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2.37k
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PASSED | 04151685f414fc1476aa42b17e71edd9 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main
{
public static void main(String args[])
{
FastScanner input = new FastScanner();
int tc = input.nextInt();
work:
while (tc-- > 0) {
int n = input.nextInt();
int k = input.nextInt();
int hold = k;
char a[] =input.next().toCharArray();
int ans[] =new int[n];
char res[] = new char[n];
if(k%2==0)
{
for (int i = 0; i <n&&k>0; i++) {
if(a[i]=='0')
{
ans[i]++;
k--;
}
}
ans[n-1]+=k;
// System.out.println(Arrays.toString(ans));
}
else
{
for (int i = 0; i <n&&k>0; i++) {
if(a[i]=='1')
{
ans[i]++;
k--;
}
}
ans[n-1]+=k;
// System.out.println(Arrays.toString(ans));
}
for (int i = 0; i <n; i++) {
if(a[i]=='1'&&(hold-ans[i])%2==0)
{
res[i] = '1';
}
else if(a[i]=='1'&&(hold-ans[i])%2==1)
{
res[i] = '0';
}
else if(a[i]=='0'&&(hold-ans[i])%2==1)
{
res[i]='1';
}
else if(a[i]=='0'&&(hold-ans[i])%2==0)
{
res[i]='0';
}
}
StringBuilder result =new StringBuilder();
result.append(res);
result.append("\n");
for (int an : ans) {
result.append(an+" ");
}
System.out.println(result);
}
}
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 38e2dd69e47909cab3fd907c3f256b8f | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int T = fs.nextInt();
for (int tt=0; tt<T; tt++) {
int n = fs.nextInt();
int k = fs.nextInt();
int K = k;
String s = fs.next();
int[] arr = new int[n];
if(k%2==0) {
int i=0;
while(k>0 && i<n) {
if(s.charAt(i)=='0') {
arr[i] = 1;
k--;
}
i++;
}
}
else {
int i=0;
while(k>0 && i<n) {
if(s.charAt(i)=='1') {
arr[i] = 1;
k--;
}
i++;
}
}
if(k>0)
arr[n-1]+=k;
StringBuilder sb = new StringBuilder();
for(int i =0; i<n; i++) {
if((K-arr[i])%2==0) {
sb.append(s.charAt(i));
}
else {
sb.append(s.charAt(i)=='1'?'0':'1');
}
}
out.println(sb);
for(int x: arr) {
out.print(x+" ");
}
out.println();
}
out.close();
}
static final Random random=new Random();
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static 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());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 864c337d3a445272dfe53d3def713dee | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.Scanner;
//B. Bit Flipping
public class B {
public static void main(String[] args) {
new B().solve();
}
public void solve() {
Scanner scanner = new Scanner(System.in);
int t = scanner.nextInt();
char[] res = null;
int[] cnt = null;
while (t-- > 0) {
int n = scanner.nextInt();
int k = scanner.nextInt();
scanner.nextLine();
res = scanner.nextLine().toCharArray();
cnt = new int[n];
int ones = 0;
int zeros = 0;
for (int i = 0; i < n; i++) {
if (res[i] == '1') ones++;
else zeros++;
}
int j = 0;
if (k % 2 == 0) {
for (int i = 0; i < n && j < k; i++) {
if (res[i] == '0') {
res[i] = '1';
j++;
cnt[i]++;
}
}
if (zeros % 2 != 0 && zeros < k) res[n - 1] = '0';
}
else {
for (int i = 0; i < n; i++) {
if (res[i] == '0') res[i] = '1';
else res[i] = '0';
}
for (int i = 0; i < n && j < k; i++) {
if (res[i] == '0') {
res[i] = '1';
j++;
cnt[i]++;
}
}
if (ones % 2 == 0 && ones < k) res[n - 1] = '0';
}
cnt[n - 1] += k - j;
System.out.println(String.valueOf(res, 0, n));
for (int i = 0; i < n; i++) {
System.out.print(cnt[i] + " ");
}
System.out.println();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f76b9e0b0abe1cbda0d256b974f29a39 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | /* package 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{
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-- > 0){
int n = scn.nextInt(), k = scn.nextInt();
String str = scn.next();
char[]ch = str.toCharArray();
int freq[] = new int[n], temp = k;
StringBuilder sb = new StringBuilder();
for(int i = 0; i < n && temp > 0; i++){
if(k % 2 == ch[i] - '0'){
freq[i] = 1;
temp--;
}
}
freq[n - 1] += temp;
for(int i = 0; i < n; i++){
if((k - freq[i]) % 2 == 1){
ch[i] = (char)('1' - (ch[i] - '0'));
}
}
System.out.println(new String(ch));
for(int ele: freq){
System.out.print(ele + " ");
