n)
{
if(n == 0)
return 1;
else if(n%2 == 0)
return fastPow(x*x,n/2);
else
return x*fastPow(x*x,(n-1)/2);
}
public static long powMod(long x, long y, long p)
{
long res = 1;
x = x % p;
while (y > 0)
{
if (y % 2 == 1)
{
res = (res * x) % p;
}
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long modInverse(long n, long p)
{
return powMod(n, p - 2, p);
}
// Returns nCr % p using Fermat's little theorem.
public static long nCrModP(long n, long r,long p)
{
if (n<r)
return 0;
if (r == 0)
return 1;
long[] fac = new long[(int)(n) + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i % p;
return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
% p * modInverse(fac[(int)(n - r)], p)
% p)
% p;
}
public static long fact(long n)
{
long[] fac = new long[(int)(n) + 1];
fac[0] = 1;
for (long i = 1; i <= n; i++)
fac[(int)(i)] = fac[(int)(i - 1)] * i;
return fac[(int)(n)];
}
public static long nCr(long n, long k)
{
long ans = 1;
for(long i = 0;i<k;i++)
{
ans *= (n-i);
ans /= (i+1);
}
return ans;
}
//Modular Operations for Addition and Multiplication.
public static long perfomMod(long x)
{
return ((x%M + M)%M);
}
public static long addMod(long a, long b)
{
return perfomMod(perfomMod(a)+perfomMod(b));
}
public static long subMod(long a, long b)
{
return perfomMod(perfomMod(a)-perfomMod(b));
}
public static long mulMod(long a, long b)
{
return perfomMod(perfomMod(a)*perfomMod(b));
}
public static boolean isPrime(long n)
{
if(n == 1)
{
return false;
}
//check only for sqrt of the number as the divisors
//keep repeating so only half of them are required. So,sqrt.
for(int i = 2;i*i<=n;i++)
{
if(n%i == 0)
{
return false;
}
}
return true;
}
public static List<Long> SieveList(int n)
{
boolean prime[] = new boolean[(int)(n+1)];
Arrays.fill(prime, true);
List<Long> l = new ArrayList<>();
for (long p = 2; p*p<=n; p++)
{
if (prime[(int)(p)] == true)
{
for(long i = p*p; i<=n; i += p)
{
prime[(int)(i)] = false;
}
}
}
for (long p = 2; p<=n; p++)
{
if (prime[(int)(p)] == true)
{
l.add(p);
}
}
return l;
}
public static int countDivisors(int x)
{
int c = 0;
for(int i = 1;i*i<=x;i++)
{
if(x%i == 0)
{
if(x/i != i)
{
c+=2;
}
else
{
c++;
}
}
}
return c;
}
public static long log2(long n)
{
long ans = (long)(log(n)/log(2));
return ans;
}
public static boolean isPow2(long n)
{
return (n != 0 && ((n & (n-1))) == 0);
}
public static boolean isSq(int x)
{
long s = (long)Math.round(Math.sqrt(x));
return s*s==x;
}
/*
*
* >= <=
0 1 2 3 4 5 6 7
5 5 5 6 6 6 7 7
lower_bound for 6 at index 3 (>=)
upper_bound for 6 at index 6(To get six reduce by one) (<=)
*/
public static int LowerBound(int a[], int x)
{
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;
}
public static int UpperBound(int a[], int x)
{
int l=-1, r=a.length;
while(l+1<r)
{
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
public static void Sort(long[] a)
{
List<Long> l = new ArrayList<>();
for (long i : a) l.add(i);
Collections.sort(l);
// Collections.reverse(l); //Use to Sort decreasingly
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
public static void ssort(char[] a)
{
List<Character> l = new ArrayList<>();
for (char i : a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
public static void main(String[] args) throws Exception
{
Reader sc = new Reader();
PrintWriter fout = new PrintWriter(System.out);
int tt = sc.nextInt();
fr: while(tt-- > 0)
{
int n = sc.nextInt(), r = sc.nextInt(), b = sc.nextInt();
StringBuilder sb = new StringBuilder();
int c = 0, x = b;
int exp2 = r%(b + 1);
while(x-- > 0)
{
int exp1 = r/(b + 1);
while(exp1-- > 0)
{
sb.append('R');
c++;
}
if(exp2-- > 0)
{
sb.append('R');
c++;
}
sb.append('B');
}
int len = (r - c);
while(len-- > 0) sb.append('R');
fout.println(sb.toString());
}
fout.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | f1a80769ac8b094409972a924c14fa58 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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 r = fr.nextInt();
int b = fr.nextInt();
int mod = r % (b + 1);
int s = r / (b + 1);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < (b + 1); i ++) {
if (i != 0) sb.append("B");
for (int j = 0; j < s; j ++) {
sb.append("R");
}
if (mod > 0) {
sb.append("R");
mod --;
}
}
System.out.println(sb.toString());
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 0a2be397b207ef8e0f5c290e4c90ef7b | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
// try {
// System.setIn(new FileInputStream("input.txt"));
// System.setOut(new PrintStream(new FileOutputStream("output.txt")));
// } catch (Exception e) {
// System.err.println("Error");
// }
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t-- > 0) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
solve(n, r, b);
System.out.println();
}
}
static void solve(int n, int r, int b) {
int after = (int)Math.ceil(r / ((b+1)* 1.0));
while(b > 0 && r > 0) {
for(int i = 1; i <= after; i++){
System.out.print("R");
r--;
}
System.out.print("B");
b--;
after = (int)Math.ceil(r / ((b+1)* 1.0));
}
while(r > 0) {
System.out.print("R");
r--;
}
while(b > 0) {
System.out.print("B");
b--;
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | a8dd1ed80e56c7161d9349cf59c48354 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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();
int r = sc.ni();
int b = sc.ni();
b++;
int q = r/b;
int rem = r%b;
b--;
int bb = b;
StringBuilder sb = new StringBuilder();
for(int i = 0; i < b; i++) {
for(int j = 0; j < q; j++) {
sb.append('R');
}
r-=q;
if(rem > 0) {
sb.append('R');
rem--;
}
if(bb > 0) {
sb.append('B');
bb--;
}
}
for(int i = 0; i < q; i++) {
sb.append('R');
}
if(rem > 0) {
sb.append('R');
}
w.p(sb.toString());
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9a69dfb3a0a1ce962d4a123726169926 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int TT = sc.nextInt();
while (TT-- > 0) {
int n = sc.nextInt(), r = sc.nextInt(), b = sc.nextInt();
int k = n / (b + 1), cnt = 0;
for (int i = 0; i < n; i++) {
if (r > 0 && cnt < k && r >= b) {
System.out.print("R");
cnt++;
r--;
} else {
System.out.print("B");
b--;
cnt = 0;
}
}
System.out.println();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 94b150542f323afde8f05f682643e89e | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
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.Comparator;
import java.util.StringTokenizer;
public class CodeForces {
static int mod = (int)1e9+7;
public static void main(String[] args) throws InterruptedException {
// PrintWriter out = new PrintWriter("output.txt");
// File input = new File("input.txt");
// FastScanner fs = new FastScanner(input);
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int testNumber =fs.nextInt();
for (int T =0;T<testNumber;T++){
int n=fs.nextInt();
int r = fs.nextInt();
int b = fs.nextInt();
String ans ="";
int reds = r/(b+1);
int extra = r%(b+1);
for (int i=0;i<b;i++){
for (int j=0;j<reds;j++){
ans+="R";
}
if(extra>0){
extra--;
ans+="R";
}
ans+="B";
}
for (int j=0;j<reds;j++){
ans+="R";
}
if(extra>0){
extra--;
ans+="R";
}
out.print(ans+"\n");
}
out.flush();
}
static long modInverse( long n, long p)
{
return FastPower(n, p - 2, p);
}
static int[] factorials(int max,int mod){
int [] ans = new int[max+1];
ans[0]=1;
for (int i=1;i<=max;i++){
ans[i]=ans[i-1]*i;
ans[i]%=mod;
}
return ans;
}
static String toBinary(int num,int bits){
String res =Integer.toBinaryString(bits);
while(res.length()<bits)res="0"+res;
return res;
}
static String toBinary(long num,int bits){
String res =Long.toBinaryString(bits);
while(res.length()<bits)res="0"+res;
return res;
}
static long LCM(long a,long b){
return a*b/gcd(a,b);
}
static long FastPower(long x,long p,long mod){
if(p==0)return 1;
long ans =FastPower(x, p/2,mod);
ans%=mod;
ans*=ans;
ans%=mod;
if(p%2==1)ans*=x;
ans%=mod;
return ans;
}
static double FastPower(double x,int p){
if(p==0)return 1.0;
double ans =FastPower(x, p/2);
ans*=ans;
if(p%2==1)ans*=x;
return ans;
}
static int FastPower(int x,int p){
if(p==0)return 1;
int ans =FastPower(x, p/2);
ans*=ans;
if(p%2==1)ans*=x;
return ans;
}
static ArrayList<Vertex> vertices = new ArrayList<>();
static class Vertex {
public ArrayList<Integer>edges = new ArrayList<>();
public boolean isEnd=false;
public Vertex (){
for(int i=0;i<26;i++){
edges.set(i, -1);
}
}
}
static class Trie {
private int root=0;
public Trie(){
vertices.add(new Vertex());
}
public void InsertWord(String s){
int current = root;
for(char c:s.toCharArray()){
int pos = c-'a';
if(vertices.get(current).edges.get(pos)==-1){
vertices.add(new Vertex());
Vertex x = vertices.get(current);
x.edges.set(pos,vertices.size()-1);
vertices.set(current, x);
}
current=vertices.get(current).edges.get(pos);
}
}
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(){
br=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
}
public FastScanner(File f){
try {
br=new BufferedReader(new FileReader(f));
st=new StringTokenizer("");
} catch(FileNotFoundException e){
br=new BufferedReader(new InputStreamReader(System.in));
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());
}
double nextDouble() {
return Double.parseDouble(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a =new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
public static long factorial(int n){
if(n==0)return 1;
return (long)n*factorial(n-1);
}
public static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
static void sort (int[]a){
ArrayList<Integer> b = new ArrayList<>();
for(int i:a)b.add(i);
Collections.sort(b);
for(int i=0;i<b.size();i++){
a[i]=b.get(i);
}
}
static void sortReversed (int[]a){
ArrayList<Integer> b = new ArrayList<>();
for(int i:a)b.add(i);
Collections.sort(b,new Comparator<Integer>(){
@Override
public int compare(Integer o1, Integer o2) {
return o2-o1;
}
});
for(int i=0;i<b.size();i++){
a[i]=b.get(i);
}
}
static void sort (long[]a){
ArrayList<Long> b = new ArrayList<>();
for(long i:a)b.add(i);
Collections.sort(b);
for(int i=0;i<b.size();i++){
a[i]=b.get(i);
}
}
static ArrayList<Integer> 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;
}
}
ArrayList<Integer> ans = new ArrayList<>();
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
ans.add(i);
}
return ans;
}
static int binarySearchSmallerOrEqual(int arr[], int key)
{
int n = arr.length;
int left = 0, right = n;
int mid = 0;
while (left < right) {
mid = (right + left) >> 1;
if (arr[mid] == key) {
while (mid + 1 < n && arr[mid + 1] == key)
mid++;
break;
}
else if (arr[mid] > key)
right = mid;
else
left = mid + 1;
}
while (mid > -1 && arr[mid] > key)
mid--;
return mid;
}
static int binarySearchSmallerOrEqual(long arr[], long key)
{
int n = arr.length;
int left = 0, right = n;
int mid = 0;
while (left < right) {
mid = (right + left) >> 1;
if (arr[mid] == key) {
while (mid + 1 < n && arr[mid + 1] == key)
mid++;
break;
}
else if (arr[mid] > key)
right = mid;
else
left = mid + 1;
}
while (mid > -1 && arr[mid] > key)
mid--;
return mid;
}
public static int binarySearchStrictlySmaller(int[] arr, int target)
{
int start = 0, end = arr.length-1;
if(end == 0) return -1;
if (target > arr[end]) return end;
int ans = -1;
while (start <= end) {
int mid = (start + end) / 2;
if (arr[mid] >= target) {
end = mid - 1;
}
else {
ans = mid;
start = mid + 1;
}
}
return ans;
}
public static int binarySearchStrictlySmaller(long[] arr, long target)
{
int start = 0, end = arr.length-1;
if(end == 0) return -1;
if (target > arr[end]) return end;
int ans = -1;
while (start <= end) {
int mid = (start + end) / 2;
if (arr[mid] >= target) {
end = mid - 1;
}
else {
ans = mid;
start = mid + 1;
}
}
return ans;
}
static int binarySearch(int arr[], int x)
{
int l = 0, r = arr.length - 1;
while (l <= r) {
int m = l + (r - l) / 2;
if (arr[m] == x)
return m;
if (arr[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
static int binarySearch(long arr[], long x)
{
int l = 0, r = arr.length - 1;
while (l <= r) {
int m = l + (r - l) / 2;
if (arr[m] == x)
return m;
if (arr[m] < x)
l = m + 1;
else
r = m - 1;
}
return -1;
}
static void init(int[]arr,int val){
for(int i=0;i<arr.length;i++){
arr[i]=val;
}
}
static void init(int[][]arr,int val){
for(int i=0;i<arr.length;i++){
for(int j=0;j<arr[i].length;j++){
arr[i][j]=val;
}
}
}
static void init(long[]arr,long val){
for(int i=0;i<arr.length;i++){
arr[i]=val;
}
}
static<T> void init(ArrayList<ArrayList<T>>arr,int n){
for(int i=0;i<n;i++){
arr.add(new ArrayList());
}
}
static int binarySearchStrictlySmaller(ArrayList<Pair> arr, int target)
{
int start = 0, end = arr.size()-1;
if(end == 0) return -1;
if (target > arr.get(end).y) return end;
int ans = -1;
while (start <= end) {
int mid = (start + end) / 2;
if (arr.get(mid).y >= target) {
end = mid - 1;
}
else {
ans = mid;
start = mid + 1;
}
}
return ans;
}
static int binarySearchStrictlyGreater(int[] arr, int target)
{
int start = 0, end = arr.length - 1;
int ans = -1;
while (start <= end) {
int mid = (start + end) / 2;
if (arr[mid] <= target) {
start = mid + 1;
}
else {
ans = mid;
end = mid - 1;
}
}
return ans;
}
public static long pow(long n, long pow) {
if (pow == 0) {
return 1;
}
long retval = n;
for (long i = 2; i <= pow; i++) {
retval *= n;
}
return retval;
}
static String reverse(String s){
StringBuffer b = new StringBuffer(s);
b.reverse();
return b.toString();
}
static String charToString (char[] arr){
String t="";
for(char c :arr){
t+=c;
}
return t;
}
int[] copy (int [] arr , int start){
int[] res = new int[arr.length-start];
for (int i=start;i<arr.length;i++)res[i-start]=arr[i];
return res;
}
static int[] swap(int [] A,int l,int r){
int[] B=new int[A.length];
for (int i=0;i<l;i++){
B[i]=A[i];
}
int k=0;
for (int i=r;i>=l;i--){
B[l+k]=A[i];
k++;
}
for (int i=r+1;i<A.length;i++){
B[i]=A[i];
}
return B;
}
static int mex (int[] d){
int [] a = Arrays.copyOf(d, d.length);
sort(a);
if(a[0]!=0)return 0;
int ans=1;
for(int i=1;i<a.length;i++){
if(a[i]==a[i-1])continue;
if(a[i]==a[i-1]+1)ans++;
else break;
}
return ans;
}
static int[] mexes(int[] arr){
int[] freq = new int [100000+7];
for (int i:arr)freq[i]++;
int maxMex =0;
for (int i=0;i<=100000+7;i++){
if(freq[i]!=0)maxMex++;
else break;
}
int []ans = new int[arr.length];
ans[arr.length-1] = maxMex;
for (int i=arr.length-2;i>=0;i--){
freq[arr[i+1]]--;
if(freq[arr[i+1]]<=0){
if(arr[i+1]<maxMex)
maxMex=arr[i+1];
ans[i]=maxMex;
} else {
ans[i]=ans[i+1];
}
}
return ans;
}
static int [] freq (int[]arr,int max){
int []b = new int[max];
for (int i:arr)b[i]++;
return b;
}
static int[] prefixSum(int[] arr){
int [] a = new int[arr.length];
a[0]=arr[0];
for (int i=1;i<arr.length;i++)a[i]=a[i-1]+arr[i];
return a;
}
static class Pair {
int x;
long y;
int extra;
public Pair(int x,long y){
this.x=x;
this.y=y;
}
public Pair(int x,long y,int extra){
this.x=x;
this.y=y;
this.extra=extra;
}
// @Override
// public boolean equals(Object o) {
// if(o instanceof Pair){
// if(o.hashCode()!=hashCode()){
// return false;
// } else {
// return x==((Pair)o).x&&y==((Pair)o).y;
// }
// }
//
// return false;
//
// }
//
//
//
//
//
// @Override
// public int hashCode() {
// return x+(int)y*2;
// }
//
//
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | a9b1f890f6ca17a37dc5f9c2fab1556e | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main implements Runnable {
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
public static void main(String[] args) {
new Thread(null, new Main(), "", 256 * (1L << 20)).start();
}
public void run() {
try {
long t1 = System.currentTimeMillis();
if (System.getProperty("ONLINE_JUDGE") != null) {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
} else {
in = new BufferedReader(new FileReader("input.txt"));
out = new PrintWriter("output.txt");
}
Locale.setDefault(Locale.US);
solve();
in.close();
out.close();
long t2 = System.currentTimeMillis();
System.err.println("Time = " + (t2 - t1));
} catch (Throwable t) {
t.printStackTrace(System.err);
System.exit(-1);
}
}
String readString() throws IOException {
while (!tok.hasMoreTokens()) {
tok = new StringTokenizer(in.readLine());
}
return tok.nextToken();
}
int readInt() throws IOException {
return Integer.parseInt(readString());
}
long readLong() throws IOException {
return Long.parseLong(readString());
}
double readDouble() throws IOException {
return Double.parseDouble(readString());
}
int[] readIntArray(int n) throws IOException {
int [] a = new int[n];
for(int i = 0; i < n; i++) {
a[i] = readInt();
}
return a;
}
long[] readLongArray(int n) throws IOException {
long [] a = new long[n];
for(int i = 0; i < n; i++) {
a[i] = readLong();
}
return a;
}
void solveTest() throws IOException {
int n = readInt();
int r = readInt();
int b = readInt();
int rGroups = r / (b+1);
int rPlusOne = r % (b+1);
if (r % (b+1) == 0) {
for (int i = 0; i < b; i++) {
for (int j = 0; j < rGroups; j++) {
out.print('R');
}
out.print('B');
}
for (int j = 0; j < rGroups; j++) {
out.print('R');
}
out.println();
} else {
for (int i = 0; i < b; i++) {
for (int j = 0; j < rGroups; j++) {
out.print('R');
}
if (rPlusOne > 0) {
out.print('R');
}
rPlusOne--;
out.print('B');
}
for (int j = 0; j < rGroups; j++) {
out.print('R');
}
out.println();
}
}
void solve() throws IOException {
int numTests = readInt();
while(numTests-->0) {
solveTest();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 430650176dd923af753e400eadca0222 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.util.*;
import java.io.*;
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 r = sc.nextInt();
int b= sc.nextInt();
int x = r/(b+1);
if(r%(b+1)!=0){
x++;
}
int extra=r%(b+1);
int i=0;
int j=0;
int ti=0;
while((i+j)<(r+b)){
if(ti%2==0){
int p=0;
while(p<x & i<r){
System.out.print("R");
p++;
i++;
}
extra--;
if(extra==0){
x--;
}
} else{
System.out.print("B");
j++;
}
ti++;
}
System.out.println();
t--;
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 3aa4018047b373a077591d576feda398 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
// "static void main" must be defined in a public class.
public class Main {
public static void main(String[] args) {
new Solution().function();
}
}
class Solution {
private Scanner sc = new Scanner(System.in);
public void function() {
int T = sc.nextInt();
while (T-- > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int t = r / (b + 1);
int m = r % (b + 1);
StringBuilder bd = new StringBuilder();
for (int i = 0; i < b + 1; i++) {
int count = t;
if (m-- > 0) {
count++;
}
for (int j = 0; j < count; j++) {
bd.append("R");
}
if (i < b) {
bd.append("B");
}
}
sout(bd.toString());
}
}
private void sout(Object o) {
System.out.println(o);
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 6427832167fd444169aafc7d40dfee7b | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | //make sure to make new file!
import java.io.*;
import java.util.*;
public class A782{
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());
int r = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
int div = (r)/(b+1);
StringJoiner sj = new StringJoiner("");
int added = 0;
for(int k = 0; k < b; k++){
for(int j = 0; j < div; j++){
sj.add("R");
}
if(k < (r%(b+1))) sj.add("R");
sj.add("B");
}
for(int k = 0; k < div; k++) sj.add("R");
//if(r%(b+1) == 0) sj.add("R");
out.println(sj.toString());
}
out.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 37a8eaed28d09016e38e092b718ba92e | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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(),a=i(),b=i();
int parts=b+1;
int max=a/parts;
int mod=a%parts;
for(int i=0; i<parts; i++)
{
for(int j=0; j<max; j++)
{
ans.append("R");
}
if(mod>0)
{
mod--;
ans.append("R");
}
if(i+1!=parts)
ans.append("B");
}
ans.append("\n");
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 8397e903885c4d5bc74f94051452b656 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | 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++){
int n = sc.nextInt();
int r= sc.nextInt();
int b= sc.nextInt();
int x = r/(b+1);
int rem= r-x*(b+1);
StringBuilder sb= new StringBuilder();
for (int i=1; i<=b+1; i++){
for (int j=0; j<x; j++){
sb.append("R");
}
if (rem>0) {
sb.append("R");
rem--;
}
if (i!=b+1)sb.append('B');
}
System.out.println(sb.toString());
}
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 4303d7378a1915ddbabccb2802b29f95 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.*;
public class A {
public static void main(String[] args) throws IOException {
FastReader in = new FastReader(System.in);
PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
int t = in.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
int ri = r / (b + 1);
int rleft = r % (b + 1);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < b; i++) {
for (int j = 0; j < ri; j++) sb.append('R');
if (rleft > 0) {
sb.append('R');
rleft--;
}
sb.append('B');
}
for (int i = 0; i < ri; i++) sb.append('R');
pw.println(sb);
}
pw.close();
}
public static long Xsum(int[][] a, int i, int j, int n, int m) {
long res = 0;
res += a[i][j];
int tempi = i, tempj = j;
tempi--;
tempj--;
while ((tempi < n && tempj < m) && (tempi >= 0 && tempj >= 0)) {
res += a[tempi][tempj];
tempi--;
tempj--;
}
tempi = i;
tempj = j;
tempi++;
tempj--;
while ((tempi < n && tempj < m) && (tempi >= 0 && tempj >= 0)) {
res += a[tempi][tempj];
tempi++;
tempj--;
}
tempi = i;
tempj = j;
tempi++;
tempj++;
while ((tempi < n && tempj < m) && (tempi >= 0 && tempj >= 0)) {
res += a[tempi][tempj];
tempi++;
tempj++;
}
tempi = i;
tempj = j;
tempi--;
tempj++;
while ((tempi < n && tempj < m) && (tempi >= 0 && tempj >= 0)) {
res += a[tempi][tempj];
tempi--;
tempj++;
}
return res;
}
static void subString(char str[], int n) {
for (int len = 1; len <= n; len++) {
for (int i = 0; i <= n - len; i++) {
int j = i + len - 1;
for (int k = i; k <= j; k++) {
System.out.print(str[k]);
}
System.out.println();
}
}
}
static long gcd(long a, long b) {
if (b == 0) return a;
else return gcd(b, a % b);
}
static boolean isPalindrom(String txt) {
return txt.equals(new StringBuilder(txt).reverse().toString());
}
public static boolean isSorted(int[] arr) {
for (int i = 1; i < arr.length; i++)
if (arr[i] < arr[i - 1]) return false;
return true;
}
static long binaryExponentiation(long x, long n) {
if (n == 0) return 1;
else if (n % 2 == 0) return binaryExponentiation(x * x, n / 2);
else return x * binaryExponentiation(x * x, (n - 1) / 2);
}
public static void Sort(int[] a) {
ArrayList<Integer> lst = new ArrayList<>();
for (int i : a) lst.add(i);
Collections.sort(lst);
for (int i = 0; i < lst.size(); i++) a[i] = lst.get(i);
}
static void debug(Object... obj) {
System.err.println(Arrays.deepToString(obj));
}
static class Pair implements Comparable<Pair> {
int first, second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public int compareTo(Pair ob) {
return first - ob.first;
}
}
static class FastReader {
InputStream is;
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
public FastReader(InputStream is) {
this.is = is;
}
public int readByte() {
if (lenbuf == -1) throw new InputMismatchException();
if (ptrbuf >= lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read(inbuf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (lenbuf <= 0) return -1;
}
return inbuf[ptrbuf++];
}
public boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b)) ;
return b;
}
public String next() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public String nextLine() {
int c = skip();
StringBuilder sb = new StringBuilder();
while (!isEndOfLine(c)) {
sb.appendCodePoint(c);
c = readByte();
}
return sb.toString();
}
public int nextInt() {
int num = 0, b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = (num << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public long nextLong() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = (num << 3) + (num << 1) + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
public double nextDouble() {
return Double.parseDouble(next());
}
public char[] next(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
public char readChar() {
return (char) skip();
}
public long[] readArrayL(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = nextLong();
return arr;
}
public int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = nextInt();
return arr;
}
public int[][] read2Darray(int n, int m) {
int[][] a = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
a[i][j] = nextInt();
}
}
return a;
}
/*public static int min(int a, int b) {
return Math.min(a, b);
}
public static int max(int a, int b) {
return Math.max(a, b);
}*/
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 20c667d730cced0f9277191d3fb2ba30 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
public class Practice {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int r=sc.nextInt();
int b=sc.nextInt();
int mod=b+1;
int place=r/mod;
int rem=r%mod;
boolean bob=false;
char [] arr=new char[n];
for(int i=0;i<n;) {
if(!bob) {
int res=0;
while(res<place) {
arr[i]='R';i++;
res++;
}
if(rem>0) {
arr[i++]='R';rem--;
}
bob=!bob;
}else {
arr[i]='B';
i++;
bob=!bob;
}
}
for(char x:arr) System.out.print(x);
System.out.println();
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | f19f99343d4785c9d39c8ea738fb77bd | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | /*
* Author: rickytsung(En Chi Tsung)
* Date: 2022/7/30
* Problem: CF Round 782
*/
import java.util.*;
import java.time.*;
import java.io.*;
import java.math.*;
public class Main {
public static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
public static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
public static long ret,ans=0,x,y;
public static int reti,rd;
public static boolean neg;
public static final int mod=998244353,mod2=1_000_000_007,ma=600005;
public static Useful us=new Useful(mod);
public static int[] A=new int[ma];
public static int[] B=new int[ma];
public static void main(String[] args) throws Exception{
//String [] in=br.readLine().split(" ");
final int t=readint();
for(int z=0;z<t;z++) {
final int n=readint();
final int r=readint();
final int b=readint();
int p=r/(b+1);
int q=r%(b+1);
for(int i=0;i<=b;i++) {
for(int j=0;j<p;j++) {
bw.write("R");
}
if(q>0) {
bw.write("R");
q--;
}
if(i==b)break;
bw.write("B");
}
bw.write("\n");
}
bw.flush();
}
public static int readint() throws Exception{
reti=0;
neg=false;
while(rd<48||rd>57) {
rd=br.read();
if(rd=='-') {
neg=true;
}
}
while(rd>47&&rd<58) {
reti*=10;
reti+=(rd&15);
rd=br.read();
}
if(neg)reti*=-1;
return reti;
}
public static long readlong() throws Exception{
ret=0;
neg=false;
while(rd<48||rd>57) {
rd=br.read();
if(rd=='-') {
neg=true;
}
}
while(rd>47&&rd<58) {
ret*=10;
ret+=(rd&15);
rd=br.read();
}
if(neg)ret*=-1;
return ret;
}
}
/*
*/
/*
*/
class Pii{
int x,y,z;
Pii(int a,int b,int c){
x=a;
y=b;
z=c;
}
@Override
public boolean equals(Object o) {
if (this==o) return true;
if (!(o instanceof Pii)) return false;
Pii key = (Pii) o;
return x==key.x&&y==key.y;
}
@Override
public int hashCode() {
long result=x;
result=31*result+y;
return (int)(result%998244353);
}
}
class Pll{
long x,y;
Pll(long a,long b){
x=a;
y=b;
}
@Override
public boolean equals(Object o) {
if (this==o) return true;
if (!(o instanceof Pll)) return false;
Pll key = (Pll) o;
return x==key.x&&y==key.y;
}
@Override
public int hashCode() {
long result=x;
result=31*result+y;
return (int)(result%998244353);
}
}
class Useful{
long mod;
Useful(long m){mod=m;}
void al(ArrayList<ArrayList<Integer>> a,int n){for(int i=0;i<n;i++) {a.add(new ArrayList<Integer>());}}
void arr(int[] a,int init) {for(int i=0;i<a.length;i++) {a[i]=init;}}
void arr(long[] a,long init) {for(int i=0;i<a.length;i++) {a[i]=init;}}
void arr(int[][] a,int init) {for(int i=0;i<a.length;i++) {for(int j=0;j<a[i].length;j++) {a[i][j]=init;}}}
void arr(long[][] a,long init) {for(int i=0;i<a.length;i++) {for(int j=0;j<a[i].length;j++) {a[i][j]=init;}}}
void arr(int[][][] a,int init) {for(int i=0;i<a.length;i++) {for(int j=0;j<a[i].length;j++) {for(int k=0;k<a[i][j].length;k++) {a[i][j][k]=init;}}}}
void arr(long[][][] a,long init) {for(int i=0;i<a.length;i++) {for(int j=0;j<a[i].length;j++) {for(int k=0;k<a[i][j].length;k++) {a[i][j][k]=init;}}}}
void arr(int[][][][] a,int init) {for(int i=0;i<a.length;i++) {for(int j=0;j<a[i].length;j++) {for(int k=0;k<a[i][j].length;k++) {Arrays.fill(a[i][j][k],init);}}}}
void arr(long[][][][] a,long init) {for(int i=0;i<a.length;i++) {for(int j=0;j<a[i].length;j++) {for(int k=0;k<a[i][j].length;k++) {Arrays.fill(a[i][j][k],init);}}}}
long fast(long x,long pw) {
if(pw<=0)return 1;
if(pw==1)return x;
long h=fast(x,pw>>1);
if((pw&1)==0) {
return h*h%mod;
}
return h*h%mod*x%mod;
}
long[][] mul(long[][] a,long[][] b){
long[][] c=new long[a.length][b[0].length];
for(int i=0;i<a.length;i++) {
for(int j=0;j<b[0].length;j++) {
for(int k=0;k<a[0].length;k++) {
c[i][j]+=a[i][k]*b[k][j];
c[i][j]%=mod;
}
}
}
return c;
}
long[][] fast(long[][] x,int pw){
if(pw==1)return x;
long[][] h=fast(x,pw>>1);
if((pw&1)==0) {
return mul(h,h);
}
else {
return mul(mul(h,h),x);
}
}
int gcd(int a,int b) {
if(a==0)return b;
if(b==0)return a;
return gcd(b,a%b);
}
long gcd(long a,long b) {
if(a==0)return b;
if(b==0)return a;
return gcd(b,a%b);
}
long lcm(long a, long b){
return a*(b/gcd(a,b));
}
double log2(int x) {
return (Math.log(x)/Math.log(2));
}
double log2(long x) {
return (Math.log(x)/Math.log(2));
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | bda9075782af4983ca61550d0cebc4f3 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m=sc.nextInt();
while(m-- > 0){
int n=sc.nextInt(),r=sc.nextInt(),b=sc.nextInt();
int mod=r%(b+1),div=r/(b+1);
StringBuilder sb=new StringBuilder();
while(sb.length()<n){
for(int j=0;j<div;j++){
sb.append("R");
}
if(mod>0){
mod--;
sb.append("R");
}
if(sb.length()<n)
sb.append("B");
}
System.out.println(sb);
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c2fdca6b08365540281001ba15006e4e | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 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();
for(int i=0;i<b;i++){
for(int j=0;j<a/(b+1);j++){
System.out.print("R");
}
if(i<a%(b+1))
System.out.print("R");
System.out.print("B");
}
for(int j=0;j<a/(b+1);j++){
System.out.print("R");
}
System.out.println();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9b1aad8040a1e4d39f99035c3cec0ad1 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
if (st.hasMoreTokens()) {
str = st.nextToken("\n");
} else {
str = br.readLine();
}
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader scn = new FastReader();
int t = scn.nextInt();
while (t-- > 0) {
int n = scn.nextInt();
int r = scn.nextInt();
int b = scn.nextInt();
int count = r/(b+1);
int rem = r%(b+1);
for(int i = 1; i <= b; i++){
for(int j = 1; j <= count; j++){
System.out.print("R");
}
System.out.print("B");
if(rem-- > 0){
System.out.print("R");
}
}
for(int j = 1; j <= count; j++){
System.out.print("R");
}
if(rem > 0){
System.out.print("R");
}
System.out.println();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 14c0d5fa86aec6e1359840acabba9297 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.io.*;
import java.util.*;
public final class Main {
//int 2e9 - long 9e18
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(0, 1), new Pair(1, 0), new Pair(0, -1)};
static int mod = (int) (1e9 + 7);
static int mod2 = 998244353;
public static void main(String[] args) {
int tt = i();
while (tt-- > 0) {
solve();
}
out.flush();
}
public static void solve() {
int n = i();
int r = i();
int b = i();
int d = r / (b + 1);
int y = r - d * (b + 1);
int x = b + 1 - y;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < x; i++) {
sb.append(repeat('R', d));
sb.append('B');
}
for (int i = 0; i < y; i++) {
sb.append(repeat('R', d + 1));
sb.append('B');
}
sb.deleteCharAt(sb.length()-1);
out.println(sb.toString());
}
// (10,5) = 2 ,(11,5) = 3
static long upperDiv(long a, long b) {
return (a / b) + ((a % b == 0) ? 0 : 1);
}
static long sum(int[] a) {
long sum = 0;
for (int x : a) {
sum += x;
}
return sum;
}
static int[] preint(int[] a) {
int[] pre = new int[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] pre(int[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] post(int[] a) {
long[] post = new long[a.length + 1];
post[0] = 0;
for (int i = 0; i < a.length; i++) {
post[i + 1] = post[i] + a[a.length - 1 - i];
}
return post;
}
static long[] pre(long[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static void print(char A[]) {
for (char c : A) {
out.print(c);
}
out.println();
}
static void print(boolean A[]) {
for (boolean c : A) {
out.print(c + " ");
}
out.println();
}
static void print(int A[]) {
for (int c : A) {
out.print(c + " ");
}
out.println();
}
static void print(long A[]) {
for (long i : A) {
out.print(i + " ");
}
out.println();
}
static void print(List<Integer> A) {
for (int a : A) {
out.print(a + " ");
}
}
static int i() {
return in.nextInt();
}
static long l() {
return in.nextLong();
}
static double d() {
return in.nextDouble();
}
static String s() {
return in.nextLine();
}
static String c() {
return in.next();
}
static int[][] inputWithIdx(int N) {
int A[][] = new int[N][2];
for (int i = 0; i < N; i++) {
A[i] = new int[]{i, in.nextInt()};
}
return A;
}
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 int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long GCD(long a, long b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long LCM(int a, int b) {
return (long) a / GCD(a, b) * b;
}
static long LCM(long a, long b) {
return a / GCD(a, b) * b;
}
// find highest i which satisfy a[i]<=x
static int lowerbound(int[] a, int x) {
int l = 0;
int r = a.length - 1;
while (l < r) {
int m = (l + r + 1) / 2;
if (a[m] <= x) {
l = m;
} else {
r = m - 1;
}
}
return l;
}
static void shuffle(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
}
static void shuffleAndSort(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static void shuffleAndSort(int[][] arr, Comparator<? super int[]> comparator) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int[] temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr, comparator);
}
static void shuffleAndSort(long[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
long temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static boolean isPerfectSquare(double number) {
double sqrt = Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static void swap(int A[], int a, int b) {
int t = A[a];
A[a] = A[b];
A[b] = t;
}
static void swap(char A[], int a, int b) {
char t = A[a];
A[a] = A[b];
A[b] = t;
}
static long pow(long a, long b, int mod) {
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 pow(long a, long b) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow *= x;
}
x = x * x;
b /= 2;
}
return pow;
}
static long modInverse(long x, int mod) {
return pow(x, mod - 2, mod);
}
static boolean isPrime(long N) {
if (N <= 1) {
return false;
}
if (N <= 3) {
return true;
}
if (N % 2 == 0 || N % 3 == 0) {
return false;
}
for (int i = 5; i * i <= N; i = i + 6) {
if (N % i == 0 || N % (i + 2) == 0) {
return false;
}
}
return true;
}
public static String reverse(String str) {
if (str == null) {
return null;
}
return new StringBuilder(str).reverse().toString();
}
public static void reverse(int[] arr) {
for (int i = 0; i < arr.length / 2; i++) {
int tmp = arr[i];
arr[arr.length - 1 - i] = tmp;
arr[i] = arr[arr.length - 1 - i];
}
}
public static String repeat(char ch, int repeat) {
if (repeat <= 0) {
return "";
}
final char[] buf = new char[repeat];
for (int i = repeat - 1; i >= 0; i--) {
buf[i] = ch;
}
return new String(buf);
}
public static int[] manacher(String s) {
char[] chars = s.toCharArray();
int n = s.length();
int[] d1 = new int[n];
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : Math.min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && chars[i - k] == chars[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
return d1;
}
public static int[] kmp(String s) {
int n = s.length();
int[] res = new int[n];
for (int i = 1; i < n; ++i) {
int j = res[i - 1];
while (j > 0 && s.charAt(i) != s.charAt(j)) {
j = res[j - 1];
}
if (s.charAt(i) == s.charAt(j)) {
++j;
}
res[i] = j;
}
return res;
}
}
class Pair {
int i;
int j;
Pair(int i, int j) {
this.i = i;
this.j = j;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Pair pair = (Pair) o;
return i == pair.i && j == pair.j;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
class ThreePair {
int i;
int j;
int k;
ThreePair(int i, int j, int k) {
this.i = i;
this.j = j;
this.k = k;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
ThreePair pair = (ThreePair) o;
return i == pair.i && j == pair.j && k == pair.k;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
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 Node {
int val;
public Node(int val) {
this.val = val;
}
}
class ST {
int n;
Node[] st;
ST(int n) {
this.n = n;
st = new Node[4 * Integer.highestOneBit(n)];
}
void build(Node[] nodes) {
build(0, 0, n - 1, nodes);
}
private void build(int id, int l, int r, Node[] nodes) {
if (l == r) {
st[id] = nodes[l];
return;
}
int mid = (l + r) >> 1;
build((id << 1) + 1, l, mid, nodes);
build((id << 1) + 2, mid + 1, r, nodes);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
void update(int i, Node node) {
update(0, 0, n - 1, i, node);
}
private void update(int id, int l, int r, int i, Node node) {
if (i < l || r < i) {
return;
}
if (l == r) {
st[id] = node;
return;
}
int mid = (l + r) >> 1;
update((id << 1) + 1, l, mid, i, node);
update((id << 1) + 2, mid + 1, r, i, node);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
Node get(int x, int y) {
return get(0, 0, n - 1, x, y);
}
private Node get(int id, int l, int r, int x, int y) {
if (x > r || y < l) {
return new Node(0);
}
if (x <= l && r <= y) {
return st[id];
}
int mid = (l + r) >> 1;
return comb(get((id << 1) + 1, l, mid, x, y), get((id << 1) + 2, mid + 1, r, x, y));
}
Node comb(Node a, Node b) {
if (a == null) {
return b;
}
if (b == null) {
return a;
}
return new Node(GCD(a.val, b.val));
}
static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | abe95e21ade51f63ecf2bdd516343ba4 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
try {
br = new BufferedReader(
new FileReader("input.rtf"));
PrintStream out = new PrintStream(new FileOutputStream("output.rtf"));
System.setOut(out);
} catch (Exception e) {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader read = new FastReader();
int tests = read.nextInt();
for(int i = 0 ; i < tests; i++){
int n = read.nextInt();
int r = read.nextInt();
int b = read.nextInt();
int p = r/(b+1);
int q = r % (b+1);
String y = "";
for(int j = 0; j < p; j++) y += "R";
String ans = "";
for(int j = 0; j < b + 1; j++){
if(j > 0) ans += "B";
ans += y;
if(q > 0) ans += "R";
q -= 1;
}
System.out.println(ans);
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 4091b24cfbb6d4cd720292deeb302132 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class RedVersusBlue {
public static String duplicateR(int min) {
StringBuilder builder = new StringBuilder();
for (int i = 0; i < min; i++)
builder.append("R");
return builder.toString();
}
public static String print(ArrayList<Integer> rVb) {
/* Formula: || r = extraOne + min * (b + 1) ||
* Which means: There are (b + 1) strings with min amount of R,
* and there are extraOne strings with an additional R char in it.
* For example, given r = 12, b = 7:
* min = 12 / (7 + 1) = 12 / 8 = 1
* extraOne = 12 % (7 + 1) = 12% 8 = 4
* This is true because 12 R and 7 B yields: RR B RR B RR B RR B R B R B R
* The min amount of R that can appear continuously is 1, and there are
* 4 (extraOne) amount of strings that appear with (min + 1) = 2 R continuously.