}
System.out.println();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 46fffeabeabdae3108a72ef5c097d5f5 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.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 k = sc.nextInt();
String s = sc.next();
int c1 = 0;
int c2 = 0;
for(int i = 0 ; i<n;i++){
if(s.charAt(i) == 0) {
c2 ++;
}else {
c1++;
}
}
int[] arr = new int[n];
if(k%2 == 1) {
int x =0;
for(int i = 0 ; i<n;i++) {
if(x == k) {
break;
}
if(s.charAt(i) == '0') {
continue;
}
arr[i] = 1;
x++;
}
int r = k - x;
arr[n-1] += r;
}else {
int x =0;
for(int i = 0 ; i<n;i++) {
if(x == k) {
break;
}
if(s.charAt(i) == '1') {
continue;
}
arr[i] = 1;
x++;
}
int r = k - x;
arr[n-1] += r;
}
StringBuilder sb1 = new StringBuilder();
StringBuilder sb2 = new StringBuilder();
for(int i = 0; i<n; i++) {
if(k%2 == 1) {
if(s.charAt(i) == '1') {
if(arr[i]%2 == 1) {
sb1.append("1" );
}else {
sb1.append("0");
}
}else {
if(arr[i]%2 == 0) {
sb1.append("1" );
}else {
sb1.append("0");
}
}
}else {
if(s.charAt(i) == '1') {
if(arr[i]%2 == 0) {
sb1.append("1" );
}else {
sb1.append("0");
}
}else {
if(arr[i]%2 == 1) {
sb1.append("1" );
}else {
sb1.append("0");
}
}
}
}
for(int i = 0 ; i<n;i++) {
sb2.append(arr[i] + " ");
}
System.out.println(sb1);
System.out.println(sb2);
}
}
//------------------------------------------------------------------------------------------------------------------------------------------------
public static boolean check(int[] arr , int n , int v , int l ) {
int x = v/2;
int y = v/2;
// System.out.println(x + " " + y);
if(v%2 == 1 ) {
x++;
}
for(int i = 0 ; i<n;i++) {
int d = l - arr[i];
int c = Math.min(d/2, y);
y -= c;
arr[i] -= c*2;
if(arr[i] > x) {
return false;
}
x -= arr[i];
}
return true;
}
public static int cnt_set(long x) {
long v = 1l;
int c =0;
int f = 0;
while(v<=x) {
if((v&x)!=0) {
c++;
}
v = v<<1;
}
return c;
}
public static int lis(int[] arr,int[] dp) {
int n = arr.length;
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(arr[0]);
dp[0]= 1;
for(int i = 1 ; i<n;i++) {
int x = al.get(al.size()-1);
if(arr[i]>x) {
al.add(arr[i]);
}else {
int v = lower_bound(al, 0, al.size(), arr[i]);
// System.out.println(v);
al.set(v, arr[i]);
}
dp[i] = al.size();
}
//return al.size();
return al.size();
}
public static int lis2(int[] arr,int[] dp) {
int n = arr.length;
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(-arr[n-1]);
dp[n-1] = 1;
// System.out.println(al);
for(int i = n-2 ; i>=0;i--) {
int x = al.get(al.size()-1);
// System.out.println(-arr[i] + " " + i + " " + x);
if((-arr[i])>x) {
al.add(-arr[i]);
}else {
int v = lower_bound(al, 0, al.size(), -arr[i]);
// System.out.println(v);
al.set(v, -arr[i]);
}
dp[i] = al.size();
}
//return al.size();
return al.size();
}
static int cntDivisors(int n){
int cnt = 0;
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i)
cnt++;
else
cnt+=2;
}
}
return cnt;
}
public static long power(long x, long y, long p){
long res = 1;
x = x % p;
if (x == 0)
return 0;
while (y > 0){
if ((y & 1) != 0)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
public static long ncr(long[] fac, int n , int r , long m) {
if(r>n) {
return 0;
}
return fac[n]*(modInverse(fac[r], m))%m *(modInverse(fac[n-r], m))%m;
}
public static int lower_bound(ArrayList<Integer> arr,int lo , int hi, int k)
{
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (arr.get(mid) <k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==arr.size())
{
return -1;
}
return s;
}
public static int upper_bound(ArrayList<Integer> arr,int lo , int hi, int k)
{
int s=lo;
int e=hi;
while (s !=e)
{
int mid = s+e>>1;
if (arr.get(mid) <=k)
{
s=mid+1;
}
else
{
e=mid;
}
}
if(s==arr.size())
{
return -1;
}
return s;
}
// -----------------------------------------------------------------------------------------------------------------------------------------------
public static long gcd(long a, long b){
if (a == 0)
return b;
return gcd(b % a, a);
}
//--------------------------------------------------------------------------------------------------------------------------------------------------------