*/
int n = rVb.get(0), r = rVb.get(1), b = rVb.get(2);
int min = r / (b + 1);
int extraOne = r % (b + 1);
String duplicateRB = duplicateR(min) + "B ";
StringBuilder builder = new StringBuilder();
for (int i = 0; i < b; i++)
builder.append(duplicateRB);
String base = builder.toString();
base = base + duplicateR(min);
String[] baseResult = base.split(" ", b + 1);
for (int i = 0; i < extraOne; i++)
baseResult[i] = "R" + duplicateR(min) + "B";
StringBuilder result = new StringBuilder();
for (int i = 0; i < b + 1; i++)
result.append(baseResult[i]);
return result.toString();
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// Scan amount of test cases
int t = input.nextInt();
int tScan = t * 3;
// Print result
ArrayList<String> result = new ArrayList<>();
for (int i = 0; i < tScan; i = i + 3) {
ArrayList<Integer> rVb = new ArrayList<>();
int n = input.nextInt();
int r = input.nextInt();
int b = input.nextInt();
rVb.add(n);
rVb.add(r);
rVb.add(b);
result.add(print(rVb));
}
for (int i = 0; i < t; i++)
System.out.println(result.get(i));
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | b2aa951c532b83de3ab9e4603dbebfe8 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.io.*;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
public class Main {
static class Reader{
BufferedReader br;
StringTokenizer st;
public Reader(boolean f) throws IOException{
if(f) {
br = new BufferedReader(new FileReader("input.txt"));
}else{
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 Writer {
private final BufferedWriter bw;
public Writer(boolean f) throws IOException {
if(f) {
this.bw = new BufferedWriter(new FileWriter("output.txt"));
}else{
this.bw = new BufferedWriter(new OutputStreamWriter(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 close() throws IOException {
bw.close();
}
}
static boolean isPowerOfTwo (long x)
{
/* First x in the below expression is
for the case when x is 0 */
return x!=0 && ((x&(x-1)) == 0);
}
static boolean checkperfectsquare(long n)
{
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.ceil((double)Math.sqrt(n)) ==
Math.floor((double)Math.sqrt(n)))
{
return true;
}
else
{
return false;
}
}
public static void main(String[] args){
try {
Reader in=new Reader(false);
Writer out =new Writer(false);
int testCase = 1;
testCase=in.nextInt();
while(testCase-- > 0) {
// write code here
long n = in.nextLong();
long r = in.nextLong();
long b = in.nextLong();
long x = r / (b + 1);
long y = r % (b + 1);
String str="";
String ans ="";
for (int i = 0; i < x; i++) {
str+='R';
}
for (int i = 0; i < b+1; i++) {
if(i>0){
ans+='B';
}
ans+=str;
if(y>0){
ans+='R';
y--;
}
}
out.println(ans);
}
out.close();
} catch (Exception e) {
return;
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 38e07a0573b154954db61a97f9530fb4 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.ArrayList;
import java.util.Scanner;
import java.util.Arrays;
import java.util.HashMap;
public class vanita {
public static void main(String[] args){
Scanner in =new Scanner(System.in);
int cases = in.nextInt();
for(int i = 0; i < cases; i++){
int nums = in.nextInt();
int reds = in.nextInt();
int blues = in.nextInt();
int counter = reds;
int places = blues + 1;
int extras = reds % places;
double min =Math.floor(reds/places);
//double max = Math.ceil((double) (reds-1)/(blues+1));
for(int j = 0; j < places;j++){
for(int k = 0; k <min; k++ ){
System.out.print("R");
}
if(extras != 0){
extras--;
System.out.print("R");
}
if(blues != 0){
System.out.print("B");
blues--;
}
}
System.out.println(" ");
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | f6eed0235669e4312e04e05f4cbc250d | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.ArrayList;
import java.util.Scanner;
import java.util.Arrays;
import java.util.HashMap;
public class vanita {
public static void main(String[] args){
Scanner in =new Scanner(System.in);
int cases = in.nextInt();
for(int i = 0; i < cases; i++){
int nums = in.nextInt();
int reds = in.nextInt();
int blues = in.nextInt();
int counter = reds;
int places = blues + 1;
int extras = reds % places;
double min =Math.floor(reds/places);
double max = Math.ceil((double) (reds-1)/(blues+1));
for(int j = 0; j < places;j++){
for(int k = 0; k <min; k++ ){
System.out.print("R");
}
if(extras != 0){
extras--;
System.out.print("R");
}
if(blues != 0){
System.out.print("B");
blues--;
}
}
System.out.println(" ");
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | e563083fa44f1f7e1cb6752f297d394a | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main
{
public static void main(String args[])
{
FastScanner input = new FastScanner();
int tc = input.nextInt();
work:
while (tc-- > 0) {
int n = input.nextInt();
int r = input.nextInt();
int b= input.nextInt();
String ans = "";
while(r>0||b>0)
{
int temp = (r+b)/(b+1);
r-=temp;
while(temp-->0)
{
ans+="R";
}
if(b>0)
{
ans+="B";
b--;
}
}
System.out.println(ans);
}
}
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 nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine() throws IOException
{
return br.readLine();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c5a6c3b372583294928598456698e38b | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class A1659 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int t=0; t<T; t++) {
int N = in.nextInt();
int R = in.nextInt();
int B = in.nextInt();
int[] strikes = new int[B+1];
for (int r=0; r<R; r++) {
strikes[r%strikes.length]++;
}
StringBuilder answer = new StringBuilder();
for (int strike : strikes) {
for (int i=0; i<strike; i++) {
answer.append('R');
}
answer.append('B');
}
answer.deleteCharAt(answer.length()-1);
System.out.println(answer);
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | cf2f060eca4b80147386098820b9ec63 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static PrintStream out = new PrintStream(System.out);
static LinkedList<LinkedList<Integer>> adj;
static boolean[] vis;
static long ans;
static int target;
static HashMap<ArrayList<Integer>, Integer> seqs;
//static ArrayList<ArrayList<Integer>> lists;
public static void main(String[] args){
FastScanner sc = new FastScanner();
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int groups = b + 1;
int rem = r % groups;
int init = r / groups;
String ans = "";
String s = "";
for(int i = 0; i < init; i++){
s += 'R';
}
while(b > 0){
ans += s;
if(rem > 0){
ans += 'R';
rem--;
}
ans += 'B';
b--;
}
ans += s;
out.println(ans);
}
}
public static String generate(char c, int freq){
String str = "";
for(int i = 0; i < freq; i++)
str += c;
return str;
}
public static void dfs(int v){
if(vis[v]) return;
vis[v] = true;
LinkedList<Integer> neighbors = adj.get(v);
Iterator<Integer> it = neighbors.iterator();
while(it.hasNext()){
int node = it.next();
if(!vis[node]){
dfs(node);
}
}
}
public static void mergeSort(int[] inputArray){
int n = inputArray.length;
// if input array is empty or contains only one element (meaning already sorted)
if(n < 2){
return;
}
// split input array
int mid = n / 2;
int[] leftHalf = new int[mid];
int[] rightHalf = new int[n - mid];
for(int i = 0; i < mid; i++){
leftHalf[i] = inputArray[i];
}
for(int i = mid; i < n; i++){
rightHalf[i - mid] = inputArray[i];
}
// merge sort both halves
mergeSort(leftHalf);
mergeSort(rightHalf);
// merge halves back into one array
merge(inputArray, leftHalf, rightHalf);
}
public static void merge(int[] inputArray, int[] leftArray, int[] rightArray){
int leftSize = leftArray.length;
int rightSize = rightArray.length;
int i = 0, j = 0, k = 0;
// get smaller element between two arrays
while(i < leftSize && j < rightSize){
if(leftArray[i] <= rightArray[j]){
inputArray[k] = leftArray[i++];
}
else{
for(int x = i ; x < leftSize; x++){
//out.print(leftArray[x] + " " + rightArray[j]);
int u = leftArray[x];
int v = rightArray[j];
adj.get(u).add(v);
adj.get(v).add(u);
}
inputArray[k] = rightArray[j++];
}
k++;
}
// check left for leftovers
while(i < leftSize){
inputArray[k++] = leftArray[i++];
}
// check right for leftovers
while(j < rightSize){
inputArray[k++] = rightArray[j++];
}
}
public static void addUndirectedEdge(int u, int v){
adj.get(u).add(v);
adj.get(v).add(u);
}
}
// custom I/O
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner()
{
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[] nextIntArray(int length) {
int[] arr = new int[length];
for (int i = 0; i < length; i++)
arr[i] = nextInt();
return arr;
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 260a6b8b3f0522bb47bc4d6d497a94ae | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class a_1659 {
public static void main(String[] args) throws Exception {
// TODO Auto-generated method stub
FastScanner in = new FastScanner(System.in);
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
int T = in.nextInt();
for(int aa = 0; aa < T; aa++) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
int amount = (r)/(b+1);
int rem = r%(b+1);
for(int i = 0; i < b+1; i++) {
for(int j = 0; j < amount; j++) {
out.print("R");
}
if(rem > 0) {
out.print("R");
rem--;
}
if(i < b) {
out.print("B");
}
}
out.println();
}
out.close();
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(InputStream stream) {
br = new BufferedReader(new InputStreamReader(stream));
st = new StringTokenizer("");
}
public FastScanner(String fileName) throws Exception {
br = new BufferedReader(new FileReader(new File(fileName)));
st = new StringTokenizer("");
}
public String next() throws Exception {
while (!st.hasMoreTokens()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
public int nextInt() throws Exception {
return Integer.parseInt(next());
}
public long nextLong() throws Exception {
return Long.parseLong(next());
}
public Double nextDouble() throws Exception {
return Double.parseDouble(next());
}
public String nextLine() throws Exception {
if (st.hasMoreTokens()) {
StringBuilder str = new StringBuilder();
boolean first = true;
while (st.hasMoreTokens()) {
if (first) {
first = false;
} else {
str.append(" ");
}
str.append(st.nextToken());
}
return str.toString();
} else {
return br.readLine();
}
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 5dca2632c1824660332c09623128705b | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.BufferedInputStream;
import java.util.Scanner;
public class Accepted {
public static void main(String[] args){
Scanner sc = new Scanner(new BufferedInputStream(System.in));
int t = sc.nextInt();
System.out.println();
while (t-- > 0) {
int n = sc.nextInt();
int r = sc.nextInt(), b = sc.nextInt();
char[] ans = new char[n];
int turn = r / (b + 1);
int res = r % (b + 1);
int idx = 0;
for (int i = 0; i < b; i++){
for (int j = 0; j < turn; j++) {
ans[idx++] = 'R';
r--;
}
if (res-- > 0){
ans[idx++] = 'R';
r--;
}
ans[idx++] = 'B';
}
while (r-- > 0){
ans[idx++] = 'R';
}
System.out.println(String.valueOf(ans));
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 687761eaf46b3ca19cf019d117cb79a3 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | //package codeforces;
import java.io.*;
import java.util.*;
import java.lang.*;
public class Stringinput {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(InputStream in) {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public boolean hasNextInt() {
return true;
}
}
// static int dx[]={0,-1,0,1};
// static int dy[]={-1,0,1,0};
public static void main(String[] args) throws IOException {
FastReader sc = new FastReader(System.in);
// PrintWriter out = new PrintWriter(System.out);
int t = sc.nextInt();
StringBuffer top = new StringBuffer();
// char alp[]={'a', 'b', 'c' , 'd' , 'e', 'f', 'g', 'h','i','j','k','l','m','n','o','p', 'q','r','s', 't', 'u','v','w','x','y','z' };
while (t-- > 0) {
ArrayList<pair> pr = new ArrayList<>();
//ArrayList<Long> al = new ArrayList<>();
// SortedSet<Integer> hs = new TreeSet<>();
HashMap<String , Long> mp = new HashMap<>();
HashSet<Integer> hs = new HashSet<>();
int n=sc.nextInt();
int r=sc.nextInt();
int b=sc.nextInt();
int ar[] = new int[n];
// int br[] = new int[n];
// for(int i=0; i<n; i++) {
// ar[i] = sc.nextInt();}
int val = r/(b+1);
int mod = r%(b+1);
String s1="";
for(int i=1; i<=val; i++){
s1+='R';
}
String s2=s1+'R';
for(int i=1; i<=mod; i++){
top.append(s2).append('B');
}
for(int i=mod+1; i<=b; i++){
top.append(s1).append('B');
}
top.append(s1).append("\n");
}
System.out.println(top);
}
static long differBit(long A, long B)
{
long XOR = A ^ B;
// Check for 1's in the binary form using
// Brian Kerninghan's Algorithm
long count = 0;
while (XOR > 0) {
XOR = XOR & (XOR - 1);
count++;
}
// return the count of different bits
return count;
}
static boolean match(String st,int it){
for(int i=0; i+it<st.length(); i++){
if(st.charAt(i)!= st.charAt(i+it)){
return false;
}
}
return true;
}
public static int upper_bound(long arr[], int N, long X) {
int mid;
// Initialise starting index and
// ending index
int low = 0;
int high = N;
// Till low is less than high
while (low < high) {
// Find the middle index
mid = low + (high - low) / 2;
// System.out.print(mid+" ");
// If X is greater than or equal
// to arr[mid] then find
// in right subarray
if (X >= arr[mid]) {
low = mid + 1;
}
// If X is less than arr[mid]
// then find in left subarray
else {
high = mid;
}
}
// if X is greater than arr[n-1]
// if(low < N && arr.get(low) <= X) {
// low++;
// }
if(arr[low]>X){
return low;
}
if(arr[high]>X){
return high;
}
// Return the upper_bound index
System.out.println("sumit");
return N;
}
static long meBinary(long[] ar,long X){
int l=0; int h=ar.length-1;
int mid=0;
while(h-l>1){
mid = (l+h)/2;
long val =ar[mid]-(long) (mid+1);
if(X>=val){
l=mid;
}else {
h=mid-1;
}
}
if(X>=ar[h]-(long) (h+1)){
X=X-(ar[h]-(long) (h+1));
return ar[h]+X;
}
if(X>=ar[l]-(long) (l+1)){
X=X-(ar[l]-(long) (l+1));
return ar[l]+X;
}
return X;
}
static int lowestprimeFactor(int num){
int lst=num;
for(int i=2; i*i<=num; i++){
if(num%i==0){
lst=i;
break;
}
}
return lst;
}
static long powerOptimised(long a, long n) {
// Stores final answer
long ans = 1;
while (n > 0)
{
long last_bit = (n & 1);
// Check if current LSB
// is set
if (last_bit > 0)
{
ans = ans * a;
}
a = a * a;
// Right shift
n = n >> 1;
}
return ans;
}
public static long lower_bound(long prf[], int N, long X) {
int mid;
// Initialise starting index and
// ending index
int low = 0;
int high = N;
// Till low is less than high
while ( high-low >1) {
mid = (high + low) / 2;
// If X is less than or equal
// to arr[mid], then find in
// left subarray
long rmg = X-prf[mid];
// If X is greater arr[mid]
// then find in right subarray
if(rmg<=prf[mid]){
high=mid;
} else{
low=mid;
}
}
long rmg = X-prf[low];
int cntr = N-low;
int cntl = low+1;
if(rmg>prf[low] && cntr<cntl) return 1;
rmg = X-prf[high];
cntr = N-high;
cntl = high+1;
if(rmg>prf[high] && cntr<cntl) return 1;
return 0;
// if X is greater than arr[n-1]
// if(low < N && arr.get(low) < X) {
// low++;
// }
// if(((low*(low+1))/2)>=X){
// return low;
// }
// if(((high*(high+1))/2)>=X){
// return high;
// }
// Return the lower_bound index
// return -1;
}
static boolean compareString(String curr,String prev,int n){
for(int i=0; i<n; i++){
if(curr.charAt(i)<prev.charAt(i)){
return true;
} else if(curr.charAt(i)>prev.charAt(i)){
return false;
}
}
return false;
}
private static boolean isLCM(long a, long b, long n) {
long lcm = ((a*b)/gcd(a,b));
if(lcm==n) return true;
return false;
}
static long highestPowerof2(long x) {
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1);
}
static int nextPowerOf2(int n) {
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
static long gcd(long a, long b) {
if (b == 0) {
return a;
}
return gcd(b, a % b);
}
static void height(int idx, ArrayList<ArrayList<Integer>> adj, int vis[], int n, long cnt, long maxh[]) {
vis[idx] = 1;
maxh[0] = Math.max(maxh[0], cnt);
for (int it : adj.get(idx)) {
cnt++;
height(it, adj, vis, n, cnt, maxh);
cnt--;
}
}
}
class pair{
int ft;
int st;
pair(int ft, int st){
this.ft=ft;
this.st=st;
}}
class change implements Comparator<pair>{
@Override
public int compare(pair m1, pair m2){
if(m1.ft==m2.ft){
return (m1.st-m2.st);
} else{
return (m1.ft-m2.ft);
}}}
class piro{
int ft;
int st;
piro(int ft,int st){
this.ft=ft;
this.st=st;
}}
class chance implements Comparator<piro>{
@Override
public int compare(piro m1, piro m2){
if(m1.ft==m2.ft){
return m1.st-m2.st;
} else{
return (m1.ft-m2.ft);
}}}
//
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 15a5eac0509dbf7a200c7a9a83c48f03 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class Main {
public static String solution(int n, int r, int b) {
int group = b+1;
int mod = r%group;
StringBuilder sb = new StringBuilder();
int each = r/group;
for(int i = 0; i < mod; i++) {
for(int j = 0; j < each+1; j++) {
sb.append('R');
}
sb.append('B');
}
for(int i = 0; i < group-mod-1; i++) {
for(int j = 0; j < each; j++) {
sb.append('R');
}
sb.append('B');
}
for(int j = 0; j < each; j++) {
sb.append('R');
}
return sb.toString();
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int i = 0; i < t; i++) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
String s = solution(n,r,b);
System.out.println(s);
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | f713eb10296857236de8386deaf86d7f | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
//A. Red Versus Blue
public class A {
public static void main(String[] args) {
new A().solve();
}
public void solve() {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
char[] res = new char[105];
while (t-- > 0) {
int length = scanner.nextInt();
int r = scanner.nextInt();
int b = scanner.nextInt();
int k = 0;
for (int i = 0; i < b + 1; i++) {
for (int j = 0; j < r / (b + 1); j++) {
res[k++] = 'R';
}
if (i < r - r / (b + 1) * (b + 1)) {
res[k++] = 'R';
}
res[k++] = 'B';
}
System.out.println(String.valueOf(res, 0, length));
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 8 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | db5a611aebc7c527a6750087b0acb0cc | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // Question -> https://codeforces.com/problemset/problem/1659/A
import java.util.*;
public class LiveQes1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int min = r / (b + 1);
int rest = r % (b + 1);
StringBuilder str = new StringBuilder();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < min; i++) {
str.append("R");
}
for (int i = 0; i < b + 1; i++) {
ans.append(str.toString());
if (rest != 0) {
ans.append("R");
rest--;
}
if (i != b) {
ans.append("B");
}
}
System.out.println(ans.toString());
t--;
}
}
// public static void main(String[] args) {
// Scanner scn = new Scanner(System.in);
// int test_cases = scn.nextInt();
// for (int i = 0; i < test_cases; i++) {
// int n = scn.nextInt();
// int r = scn.nextInt();
// int b = scn.nextInt();
// int numberOfConssecutiveR = r / (b + 1);
// int numberOfRemainingR = r % (b + 1);
// // System.out.println(numberOfConssecutiveR + " " + numberOfRemainingR);
// //debug
// String consecutiveR = "";
// for (int k = 0; k < numberOfConssecutiveR; k++) {
// consecutiveR += "R";
// }
// /* WARNING!! dumb codes in this (slash star) comment section
// String remainingR = "";
// for (int k = 0; k < numberOfRemainingR; k++) {
// remainingR += "R";
// }
// String ans = "";
// // gaps = b+1 or n-b
// for (int j = 0; j < b; j++) {
// ans = consecutiveR + "B" + ans;
// }
// System.out.println(ans + remainingR);
// */
// String ans = "";
// for (int j = 0; j < b; j++) {
// ans += consecutiveR;
// if (numberOfRemainingR > 0) {
// ans += "R";
// numberOfRemainingR--;
// }
// ans += "B";
// }
// System.out.println(ans + consecutiveR);
// }
// scn.close();
// }
// }
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 1577263f28f1a238b7d4174596c56499 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // Question -> https://codeforces.com/problemset/problem/1659/A
import java.util.*;
public class test {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int min = r / (b + 1);
int rest = r % (b + 1);
StringBuilder str = new StringBuilder();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < min; i++) {
str.append("R");
}
for (int i = 0; i < b + 1; i++) {
ans.append(str.toString());
if (rest != 0) {
ans.append("R");
rest--;
}
if (i != b) {
ans.append("B");
}
}
System.out.println(ans.toString());
t--;
}
// sc.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 6d969a17ba73ec1b928265ef3c6c03b1 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // Question -> https://codeforces.com/problemset/problem/1659/A
import java.util.*;
public class test {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int min = r / (b + 1);
int rest = r % (b + 1);
StringBuilder str = new StringBuilder();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < min; i++) {
str.append("R");
}
for (int i = 0; i < b + 1; i++) {
ans.append(str.toString());
if (rest != 0) {
ans.append("R");
rest--;
}
if (i != b) {
ans.append("B");
}
}
System.out.println(ans.toString());
t--;
}
sc.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 4b940edf2924c68fa70d7fb3ec0bc4b4 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // Question -> https://codeforces.com/problemset/problem/1659/A
import java.util.*;
public class test {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int min = r / (b + 1);
int rest = r % (b + 1);
StringBuilder str = new StringBuilder();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < min; i++) {
str.append("R");
}
for (int i = 0; i < b + 1; i++) {
ans.append(str);
if (rest != 0) {
ans.append("R");
rest--;
}
if (i != b) {
ans.append("B");
}
}
System.out.println(ans.toString());
t--;
}
sc.close();
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 57b48f0dad159f491c5f88b3b76c1b27 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | // Question -> https://codeforces.com/problemset/problem/1659/A
import java.util.*;
public class LiveQes1 {
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
int test_cases = scn.nextInt();
for (int i = 0; i < test_cases; i++) {
int n = scn.nextInt();
int r = scn.nextInt();
int b = scn.nextInt();
int numberOfConssecutiveR = r / (b + 1);
int numberOfRemainingR = r % (b + 1);
// System.out.println(numberOfConssecutiveR + " " + numberOfRemainingR); //debug
String consecutiveR = "";
for (int k = 0; k < numberOfConssecutiveR; k++) {
consecutiveR += "R";
}
// String remainingR = "";
// for (int k = 0; k < numberOfRemainingR; k++) {
// remainingR += "R";
// }
// String ans = "";
// // gaps = b+1 or n-b
// for (int j = 0; j < b; j++) {
// ans = consecutiveR + "B" + ans;
// }
// System.out.println(ans + remainingR);
String ans = "";
for (int j = 0; j < b; j++) {
ans += consecutiveR;
if (numberOfRemainingR > 0) {
ans += "R";
numberOfRemainingR--;
}
ans += "B";
}
System.out.println(ans + consecutiveR);
}
scn.close();
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | dfef1940d8be5f612ab0b60fe90fca87 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class scratch
{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t>0){
t--;
int n=sc.nextInt();
int r=sc.nextInt();
int b=sc.nextInt();
String s="";
while(b>0){
int divide=r/(b+1);
if(divide*(b+1)!=r){
divide=divide+1;
}
while(divide>0){
s+="R";
divide--;
r--;
}
s+='B';
b--;
}
while(r>0){
s+='R';
r--;
}
System.out.println(s);
}
}
public static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 6342796524f25a398804ef63687e09ca | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class Solution{
public static void main(String[] args){
Scanner y=new Scanner(System.in);
int t=y.nextInt();
while(t!=0)
{
--t;
int n=y.nextInt();
int r=y.nextInt();
int b=y.nextInt();
int m=r/(b+1);
int w=r%(b+1);
String s="";
for(int i=0;i<m;i++)
{
s=s+'R';
}
for(int j=0;j<b;j++)
{
if(w>0)
{
s=s+'R';
--w;
}
s=s+'B';
for(int i=0;i<m;i++)
{
s=s+'R';
}
}
System.out.println(s);
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | a8d58b38da47c7e228639e3235392e46 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
import static java.lang.Integer.*;
public class codeforces {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t =scan.nextInt();
for (int tc = 1; tc <= t; tc++) {
int n=scan.nextInt();
int r=scan.nextInt();
int b= scan.nextInt();
int groups=r/(b+1);
int remain=r%(b+1);
StringBuilder sb=new StringBuilder();
for (int i = 0; i < n; i++) {
for (int j = 0; j < groups; j++) {
sb.append("R");
i++;
}
if(remain>0){
remain--;
sb.append("R");
i++;
}
if(b>0){
b--;
sb.append("B");
i++;
}
i--;
}
System.out.println(sb.toString());
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 931565255ffdb5b09e17ac8977646b63 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class Main
{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int i=0; i<t; i++)
{ int n=sc.nextInt(); int r=sc.nextInt();
int b=sc.nextInt();
String ans="";
int y=0; int k=0; int p=0;
if(r%(b+1)==0) { p=r/(b+1);
}
else { p=((r/(b+1))+1); }
while(y<n)
{ if(k==p) { ans+='B'; k=0; r-=p; b--; y++;
if(r%(b+1)==0) { p=r/(b+1);
}
else { p=((r/(b+1))+1); }
}
else{ ans+='R'; k++; y++; }
}
// 8 2 ==> 3B3B2
// 6 4 ==> 6/5=2; RRB
System.out.println(ans);
}
// it's same problem as
// let's say we have r and we have to divide r into (b+1) number as sums
// and we need to minimize max of this b+1
// minimized maximum is ceil(r/(b+1))
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 232fda9603e278ea4fd48312fe320a7b | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
import java.util.regex.*;
import java.util.stream.*;