public static long modInverse(long a, long m){
long m0 = m;
long y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1) {
// q is quotient
long q = a / m;
long t = m;
// m is remainder now, process
// same as Euclid's algo
m = a % m;
a = t;
t = y;
// Update x and y
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
//_________________________________________________________________________________________________________________________________________________________________
// private static int[] parent;
// private static int[] size;
public static int find(int[] parent, int u) {
while(u != parent[u]) {
parent[u] = parent[parent[u]];
u = parent[u];
}
return u;
}
private static void union(int[] parent,int[] size,int u, int v) {
int rootU = find(parent,u);
int rootV = find(parent,v);
if(rootU == rootV) {
return;
}
if(size[rootU] < size[rootV]) {
parent[rootU] = rootV;
size[rootV] += size[rootU];
} else {
parent[rootV] = rootU;
size[rootU] += size[rootV];
}
}
//-----------------------------------------------------------------------------------------------------------------------------------
//segment tree
//for finding minimum in range
public static void build(int [] seg,int []arr,int idx, int lo , int hi) {
if(lo == hi) {
seg[idx] = arr[lo];
return;
}
int mid = (lo + hi)/2;
build(seg,arr,2*idx+1, lo, mid);
build(seg,arr,idx*2+2, mid +1, hi);
// seg[idx] = Math.min(seg[2*idx+1],seg[2*idx+2]);
// seg[idx] = seg[idx*2+1]+ seg[idx*2+2];
// seg[idx] = Math.min(seg[idx*2+1], seg[idx*2+2]);
seg[idx] = seg[idx*2+1] * seg[idx*2+2];
}
//for finding minimum in range
public static int query(int[]seg,int idx , int lo , int hi , int l , int r) {
if(lo>=l && hi<=r) {
return seg[idx];
}
if(hi<l || lo>r) {
return 1;
}
int mid = (lo + hi)/2;
int left = query(seg,idx*2 +1, lo, mid, l, r);
int right = query(seg,idx*2 + 2, mid + 1, hi, l, r);
// return Math.min(left, right);
//return gcd(left, right);
// return Math.min(left, right);
return left * right;
}
// // for sum
//
//public static void update(int[]seg,int idx, int lo , int hi , int node , int val) {
// if(lo == hi) {
// seg[idx] += val;
// }else {
//int mid = (lo + hi )/2;
//if(node<=mid && node>=lo) {
// update(seg, idx * 2 +1, lo, mid, node, val);
//}else {
// update(seg, idx*2 + 2, mid + 1, hi, node, val);
//}
//seg[idx] = seg[idx*2 + 1] + seg[idx*2 + 2];
//
//}
//}
//---------------------------------------------------------------------------------------------------------------------------------------
//
//static void shuffleArray(long[] ar)
//{
// // If running on Java 6 or older, use `new Random()` on RHS here
// Random rnd = ThreadLocalRandom.current();
// for (int i = ar.length - 1; i > 0; i--)
// {
// int index = rnd.nextInt(i + 1);
// // Simple swap
// int a = ar[index];
// ar[index] = ar[i];
// ar[i] = a;
// }
//}
// static void shuffleArray(coup[] ar)
// {
// // If running on Java 6 or older, use `new Random()` on RHS here
// Random rnd = ThreadLocalRandom.current();
// for (int i = ar.length - 1; i > 0; i--)
// {
// int index = rnd.nextInt(i + 1);
// // Simple swap
// coup a = ar[index];
// ar[index] = ar[i];
// ar[i] = a;
// }
// }
//-----------------------------------------------------------------------------------------------------------------------------------------------------------
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
//------------------------------------------------------------------------------------------------------------------------------------------------------------
//-----------------------------------------------------------------------------------------------------------------------------------------------------------
class coup{
int a;
int b;
public coup(int a , int b) {
this.a = a;
this.b = b;
}
}
class tripp{
int a;
int b;
double c;
public tripp(int a , int b, double c) {
this.a = a;
this.b = b;
this.c = c;