public class A {
static FastScanner fs = new FastScanner();
public static void main(String[] args) throws IOException {
int T = fs.nextInt();
int N, r, b;
while (T-- > 0) {
N = fs.nextInt();
r = fs.nextInt();
b = fs.nextInt();
int q, rr;
q = r/(b + 1);
rr = r - (b + 1) * q;
List<Character> result = new ArrayList<>();
List<Character> pattern = new ArrayList<>();
for (int i = 0; i < q; i++)
pattern.add('R');
result.addAll(pattern);
while(rr-- > 0) {
result.add('R');
result.add('B');
result.addAll(pattern);
b--;
}
while (b-- > 0) {
result.add('B');
result.addAll(pattern);
}
System.out.println(result.stream()
.map(Object::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 {
if (stk == null || !stk.hasMoreTokens())
return br.readLine().trim();
else
return Collections.list(stk).stream()
.map(Object::toString)
.collect(Collectors.joining(" "));
}
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 | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c9cdc026359ae79d710c034d364d0145 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
int test = sc.nextInt();
for(int t = 0; t<test; ++t){
long n = sc.nextLong();
long r = sc.nextLong();
long b = sc.nextLong();
long common_in_all_reds = r/(b+1);
long extra_red_remaining = r%(b+1);
long i = 0;
while(i < n){
for(int j = 0; j<common_in_all_reds; ++j){
System.out.print("R");
}
i += common_in_all_reds;
if(i == n){
break;
}
if(extra_red_remaining > 0){
System.out.print("R");
extra_red_remaining--;
i++;
}
if(i == n){
break;
}
System.out.print("B");
i++;
}
System.out.println();
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 72b1c2569122c16f2ea7d32051c347a4 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.io.*;
import java.util.*;
public class BroCoders {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int min = r / (b + 1);//minimum r in each space among b+1 spaces
int rest = r % (b + 1);//extra r
StringBuilder str = new StringBuilder();
StringBuilder ans = new StringBuilder();
for (int i = 0; i < min; i++) {
str.append("R");
}
for (int i = 0; i < b + 1; i++) {
ans.append(str);
if (rest != 0) {
ans.append("R");
rest--;
}
if (i != b) {
ans.append("B");
}
}
System.out.println(ans.toString());
t--;
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c3a8a13b2ab4083dc4cc527945b73ec4 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.*;
public class text1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int cs = sc.nextInt();
for(int dz = 0;dz < cs;dz ++) {
int length = sc.nextInt(), r = sc.nextInt(), b = sc.nextInt() + 1;
while(r != 0 && b != 0){
for(int dz1 = 0;dz1 < r / b;dz1 ++)
System.out.print('R');
r -= r / b;
b --;
if(b > 0)
System.out.print('B');
}
System.out.println();
}
return ;
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 9362c0e96c3dcd00f41f47104071466b | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Scanner;
public class Bishop {
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int q = sc.nextInt();
for(int i=0; i<q; i++){
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
if(b==1){
double x = Math.floor(((double)r)/2);
double y = Math.ceil(((double)r)/2);
long x1 = (long)x;
long y1 = (long)y;
for(int j=0; j<y1; j++){
System.out.print("R");
}
System.out.print("B");
for(int j=0; j<x1; j++){
System.out.print("R");
}
System.out.println();
}
else{
double f = Math.floor(((double)(r))/(b+1));
double c = Math.ceil(((double)(r))/(b+1));
if(f==c){
for(int k=0; k<b; k++){
for(int j=0; j<f; j++){
System.out.print("R");
}
System.out.print("B");
}
for(int j=0; j<f; j++){
System.out.print("R");
}
System.out.println();
}
else{
double diff = c-f;
double u = r-((b+1)*f);
double v = ((b+1)*c)-r;
double count1 = u/diff;
double count2 = v/diff;
String s1 = "";
String s2 = "";
for(int k=0; k<c; k++){
s1 = s1+"R";
}
for(int k=0; k<f; k++){
s2 = s2+"R";
}
for(int k=0; k<count1; k++){
System.out.print(s1);
System.out.print("B");
}
for(int k=0; k<count2; k++){
if(k==count2-1){
System.out.print(s2);
}
else{
System.out.print(s2);
System.out.print("B");
}
}
System.out.println();
}
}
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 92266b572cdda4c399984af4046c32ee | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class test {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int b = sc.nextInt();
int r = sc.nextInt();
int x = b / (r + 1);
int y = b % (r + 1);
for (int i = 0; i < x; i++)
System.out.print('R');
for (int i = 0; i < r; i++) {
if (y > 0) {
char ch = 'R';
System.out.print(ch);
y--;
}
char ch = 'B';
System.out.print(ch);
for (int j = 0; j < x; j++)
System.out.print('R');
}
System.out.println();
}
}
} | Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | 6fa36b3eec426460c262367e85dd1709 | train_110.jsonl | 1650206100 | Team Red and Team Blue competed in a competitive FPS. Their match was streamed around the world. They played a series of $$$n$$$ matches.In the end, it turned out Team Red won $$$r$$$ times and Team Blue won $$$b$$$ times. Team Blue was less skilled than Team Red, so $$$b$$$ was strictly less than $$$r$$$.You missed the stream since you overslept, but you think that the match must have been neck and neck since so many people watched it. So you imagine a string of length $$$n$$$ where the $$$i$$$-th character denotes who won the $$$i$$$-th match — it is R if Team Red won or B if Team Blue won. You imagine the string was such that the maximum number of times a team won in a row was as small as possible. For example, in the series of matches RBBRRRB, Team Red won $$$3$$$ times in a row, which is the maximum.You must find a string satisfying the above conditions. If there are multiple answers, print any. | 256 megabytes |
import java.util.Arrays;
import java.util.Scanner;
public class test {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int b = sc.nextInt();
int r = sc.nextInt();
int x = b / (r + 1);
int y = b % (r + 1);
for (int i = 0; i < x; i++)
System.out.print('R');
for (int i = 0; i < r; i++) {
if (y > 0) {
char ch = 'R';
System.out.print(ch);
y--;
}
char ch = 'B';
System.out.print(ch);
for (int j = 0; j < x; j++)
System.out.print('R');
}
System.out.println();
}
}
}
| Java | ["3\n7 4 3\n6 5 1\n19 13 6", "6\n3 2 1\n10 6 4\n11 6 5\n10 9 1\n10 8 2\n11 9 2"] | 1 second | ["RBRBRBR\nRRRBRR\nRRBRRBRRBRRBRRBRRBR", "RBR\nRRBRBRBRBR\nRBRBRBRBRBR\nRRRRRBRRRR\nRRRBRRRBRR\nRRRBRRRBRRR"] | NoteThe first test case of the first example gives the optimal answer for the example in the statement. The maximum number of times a team wins in a row in RBRBRBR is $$$1$$$. We cannot minimize it any further.The answer for the second test case of the second example is RRBRBRBRBR. The maximum number of times a team wins in a row is $$$2$$$, given by RR at the beginning. We cannot minimize the answer any further. | Java 17 | standard input | [
"constructive algorithms",
"greedy",
"implementation",
"math"
] | 7cec27879b443ada552a6474c4e45c30 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has a single line containing three integers $$$n$$$, $$$r$$$, and $$$b$$$ ($$$3 \leq n \leq 100$$$; $$$1 \leq b < r \leq n$$$, $$$r+b=n$$$). | 1,000 | For each test case, output a single line containing a string satisfying the given conditions. If there are multiple answers, print any. | standard output | |
PASSED | c4a5eed8f62a1222e561c4f2691469dc | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | import java.util.*;
import java.io.*;
public class ReverseSortSum {
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();
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = in.nextInt();
}
int M = 1;
while (M < N) {
M <<= 1;
}
long[] tree = new long[M * 2];
for (int i = 0; i < N; i++) {
tree[i + M] = arr[i];
}
long ones = 0;
for (int i : arr) {
ones += i;
}
ones /= N;
int[] res = new int[N];
for (int i = N - 1; i >= 0; i--) {
int l = i + 1 - (int)ones + M, r = i + M;
while (l <= r) {
if (l % 2 == 1) {
tree[l]--;
l >>= 1;
l++;
} else {
l >>= 1;
}
if (r % 2 == 0) {
tree[r]--;
r >>= 1;
r--;
} else {
r >>= 1;
}
}
int current = i + M;
int sum = 0;
while (current > 0) {
sum += tree[current];
current >>= 1;
}
if (ones > 0 && sum == i) {
res[i] = 1;
ones--;
int l2 = i + M, r2 = i + M;
while (l2 <= r2) {
if (l2 % 2 == 1) {
tree[l2] -= i;
l2 >>= 1;
l2++;
} else {
l2 >>= 1;
}
if (r2 % 2 == 0) {
tree[r2] -= i;
r2 >>= 1;
r2--;
} else {
r2 >>= 1;
}
}
}
}
for (int i = 0; i < N; i++) {
out.print(res[i] + (i == N - 1 ? "\n" : " "));
}
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 98fbac0aae31a8cff9b947741b2e9e8c | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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();
int[] arr = sc.na(n);
long sum = 0;
for(int i = 0; i < n; i++) sum += arr[i];
long onnes = sum/n;
int ones = (int)onnes;
int[] res = new int[n];
int[] suffZeroes = new int[n];
suffZeroes[n-1] = (arr[n-1]==n?0:1);
if(suffZeroes[n-1] == 0) {
res[n-1] = 1;
ones--;
}
for(int i = n-2; i >= 0 && ones > 0; i--) {
int currSum = 0;
if(suffZeroes[i+1] < ones) {
currSum = n-i;
} else {
int bs = suffZeroes[i+1]-ones;
int id = binSearch(suffZeroes, i+1, bs);
id--;
currSum = id-i;
}
if(currSum < arr[i]) {
res[i] = 1;
suffZeroes[i] = suffZeroes[i+1];
ones--;
} else {
suffZeroes[i] = suffZeroes[i+1]+1;
}
}
if(ones > 0) res[0] = 1;
for(int i: res) w.pr(i+" ");
w.pl();
}
static int binSearch(int[] arr, int ll, int elem) {
int l = ll; // l > elem
int r = arr.length; // r <= elem
while(l < r-1) {
int mid = (l+r)/2;
if(arr[mid] > elem) l = mid;
else r = mid;
}
return r;
}
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | df3fc6a00ffacddb4a7d085715bba5e2 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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();
int arr[] = new int[n];
for(int i=0; i<n; i++){
arr[i] = sc.nextInt();
}
long sum = 0;
for(int i=0; i<n; i++){
sum += arr[i];
}
int ones = (int)(sum/n);
FenwickTree tree = new FenwickTree(n);
int ans[] = new int[n];
tree.update(n-1, 1);
tree.update(n-1-ones, -1);
// for(int i=n-1; i>n-1-ones; i--){
// arr[i]--;
// }
// for(int j=0; j<n; j++){
// System.out.print(tree.getSumOfRange(j, n-1));
// }
// System.out.println();
for(int i=n-1; i>=0 && ones > 0; i--){
if(arr[i] == i + tree.getSumOfRange(i, n-1)){
ans[i] = 1;
ones--;
}
// for(int j=0; j<n; j++){
// System.out.print(tree.getSumOfRange(j, n-1));
// }
// System.out.println();
// for(int j=i-1; j>=i-ones && j>=0; j--){
// arr[j]--;
// }
tree.update(i-1, 1);
tree.update(i-ones-1, -1);
}
for(int i=0; i<n; i++){
System.out.print(ans[i] + " ");
}
System.out.println();
}
}
}
class FenwickTree {
private int arr[];
private int size;
public FenwickTree(int n){
this.arr = new int[n+1];
this.size = n;
}
public int getSumOfRange(int i, int j){
return getSum(j+1) - getSum(i);
}
public int getSum(int i){
int res = 0;
while(i > 0){
res += arr[i];
i = i - (i & (-i));
}
return res;
}
public void update(int idx, int val){
idx++;
if(idx == 0) return;
while(idx <= size){
arr[idx] += val;
idx += idx & (-idx);
}
}
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 4afa1a4917340d8ca6d30c0626865c66 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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();
int arr[] = new int[n];
for(int i=0; i<n; i++){
arr[i] = sc.nextInt();
}
long sum = 0;
for(int i=0; i<n; i++){
sum += arr[i];
}
int ones = (int)(sum/n);
int zeros = n-ones;
Queue<Integer> q = new LinkedList<>();
int count = 0;
int idx = 0;
int ans[] = new int[n];
Arrays.fill(ans, 1);
while(count < zeros && idx < n){
if(arr[idx] == 0){
ans[idx] = 0;
count++;
idx++;
}
else{
if(!q.isEmpty() && q.peek() == idx){
q.add(arr[idx] + idx);
int element = q.poll();
ans[element] = 0;
}
else{
q.add(arr[idx]);
}
count++;
idx++;
}
}
// System.out.println(q);
while(!q.isEmpty()){
int element = q.poll();
ans[element] = 0;
}
for(int i=0; i<n; i++){
System.out.print(ans[i] + " ");
}
System.out.println();
}
}
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 3156c129d60b000ab53e5f33f30847b0 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | import java.io.*;
import java.util.*;
public class problemD {
public static void main(String[] args)throws IOException {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int t = Integer.parseInt(br.readLine());
for(int i=0; i<t; i++) {
int n = Integer.parseInt(br.readLine());
int[] arr = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for(int j=0; j<n; j++) arr[j] = Integer.parseInt(st.nextToken());
long k = 0;
for(int j=0; j<n; j++) k+= arr[j];
int add = 0;
k/=n;
int[] ans = new int[n];
int[] prefix = new int[n];
int curr = 0;
for(int j=n-1; j>-1; j--) {
curr+=prefix[j];
int val = arr[j] + add+curr;
if(val==j+1)
ans[j] = 1;
else if(val==1)
ans[j] = 0;
if(val==0)
break;
if(k>1) {
prefix[j-1]-=1;
if(j-k>=0)
prefix[j-(int)k]+=1;
}
if(val==j+1)k--;
}
for(int j=0; j<n; j++)out.write(ans[j]+" ");
out.write("\n");
}
out.flush();
out.close();
}
}
| Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 751051678aff4a8bcb16d6ec42b8bfbf | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes |
import java.io.*;
import java.util.*;
public final class Main {
//int 2e9 - long 9e18
static PrintWriter out = new PrintWriter(System.out);
static FastReader in = new FastReader();
static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(0, 1), new Pair(1, 0), new Pair(0, -1)};
static int mod = (int) (1e9 + 7);
static int mod2 = 998244353;
public static void main(String[] args) {
int tt = i();
while (tt-- > 0) {
solve();
}
out.flush();
}
public static void solve() {
int n = i();
int[] a = input(n);
long sum = sum(a);
int cnt = (int) (sum / n);
int[] ans = new int[n];
PriorityQueue<Integer> pq = new PriorityQueue<>(Comparator.comparingInt(i -> -i));
for (int i = n - 1; i >= 0; i--) {
while (pq.size() > 0 && i < pq.peek()) {
pq.poll();
}
if (a[i] - pq.size() == i + 1) {
ans[i] = 1;
pq.offer(i - cnt + 1);
cnt--;
} else {
pq.offer(i - cnt + 1);
}
}
print(ans);
}
// (10,5) = 2 ,(11,5) = 3
static long upperDiv(long a, long b) {
return (a / b) + ((a % b == 0) ? 0 : 1);
}
static long sum(int[] a) {
long sum = 0;
for (int x : a) {
sum += x;
}
return sum;
}
static int[] preint(int[] a) {
int[] pre = new int[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] pre(int[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static long[] post(int[] a) {
long[] post = new long[a.length + 1];
post[0] = 0;
for (int i = 0; i < a.length; i++) {
post[i + 1] = post[i] + a[a.length - 1 - i];
}
return post;
}
static long[] pre(long[] a) {
long[] pre = new long[a.length + 1];
pre[0] = 0;
for (int i = 0; i < a.length; i++) {
pre[i + 1] = pre[i] + a[i];
}
return pre;
}
static void print(char A[]) {
for (char c : A) {
out.print(c);
}
out.println();
}
static void print(boolean A[]) {
for (boolean c : A) {
out.print(c + " ");
}
out.println();
}
static void print(int A[]) {
for (int c : A) {
out.print(c + " ");
}
out.println();
}
static void print(long A[]) {
for (long i : A) {
out.print(i + " ");
}
out.println();
}
static void print(List<Integer> A) {
for (int a : A) {
out.print(a + " ");
}
}
static int i() {
return in.nextInt();
}
static long l() {
return in.nextLong();
}
static double d() {
return in.nextDouble();
}
static String s() {
return in.nextLine();
}
static String c() {
return in.next();
}
static int[][] inputWithIdx(int N) {
int A[][] = new int[N][2];
for (int i = 0; i < N; i++) {
A[i] = new int[]{i, in.nextInt()};
}
return A;
}
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 int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long GCD(long a, long b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
static long LCM(int a, int b) {
return (long) a / GCD(a, b) * b;
}
static long LCM(long a, long b) {
return a / GCD(a, b) * b;
}
// find highest i which satisfy a[i]<=x
static int lowerbound(int[] a, int x) {
int l = 0;
int r = a.length - 1;
while (l < r) {
int m = (l + r + 1) / 2;
if (a[m] <= x) {
l = m;
} else {
r = m - 1;
}
}
return l;
}
static void shuffle(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
}
static void shuffleAndSort(int[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static void shuffleAndSort(int[][] arr, Comparator<? super int[]> comparator) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
int[] temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr, comparator);
}
static void shuffleAndSort(long[] arr) {
for (int i = 0; i < arr.length; i++) {
int rand = (int) (Math.random() * arr.length);
long temp = arr[rand];
arr[rand] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static boolean isPerfectSquare(double number) {
double sqrt = Math.sqrt(number);
return ((sqrt - Math.floor(sqrt)) == 0);
}
static void swap(int A[], int a, int b) {
int t = A[a];
A[a] = A[b];
A[b] = t;
}
static void swap(char A[], int a, int b) {
char t = A[a];
A[a] = A[b];
A[b] = t;
}
static long pow(long a, long b, int mod) {
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 pow(long a, long b) {
long pow = 1;
long x = a;
while (b != 0) {
if ((b & 1) != 0) {
pow *= x;
}
x = x * x;
b /= 2;
}
return pow;
}
static long modInverse(long x, int mod) {
return pow(x, mod - 2, mod);
}
static boolean isPrime(long N) {
if (N <= 1) {
return false;
}
if (N <= 3) {
return true;
}
if (N % 2 == 0 || N % 3 == 0) {
return false;
}
for (int i = 5; i * i <= N; i = i + 6) {
if (N % i == 0 || N % (i + 2) == 0) {
return false;
}
}
return true;
}
public static String reverse(String str) {
if (str == null) {
return null;
}
return new StringBuilder(str).reverse().toString();
}
public static void reverse(int[] arr) {
for (int i = 0; i < arr.length / 2; i++) {
int tmp = arr[i];
arr[arr.length - 1 - i] = tmp;
arr[i] = arr[arr.length - 1 - i];
}
}
public static String repeat(char ch, int repeat) {
if (repeat <= 0) {
return "";
}
final char[] buf = new char[repeat];
for (int i = repeat - 1; i >= 0; i--) {
buf[i] = ch;
}
return new String(buf);
}
public static int[] manacher(String s) {
char[] chars = s.toCharArray();
int n = s.length();
int[] d1 = new int[n];
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : Math.min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && chars[i - k] == chars[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
return d1;
}
public static int[] kmp(String s) {
int n = s.length();
int[] res = new int[n];
for (int i = 1; i < n; ++i) {
int j = res[i - 1];
while (j > 0 && s.charAt(i) != s.charAt(j)) {
j = res[j - 1];
}
if (s.charAt(i) == s.charAt(j)) {
++j;
}
res[i] = j;
}
return res;
}
}
class Pair {
int i;
int j;
Pair(int i, int j) {
this.i = i;
this.j = j;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
Pair pair = (Pair) o;
return i == pair.i && j == pair.j;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
class ThreePair {
int i;
int j;
int k;
ThreePair(int i, int j, int k) {
this.i = i;
this.j = j;
this.k = k;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
ThreePair pair = (ThreePair) o;
return i == pair.i && j == pair.j && k == pair.k;
}
@Override
public int hashCode() {
return Objects.hash(i, j);
}
}
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 Node {
int val;
public Node(int val) {
this.val = val;
}
}
class ST {
int n;
Node[] st;
ST(int n) {
this.n = n;
st = new Node[4 * Integer.highestOneBit(n)];
}
void build(Node[] nodes) {
build(0, 0, n - 1, nodes);
}
private void build(int id, int l, int r, Node[] nodes) {
if (l == r) {
st[id] = nodes[l];
return;
}
int mid = (l + r) >> 1;
build((id << 1) + 1, l, mid, nodes);
build((id << 1) + 2, mid + 1, r, nodes);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
void update(int i, Node node) {
update(0, 0, n - 1, i, node);
}
private void update(int id, int l, int r, int i, Node node) {
if (i < l || r < i) {
return;
}
if (l == r) {
st[id] = node;
return;
}
int mid = (l + r) >> 1;
update((id << 1) + 1, l, mid, i, node);
update((id << 1) + 2, mid + 1, r, i, node);
st[id] = comb(st[(id << 1) + 1], st[(id << 1) + 2]);
}
Node get(int x, int y) {
return get(0, 0, n - 1, x, y);
}
private Node get(int id, int l, int r, int x, int y) {
if (x > r || y < l) {
return new Node(0);
}
if (x <= l && r <= y) {
return st[id];
}
int mid = (l + r) >> 1;
return comb(get((id << 1) + 1, l, mid, x, y), get((id << 1) + 2, mid + 1, r, x, y));
}
Node comb(Node a, Node b) {
if (a == null) {
return b;
}
if (b == null) {
return a;
}
return new Node(GCD(a.val, b.val));
}
static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 1f4d72f07333db8034ae728f650edde8 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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[]=new int[n];
long sum=0;
for(int i=0;i<n;i++)
{
a[i]=input.nextInt();
sum+=a[i];
}
int x=(int)(sum/n);
int arr[]=new int[n];
int end[]=new int[n];
for(int i=n-1;i>=n-x;i--)
{
if(a[i]==n)
{
arr[i]=1;
}
end[i]=a[i];
}
int j=n-1;
int i=n-x-1;
while(i>=0)
{
if(a[i]==0)
{
break;
}
else
{
if(end[j]==n) j--;
else
{
int e=end[j]+a[i];
if(e==n) arr[i]=1;
end[i]=e;
i--;
j--;
}
}
}
for(i=0;i<n;i++)
{
out.print(arr[i]+" ");
}
out.println();
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 453ce3710d973a227e0fb4db35336796 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class D_Reverse_Sort_Sum {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int n, testCases;
static long a[];
static void solve(int t) {
long sum = detect_sum(0, a, 0);
long avg = sum / n;
int current = 0;
int rem[] = new int[n];
int b[] = new int[n];
for(int i = n - 1; i >= 0; --i) {
current -= rem[i];
if(avg >= 1) {
++current;
if(i - avg >= 0) {
rem[i - (int)avg]++;
}
}
a[i] -= current;
if(a[i] == i && avg > 0) {
b[i] = 1;
--avg;
}
}
for(int i : b) {
ans.append(i).append(" ");
}
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();
a = new long[n];
for (int i = 0; i < n; ++i) {
a[i] = in.nextLong();
}
solve(t + 1);
}
out.print(ans.toString());
out.flush();
in.close();
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | ca513da3d67a278642b0267b8df62139 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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();
int c[]=new int [n];
int a[]=new int [n];
long sum = 0;
for(int i = 0;i<n;i++) {
c[i]=in.nextInt();
sum+=c[i];
}
int num=(int)(sum/n);
int[] cf=new int [n];
long now = 0;
int cota=0;
for(int i = n-1;i>=1;i--) {
cf[i] += 1;
if(i - num >=0)
cf[i - num] -= 1;
now += cf[i];
if(c[i] - now == 0) {
a[i] = 0;
}else {
a[i] = 1;
cota++;
num--;
}
}
if(cota<sum/n) a[0]=1;
StringBuilder ans=new StringBuilder();
for(int i=0;i<n;i++) {
ans.append(a[i]);
ans.append(' ');
}
out.println(ans.toString());
}
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | eca25167bd3086a348a3650322dbc5f7 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class D_Reverse_Sort_Sum {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int n, testCases;
static long a[];
static void solve(int t) {
long sum = detect_sum(0, a, 0);
long reminder[] = new long[n];
long ans1[] = new long[n];
long avg = sum / n, current = 0;
for (int i = n - 1; i >= 0; --i) {
current -= reminder[i];
if (avg >= 1) {
++current;
if (i - avg >= 0) {
++reminder[i - (int) avg];
}
}
a[i] -= current;
if (a[i] == i && avg >= 1) {
ans1[i] = 1;
--avg;
}
}
for (long i : ans1) {
ans.append(i).append(" ");
}
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();
a = new long[n];
for (int i = 0; i < n; ++i) {
a[i] = in.nextLong();
}
solve(t + 1);
}
out.print(ans.toString());
out.flush();
in.close();
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 4a5bf35849f10737b94fcd579c1ae278 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | /* || श्री राम समर्थ ||
|| जय जय रघुवीर समर्थ ||
*/
import java.io.*;
import java.util.*;
import static java.util.Arrays.sort;
public class CodeforcesTemp {
static Reader scan = new Reader();
static FastPrinter out = new FastPrinter();
static int stsize;
public static void main(String[] args) throws IOException {
int tt = scan.nextInt();
// int tt = 1;
StringBuilder sb=new StringBuilder();
outer:
for (int tc = 1; tc <= tt; tc++) {
int n= scan.nextInt();
int[] arr= scan.nextIntArray(n);
stsize=0;
SegmentTree st=new SegmentTree(n);
st.build(0,0,stsize-1);
long sum=0;
for (int i = 0; i < n; i++) sum+=arr[i];
int k= (int) (sum/(1L*n));
int[] ans=new int[n];
if(arr[n-1]==n)ans[n-1]=1;
else if(arr[n-1]==0){
for (int i = 0; i < n; i++) {
sb.append("0"+" ");
}
sb.append("\n");continue outer;
}else{
ans[n-1]=0;
}
if(arr[0]==0)ans[0]=0;
else ans[0]=1;
if(n==1){
sb.append(ans[0]);sb.append("\n");
continue outer;
}
for (int i = n-1; i >=1 ; i--) {
if(k==0){
ans[i]=0;
continue ;
}
if(i!=n-1){
long res=st.query(0,0,stsize-1,i,i);
if(res==arr[i]){
ans[i]=0;
}else ans[i]=1;
}
if(ans[i]==1){
k--;
if(k>0){
// 0 1 2 3 4
st.update(0,0,stsize-1,(i)-k,i-1,1);
}
}else {
if(k>1){
// 0 1 2 3 4
// 4-1 3
// 4-2=2+1
st.update(0,0,stsize-1,((i)-(k))+1,i-1,1);
}
}
}
for (int i = 0; i < n; i++) {
sb.append(ans[i]+" ");
}
sb.append("\n");
}
out.println(sb.toString());
out.flush();
out.close();
}
static class SegmentTree{
static long[] seg;
static long[] lazy;
static int size;
private SegmentTree(int n) {
size=1;
while (size<n)size*=2;
seg=new long[2*size];
lazy=new long[2*size];
stsize=size;
}
private void build(int node, int st, int en) {
if (st == en) {
// left node ,string the single array element
seg[node] = 1;
lazy[node] = 0;
return;
}
int mid = (st + en) / 2;
// recursively call for left child
build((2 * node) + 1, st, mid);
// recursively call for the right child
build((2 * node) + 2, mid + 1, en);
// Updating the parent with the values of the left and right child.
seg[node] = seg[(2 * node) + 1] + seg[ (2 * node) + 2];
}
private void update(int node, int st, int en, int l, int r, int val) {
if (lazy[node] != 0) // if node is lazy then update it
{
seg[node] += (en - st + 1) * lazy[node];
if (st != en) // if its children exist then mark them lazy
{
lazy[ (2 * node) + 1 ] += lazy[node];
lazy[ (2 * node)+ 2] += lazy[node];
}
lazy[node] = 0; // No longer lazy
}
if ((en < l) || (st > r)) // case 1
{
return;
}
if (st >= l && en <= r) // case 2
{
seg[node] += (en - st + 1) * val;
if (st != en) {
lazy[ (2 * node) + 1] += val; // mark its children lazy
lazy[ (2 * node) + 2] += val;
}
return;
}
// case 3
int mid = (st + en) / 2;
// recursively call for updating left child
update( (2 * node) + 1, st, mid, l, r, val);
// recursively call for updating right child
update( (2 * node) + 2, mid + 1, en, l, r, val);
// Updating the parent with the values of the left and right child.
seg[node] = (seg[ (2 * node) + 1] + seg[ (2 * node) + 2]);
}
private long query(int node, int st, int en, int l, int r) {
/*If the node is lazy, update it*/
if (lazy[node] != 0) {
seg[node] += (en - st + 1) * lazy[node];
if (st != en) //Check if the child exist
{
// mark both the child lazy
lazy[ (2 * node) + 1] += lazy[node];
lazy[ (2 * node) + 2] += lazy[node];
}
// no longer lazy
lazy[node] = 0;
}
// case 1
if (en < l || st > r) {
return 0;
}
// case 2
if ((l <= st) && (en <= r)) {
return seg[node];
}
int mid = (st + en) / 2;
//query left child
long q1 = query( (2 * node) + 1, st, mid, l, r);
// query right child
long q2 = query( (2 * node) + 2, mid + 1, en, l, r);
return (q1 + q2);
}
}
static class Reader {
private final InputStream in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private static final long LONG_MAX_TENTHS = 922337203685477580L;
private static final int LONG_MAX_LAST_DIGIT = 7;
private static final int LONG_MIN_LAST_DIGIT = 8;
public Reader(InputStream in) {
this.in = in;
}
public Reader() {
this(System.in);
}
private boolean hasNextByte() {
if (ptr < buflen) {
return true;
} else {
ptr = 0;
try {
buflen = in.read(buffer);
} catch (IOException e) {
e.printStackTrace();
}
if (buflen <= 0) {
return false;
}
}
return true;
}
private int readByte() {
if (hasNextByte()) return buffer[ptr++];
else return -1;
}
private static boolean isPrintableChar(int c) {
return 33 <= c && c <= 126;
}
public boolean hasNext() {
while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
return hasNextByte();
}
public String next() {
if (!hasNext())
throw new NoSuchElementException();
StringBuilder sb = new StringBuilder();
int b = readByte();
while (isPrintableChar(b)) {
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
public long nextLong() {
if (!hasNext())
throw new NoSuchElementException();
long n = 0;
boolean minus = false;
int b = readByte();
if (b == '-') {
minus = true;
b = readByte();
}
if (b < '0' || '9' < b) {
throw new NumberFormatException();
}
while (true) {
if ('0' <= b && b <= '9') {
int digit = b - '0';
if (n >= LONG_MAX_TENTHS) {
if (n == LONG_MAX_TENTHS) {
if (minus) {
if (digit <= LONG_MIN_LAST_DIGIT) {
n = -n * 10 - digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("not number"));
}
}
} else {
if (digit <= LONG_MAX_LAST_DIGIT) {
n = n * 10 + digit;
b = readByte();
if (!isPrintableChar(b)) {
return n;
} else if (b < '0' || '9' < b) {
throw new NumberFormatException(
String.format("not number"));
}
}
}
}
throw new ArithmeticException(
String.format(" overflows long."));
}
n = n * 10 + digit;
} else if (b == -1 || !isPrintableChar(b)) {
return minus ? -n : n;
} else {
throw new NumberFormatException();
}
b = readByte();
}
}
public int nextInt() {
long nl = nextLong();
if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
throw new NumberFormatException();
return (int) nl;
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long[] nextLongArray(long length) {
long[] array = new long[(int) length];
for (int i = 0; i < length; i++)
array[i] = this.nextLong();
return array;
}
public long[] nextLongArray(int length, java.util.function.LongUnaryOperator map) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsLong(this.nextLong());
return array;
}
public int[] nextIntArray(int length) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = this.nextInt();
return array;
}
public Integer[] nextIntegerArray(int length) {
Integer[] array = new Integer[length];
for (int i = 0; i < length; i++)
array[i] = this.nextInt();
return array;
}
public int[][] nextIntArrayMulti(int length, int width) {
int[][] arrays = new int[width][length];
for (int i = 0; i < length; i++) {
for (int j = 0; j < width; j++)
arrays[j][i] = this.nextInt();
}
return arrays;
}
public int[] nextIntArray(int length, java.util.function.IntUnaryOperator map) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsInt(this.nextInt());
return array;
}
public double[] nextDoubleArray(int length) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = this.nextDouble();
return array;
}
public double[] nextDoubleArray(int length, java.util.function.DoubleUnaryOperator map) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsDouble(this.nextDouble());
return array;
}
public long[][] nextLongMatrix(int height, int width) {
long[][] mat = new long[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextLong();
}
return mat;
}
public int[][] nextIntMatrix(int height, int width) {
int[][] mat = new int[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextInt();
}
return mat;
}
public double[][] nextDoubleMatrix(int height, int width) {
double[][] mat = new double[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextDouble();
}
return mat;
}
public char[][] nextCharMatrix(int height, int width) {
char[][] mat = new char[height][width];
for (int h = 0; h < height; h++) {
String s = this.next();
for (int w = 0; w < width; w++) {
mat[h][w] = s.charAt(w);
}
}
return mat;
}
}
static class FastPrinter extends PrintWriter {
public FastPrinter(PrintStream stream) {
super(stream);
}
public FastPrinter() {
super(System.out);
}
private static String dtos(double x, int n) {
StringBuilder sb = new StringBuilder();
if (x < 0) {
sb.append('-');
x = -x;
}
x += Math.pow(10, -n) / 2;
sb.append((long) x);
sb.append(".");
x -= (long) x;
for (int i = 0; i < n; i++) {
x *= 10;
sb.append((int) x);
x -= (int) x;
}
return sb.toString();
}
@Override
public void print(float f) {
super.print(dtos(f, 20));
}
@Override
public void println(float f) {
super.println(dtos(f, 20));
}
@Override
public void print(double d) {
super.print(dtos(d, 20));
}
@Override
public void println(double d) {
super.println(dtos(d, 20));
}
public void printArray(int[] array, String separator) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(array[i]);
super.print(separator);
}
super.println(array[n - 1]);
}
public void printArray(int[] array) {
this.printArray(array, " ");
}
public void printArray(int[] array, String separator, java.util.function.IntUnaryOperator map) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(map.applyAsInt(array[i]));
super.print(separator);
}
super.println(map.applyAsInt(array[n - 1]));
}
public void printArray(int[] array, java.util.function.IntUnaryOperator map) {
this.printArray(array, " ", map);
}
public void printArray(long[] array, String separator) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(array[i]);
super.print(separator);
}
super.println(array[n - 1]);
}
public void printArray(long[] array) {
this.printArray(array, " ");
}
public void printArray(long[] array, String separator, java.util.function.LongUnaryOperator map) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
super.print(map.applyAsLong(array[i]));
super.print(separator);
}
super.println(map.applyAsLong(array[n - 1]));
}
public void printArray(long[] array, java.util.function.LongUnaryOperator map) {
this.printArray(array, " ", map);
}
public void printMatrix(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
this.printArray(arr[i]);
}
}
public void printCharMatrix(char[][] arr, int n, int m) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
this.print(arr[i][j] + " ");
}
this.println();
}
}
}
static Random __r = new Random();
static int randInt(int min, int max) {
return __r.nextInt(max - min + 1) + min;
}
static void reverse(int[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
int swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(long[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
long swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(double[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
double swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] a) {
for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
char swap = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = swap;
}
}
static void reverse(char[] arr, int i, int j) {
while (i < j) {
char temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
i++;
j--;
}
}
static void reverse(int[] arr, int i, int j) {
while (i < j) {
int temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
i++;
j--;
}
}
static void reverse(long[] arr, int i, int j) {
while (i < j) {
long temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
i++;
j--;
}
}
static void shuffle(int[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
int swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void shuffle(long[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
long swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
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;
}
private static int getlowest(int l) {
if (l >= 1000000000) return 1000000000;
if (l >= 100000000) return 100000000;
if (l >= 10000000) return 10000000;
if (l >= 1000000) return 1000000;
if (l >= 100000) return 100000;
if (l >= 10000) return 10000;
if (l >= 1000) return 1000;
if (l >= 100) return 100;
if (l >= 10) return 10;
return 1;
}
static void shuffle(double[] a) {
int n = a.length - 1;
for (int i = 0; i < n; ++i) {
int ind = randInt(i, n);
double swap = a[i];
a[i] = a[ind];
a[ind] = swap;
}
}
static void rsort(int[] a) {
shuffle(a);
sort(a);
}
static void rsort(long[] a) {
shuffle(a);
sort(a);
}
static void rsort(double[] a) {
shuffle(a);
sort(a);
}
static int[] copy(int[] a) {
int[] ans = new int[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static long[] copy(long[] a) {
long[] ans = new long[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static double[] copy(double[] a) {
double[] ans = new double[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
static char[] copy(char[] a) {
char[] ans = new char[a.length];
for (int i = 0; i < a.length; ++i) ans[i] = a[i];
return ans;
}
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | d4cef16718a3aab9cc70f09981e6cfd1 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | import java.io.*;
import java.util.*;
public class p5
{
BufferedReader br;
StringTokenizer st;
BufferedWriter bw;
int x1,y1;
int x2,y2;
int l1,l2;
int n;
public static void main(String[] args)throws Exception
{
new p5().run();
}
void run()throws IOException
{
br = new BufferedReader(new InputStreamReader(System.in));
bw=new BufferedWriter(new OutputStreamWriter(System.out));
solve();
}
void solve() throws IOException
{
int t=ni();
while(t-->0)
{
int n=ni();
int a[]=nai(n);
long sum=0;
for(int i=-1;++i<n;)
sum+=a[i];
sum/=n;
int c[]=new int[n];
int one=(int)sum;
int zero=n-one;
int z2=zero;
int z[]=new int[zero];
Arrays.fill(z, -1);
for(int i=n;--i>=z2;)
{
if(a[i]==n)
{
c[i]=1;
one--;
}
else
{
zero--;
z[zero]=i;
}
}
for(int i=z2;--i>=0;)
{
if(z[i]!=-1)
{
if(a[i]==z[i])
c[i]=1;
else
{
zero--;
z[zero]=i;
}
}
}
for(int i=-1;++i<n;)
System.out.print(c[i]+" ");
System.out.println();
}
}
/////////////////////////////////////// FOR INPUT ///////////////////////////////////////
int[] nai(int n) { int a[]=new int[n]; for(int i=-1;++i<n;)a[i]=ni(); return a;}
Integer[] naI(int n) { Integer a[]=new Integer[n]; for(int i=-1;++i<n;)a[i]=ni(); return a;}
long[] nal(int n) { long a[]=new long[n]; for(int i=-1;++i<n;)a[i]=nl(); return a;}
char[] nac() {char c[]=nextLine().toCharArray(); return c;}
char [][] nmc(int n) {char c[][]=new char[n][]; for(int i=-1;++i<n;)c[i]=nac(); return c;}
int[][] nmi(int r, int c) {int a[][]=new int[r][c]; for(int i=-1;++i<r;)a[i]=nai(c); return a;}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {st = new StringTokenizer(br.readLine());}
catch (IOException e) {e.printStackTrace();}
}
return st.nextToken();
}
int ni() { return Integer.parseInt(next()); }
byte nb() { return Byte.parseByte(next()); }
short ns() { return Short.parseShort(next()); }
long nl() { return Long.parseLong(next()); }
double nd() { return Double.parseDouble(next()); }
String nextLine()
{
String str = "";
try {str = br.readLine();}
catch (IOException e) {e.printStackTrace();}
return str;
}
}
| Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | d0eae49c17dd1eba57e46df015ac1ae2 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | import java.util.*;
import java.io.*;
// res.append("Case #"+(p+1)+": "+hh+" \n");
////***************************************************************************
/* 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 D_Reverse_Sort_Sum{
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 array[]= new long[n];
long sum=0;
for(int i=0;i<n;i++){
array[i]=s.nextLong();
sum+=array[i];
}
long rem=sum/n;
long nice[]= new long[n];
//just a modified prefix array sort of thing
long well=0;
long ans[]= new long[n];
for(int i=n-1;i>=0;i--){
well+=nice[i];
long yoyo=array[i]-well;
if(yoyo==0){
break;
}
if(rem>1){
nice[i-1]+=1;
if((i-rem)>=0){
nice[(int)(i-rem)]-=1;
}
}
long lim=i+1;
if(yoyo==lim){
ans[i]=1;
rem--;
}
}
for(int i=0;i<n;i++){
res.append(ans[i]+" ");
}
res.append(" \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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | c36536602d872126c5dbbc82860fc775 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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[] arr = new int[n];
for(int i = 0 ; i<n;i++) {
arr[i] = sc.nextInt();
}
Queue<Integer> q = new LinkedList<Integer>();
int[] ans = new int[n];
int l = n-1;
for(int i = 0 ; i<n;i++) {
// System.out.println(i + " " + q.peek());
if(q.isEmpty()){
if(arr[i] == 0) {
ans[i] = 0;
}else {
ans[i] = 1;
if(arr[i]<n) {
q.add(arr[i]);
}
}
}else {
if(i == q.peek()) {
ans[i] = 0;
if(i + arr[i] <n) {
// System.out.println(i + arr[i]);
q.add(i + arr[i]);
}
q.remove();
}else {
ans[i] = 1;
if(arr[i]<n) {
q.add(arr[i]);
}
}
}
}
for(int i = 0 ; i<n;i++) {
System.out.print(ans[i] + " ");
}
System.out.println();
}
}
//------------------------------------------------------------------------------------------------------------------------------------------------
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(int[] 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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | a39230b075ca3d8c9cd719db274b7e1e | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | //make sure to make new file!
import java.io.*;
import java.util.*;
//upsolve
public class D782b{
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++){
int n = Integer.parseInt(f.readLine());
StringTokenizer st = new StringTokenizer(f.readLine());
int[] array = new int[n];
for(int k = 0; k < n; k++){
array[k] = Integer.parseInt(st.nextToken());
}
int[] answer = new int[n];
Arrays.fill(answer,-1);
int r = 0;
int added = 0;
for(int k = 0; r < n && k < n; k++){
int i = array[k];
if(answer[k] == -1){
if(i == 0){
answer[k] = 0;
r++;
} else {
i -= k;
for(int j = 0; j < i; j++){
answer[r] = 1;
added++;
r++;
}
if(r < n){
answer[r] = 0;
r++;
}
}
} else {
if(answer[k] == 1) i -= k;
int top = i-added;
for(int j = 0; j < top; j++){
answer[r] = 1;
added++;
r++;
}
if(r < n){
answer[r] = 0;
r++;
}
}
}
StringJoiner sj = new StringJoiner(" ");
for(int k = 0; k < n; k++){
sj.add("" + answer[k]);
}
out.println(sj.toString());
}
out.close();
}
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 2f03e41f7d42d5413e93ed83f59fc00b | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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 StringBuilder str = new StringBuilder();
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();
int[] c = input(n);
int[] a = new int[n + 1];
Arrays.fill(a,1);
int know = -1;
for(int i = 0;i < n && know < n;i++){
if(know < i){
if(c[i] == 0){
a[i] = 0;
}else{
a[i] = 1;
}
know = i;
}
if(a[i] == 1){
int index = Math.min(n,c[i]);
a[index] = 0;
know = index;
}
if(a[i] == 0){
int index = Math.min(n,i + c[i]);
a[index] = 0;
know = index;
}
}
for(int i = 0;i < n;i++){
out.print(a[i] + " ");
}
out.println();
}
out.close();
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 95d156b43c59b80fb0eee846ec60d399 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class D {
public static void main(String[]args) throws IOException {
Scanner sc=new Scanner(System.in);
PrintWriter out=new PrintWriter(System.out);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
// int[]a=new int[n];
int[]ans=new int[n];
int[]a=new int[n];
long sum=0l;
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
sum+=a[i];
}
int k=(int) (sum/n);
int left=n-k;
int[]b=new int[n];
for(int i=left;i<n;i++)b[i]=n-1;
for(int i=n-1;i>=0&&i>=left;i--) {
int v=a[i]-(b[i]-i);
if(v==i+1) {
ans[i]=1;
}else if(v==1) {
left--;
b[left]=i-1;
}
}
for(int x:ans)out.print(x+" ");
out.println();
}
// for(int i=0;i<100;i++) {
// int n=(int)(Math.random()*15 +1);
// out.println(n);
// int[]a=new int[n];
// int[][]aa=new int[n][n];
// for(int j=0;j<n;j++) {
// a[j]=(int)Math.round(Math.random());
// out.print(a[j]+" ");
// aa[0][j]=a[j];
// }
// out.println();
// ArrayList<Integer>arr=new ArrayList<Integer>();
// arr.add(a[0]);
// for(int j=1;j<n;j++) {
// arr.add(a[j]);
// Collections.sort(arr);
// for(int k=0;k<arr.size();k++) {
// aa[j][k]=arr.get(k);
// }
// for(int k=arr.size();k<n;k++)aa[j][k]=a[k];
// }
// for(int j=0;j<n;j++) {
// int sum=0;
// for(int k=0;k<n;k++) {
// sum+=aa[k][j];
// }
// out.print(sum+" ");
// }
// out.println("\n\n");
// }
out.close();
}
static class Scanner
{
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s){ br = new BufferedReader(new InputStreamReader(s));}
public String next() throws IOException
{
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public boolean hasNext() {return st.hasMoreTokens();}
public int nextInt() throws IOException {return Integer.parseInt(next());}
public double nextDouble() throws IOException {return Double.parseDouble(next());}
public long nextLong() throws IOException {return Long.parseLong(next());}
public String nextLine() throws IOException {return br.readLine();}
public boolean ready() throws IOException {return br.ready(); }
}
}
| Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | 0418425fab587cb8607137f4d350d65c | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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++){
int n = sc.nextInt();
int arr[]= sc.readArray(n);
int ans []= new int [n];
long num=0;
for (int i=0; i<n; i++){
num+=arr[i];
}
num/=n;
Deque<Integer> q= new ArrayDeque<>();
for (int i=n-1; i>=0; i--){
while (!q.isEmpty() && q.peek()>i) q.removeFirst();
arr[i]-= q.size();
q.add(i+1-(int)num);
if (arr[i]==i+1){
ans[i]=1;
num--;
}
else ans[i]=0;
}
for (int i=0; i<n; i++){
System.out.print(ans[i]+" ");
}
System.out.println();
}
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | cef2e22edfe3a39a65136e955aad225c | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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) {
StringBuilder sb = new StringBuilder();
int n = sc.nextInt();
int[] arr = new int[n+1], sum = new int[n+1];
long total = 0;
for(int i=1; i<=n; i++)
total += arr[i] = sc.nextInt();
int cnt = (int)(total/n);
for(int i=n; i>0; i--) {
char c = arr[i]-sum[i] == i ? '1' : '0';
sum[i]++;
sum[i-cnt]--;
sum[i-1] += sum[i];
if(c == '1') cnt--;
sb.append(c + " ");
}
System.out.println(sb.reverse().toString().trim());
}
}
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | cd52803dc3dd50efe6fa8f43cf9663ac | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws Exception {
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
long[]a=sc.nextlongArray(n);
SegmentTree sg=new SegmentTree(a);
long[]ans=new long[n];
int ind=0;
int prev=0;
while(ind+prev<n){
long x=sg.query(ind,ind);
if(x==0){
ind++;
}else{
sg.update_range(ind,n-1-prev,-1);
sg.update_point(ind+prev,-(ind+prev));
ans[ind+prev]=1;
prev++;
}
}
for(long i:ans)pw.print(i+" ");
pw.println();
}
pw.close();
}
public static class SegmentTree { // 1-based DS, OOP
int N; // the number of elements in the array as a power of 2 (i.e. after padding)
long[] array;
long[] sTree;
long[] lazy;
SegmentTree(long[] a) {
int nn = a.length;
int NN = 1;
while (NN < nn)
NN <<= 1; // padding
long[] in = new long[NN + 1];
for (int i = 1; i <= nn; i++)
in[i] = a[i - 1];
array = in;
N = in.length - 1;
sTree = new long[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
lazy = new long[N << 1];
build(1, 1, N);
}
void build(int node, int b, int e) // O(n)
{
if(b == e)
sTree[node] = array[b];
else
{
int mid = b + e >> 1;
build(node<<1,b,mid);
build(node<<1|1,mid+1,e);
sTree[node] = sTree[node<<1]+sTree[node<<1|1];
}
}
void update_point(int index, int val) // O(log n)
{
index += N ;
sTree[index] += val;
while(index>1)
{
index >>= 1;
sTree[index] = sTree[index<<1] + sTree[index<<1|1];
}
}
void update_range(int i, int j, int val) // O(log n)
{
update_range(1,1,N,i+1,j+1,val);
}
void update_range(int node, int b, int e, int i, int j, int val)
{
if(i > e || j < b)
return;
if(b >= i && e <= j)
{
sTree[node] += (e-b+1)*val;
lazy[node] += val;
}
else
{
int mid = b + e >> 1;
propagate(node, b, mid, e);
update_range(node<<1,b,mid,i,j,val);
update_range(node<<1|1,mid+1,e,i,j,val);
sTree[node] = sTree[node<<1] + sTree[node<<1|1];
}
}
void propagate(int node, int b, int mid, int e)
{
lazy[node<<1] += lazy[node];
lazy[node<<1|1] += lazy[node];
sTree[node<<1] += (mid-b+1)*lazy[node];
sTree[node<<1|1] += (e-mid)*lazy[node];
lazy[node] = 0;
}
long query(int i, int j)
{
return query(1,1,N,i+1,j+1);
}
long query(int node, int b, int e, int i, int j) // O(log n)
{
if(i>e || j <b)
return 0;
if(b>= i && e <= j)
return sTree[node];
int mid = b + e >> 1;
propagate(node, b, mid, e);
long q1 = query(node<<1,b,mid,i,j);
long q2 = query(node<<1|1,mid+1,e,i,j);
return q1 + q2;
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
long x;
long y;
public pair(long x, long y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return Long.compare(this.y, other.y);
}
return Long.compare(this.x, other.x);
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
static long mod = 1000000007;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | c9b819c48be2281cd1cbbdd508329951 | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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 NewTimeD
{
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());
int[] arr = readArr(N, infile, st);
int[] res = new int[N];
long sum = 0;
for(int x: arr)
sum += x;
int K = (int)(sum/N);
FenwickTree bit = new FenwickTree(N+3);
int pointer = N-K;
for(int i=N-1; i > 0; i--)
{
bit.add(pointer, 1);
int val = arr[i]-bit.find(0, i);
if(val == 0)
pointer--;
else
{
res[i] = 1;
K--;
}
}
res[0] = K;
for(int x: res)
sb.append(x+" ");
sb.append("\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;
}
}
/*
Special hack format?
arr[N-1] = 1 if and only if crr[N-1] == N
the total # of bits in arr = sum(crr)/N
first non-zero element of crr is where first bit of arr is located
*/
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)
{
i++;
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
i++;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
} | Java | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output | |
PASSED | ead534cc45b733b51ac6942a284ce40b | train_110.jsonl | 1650206100 | Suppose you had an array $$$A$$$ of $$$n$$$ elements, each of which is $$$0$$$ or $$$1$$$.Let us define a function $$$f(k,A)$$$ which returns another array $$$B$$$, the result of sorting the first $$$k$$$ elements of $$$A$$$ in non-decreasing order. For example, $$$f(4,[0,1,1,0,0,1,0]) = [0,0,1,1,0,1,0]$$$. Note that the first $$$4$$$ elements were sorted.Now consider the arrays $$$B_1, B_2,\ldots, B_n$$$ generated by $$$f(1,A), f(2,A),\ldots,f(n,A)$$$. Let $$$C$$$ be the array obtained by taking the element-wise sum of $$$B_1, B_2,\ldots, B_n$$$.For example, let $$$A=[0,1,0,1]$$$. Then we have $$$B_1=[0,1,0,1]$$$, $$$B_2=[0,1,0,1]$$$, $$$B_3=[0,0,1,1]$$$, $$$B_4=[0,0,1,1]$$$. Then $$$C=B_1+B_2+B_3+B_4=[0,1,0,1]+[0,1,0,1]+[0,0,1,1]+[0,0,1,1]=[0,2,2,4]$$$.You are given $$$C$$$. Determine a binary array $$$A$$$ that would give $$$C$$$ when processed as above. It is guaranteed that an array $$$A$$$ exists for given $$$C$$$ in the input. | 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[]=inputLong(N);
long ones=0;
for(long a:A)ones+=a;
ones/=N; //total no of ones
int f[]=new int[N];
// Arrays.fill(f, -1);
int c=0;
for(int i=1; i<=N; i++)if(A[i-1]>=i)c++;
if(c==ones)
{
for(int i=1; i<=N; i++) {
if(A[i-1]>=i)f[i-1]=1;
else f[i-1]=0;
}
}
else
{
c=0;
int X[]=new int[N];
for(int i=N-1; i>=0; i--)
{
c-=X[i];
if(ones>0)
{
c++;
if(i-ones>=0)
{
X[i-(int)ones]++;
}
}
A[i]-=c;
if(A[i]==i && ones>0)
{
f[i]=1;
ones--;
}
}
}
for(int a:f)ans.append(a+" ");
ans.append("\n");
}
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 | ["5\n4\n2 4 2 4\n7\n0 3 4 2 3 2 7\n3\n0 0 0\n4\n0 0 0 4\n3\n1 2 3"] | 2 seconds | ["1 1 0 1 \n0 1 1 0 0 0 1 \n0 0 0 \n0 0 0 1 \n1 0 1"] | NoteHere's the explanation for the first test case. Given that $$$A=[1,1,0,1]$$$, we can construct each $$$B_i$$$: $$$B_1=[\color{blue}{1},1,0,1]$$$; $$$B_2=[\color{blue}{1},\color{blue}{1},0,1]$$$; $$$B_3=[\color{blue}{0},\color{blue}{1},\color{blue}{1},1]$$$; $$$B_4=[\color{blue}{0},\color{blue}{1},\color{blue}{1},\color{blue}{1}]$$$ And then, we can sum up each column above to get $$$C=[1+1+0+0,1+1+1+1,0+0+1+1,1+1+1+1]=[2,4,2,4]$$$. | Java 8 | standard input | [
"constructive algorithms",
"data structures",
"greedy",
"implementation",
"math",
"two pointers"
] | 9dc1bee4e53ced89d827826f2d83dabf | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Each test case has two lines. The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$). The second line contains $$$n$$$ integers $$$c_1, c_2, \ldots, c_n$$$ ($$$0 \leq c_i \leq n$$$). It is guaranteed that a valid array $$$A$$$ exists for the given $$$C$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,900 | For each test case, output a single line containing $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i$$$ is $$$0$$$ or $$$1$$$). If there are multiple answers, you may output any of them. | standard output |
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