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | c9f606e365d67c6b8fc4c28dbdf2a148 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | /* Author _trevorphillips_ */
import java.io.*;
import java.util.*;
public class B_Bit_Flipping {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader f = new FastReader();
StringBuilder sb = new StringBuilder();
int t = f.nextInt();
while (t-- > 0) {
int n = f.nextInt();
int k = f.nextInt();
char[] s = f.nextLine().toCharArray();
if (k % 2 == 1) {
for (int g = 0; g < s.length; g++) {
if (s[g] == '1') {
s[g] = '0';
} else {
s[g] = '1';
}
}
}
int[] arr = new int[n];
for (int i = 0; i < s.length; i++) {
if (s[i] == '0' && k > 0) {
k--;
s[i] = '1';
arr[i]++;
}
}
arr[n - 1] += k;
if (k % 2 == 1) {
s[n - 1] = '0';
}
for (int i = 0; i < s.length; i++) {
sb.append(s[i]);
}
sb.append("\n");
for (int i = 0; i < arr.length; i++) {
sb.append(arr[i] + " ");
}
sb.append("\n");
}
System.out.println(sb);
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 05180992f369ea1e1eff74f4e816e35d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class B {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = in.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = in.nextInt(), k = in.nextInt();
char[] s = in.next().toCharArray();
int[] res = new int[n];
int tempK = k;
for (int i = 0; i < n && tempK > 0; i++) {
if (k % 2 == s[i] - '0') {
res[i] = 1;
tempK--;
}
}
res[n - 1] += tempK;
for (int i = 0; i < n; i++) {
if ((k - res[i]) % 2 == 1){
// debug(s[i]);
s[i] = (char)((int)'1' - (int)(s[i] - '0'));
}
}
for (char c: s) pw.print(c);
pw.println();
for (int v: res) pw.print(v + " ");
pw.println();
}
pw.close();
}
static void debug(Object... obj) {
System.err.println(Arrays.deepToString(obj));
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 34e8666cee877103f46c61ea7b570513 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.util.*;
import java.lang.Math.*;
public class Inheritance{
public static void main(String[] args){
Scanner s=new Scanner(System.in);
long t=s.nextLong();
while(t-->0)
{
int n=s.nextInt();
long k=s.nextLong();
String s1=s.next();
StringBuffer ans=new StringBuffer("");
long arr[]=new long[(int)n];
int i1=0;
int i=0;
if(k%2==0)
{
for(;i<n;i++)
{
if(s1.charAt(i)=='0' && k!=0)
{
arr[i1]=1;
i1++;
k=k-1;
ans.append("1");
}
else if(s1.charAt(i)=='1' && k!=0)
{
arr[i1]=0;
i1++;
ans.append("1");;
}
else if (k==0)
{
if(s1.charAt(i)=='0')
ans.append("0");
else
ans.append("1");
arr[i1]=0;
i1++;
}
}
if(k!=0)
{
k=k+arr[n-1];
char x=s1.charAt(n-1);
ans.setCharAt(n-1, x);;
if(k%2==0)
{
arr[(int)n-1]=k;
}
else
{
arr[(int)n-1]=k;
if(ans.charAt(n-1)=='0')
ans.setCharAt((int)n-1, '1');
else
ans.setCharAt((int)n-1, '0');
}
}
}
else
{
for(;i<n;i++)
{
if(s1.charAt(i)=='0' && k!=0)
{
arr[i1]=0;
i1++;
ans.append("1");
}
else if(s1.charAt(i)=='1' && k!=0)
{
arr[i1]=1;
k=k-1;
i1++;
ans.append("1");;
}
else if (k==0)
{
if(s1.charAt(i)=='0')
ans.append("1");
else
ans.append("0");
arr[i1]=0;
i1++;
}
}
if(k!=0)
{
k=k+arr[n-1];
char x=s1.charAt(n-1);
ans.setCharAt(n-1, x);;
if(k%2!=0)
{
arr[(int)n-1]=k;
}
else
{
arr[(int)n-1]=k;
if(ans.charAt(n-1)=='0')
ans.setCharAt((int)n-1, '1');
else
ans.setCharAt((int)n-1, '0');
}
}
}
System.out.println(ans);
for(int i2=0;i2<n;i2++)
{
System.out.print(arr[i2]+" ");
}
System.out.println();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | e0814a5a0338397dc78ed8b4926343df | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Codeforces782{
static long mod = 1000000007L;
static MyScanner sc = new MyScanner();
static void solve() {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int arr[] = new int[2 * b + 1];
Arrays.fill(arr, 1);
r = r - b;
r--;
while (r > 0)
for (int i = 0; i < arr.length&& r > 0; i += 2) {
arr[i]++;
r--;
}
for(int i = 0;i<arr.length;i++){
if(i==0 || i%2==0){
for(int j = 0;j<arr[i];j++){
out.print('R');
}
}else{
out.print('B');
}
}
out.println();
}
static void solve2(){
int n = sc.nextInt();
int k = sc.nextInt();
char arr[] = sc.nextLine().toCharArray();
if(k%2!=0){
for(int i = 0;i<n;i++){
if(arr[i]=='1') arr[i] = '0';
else arr[i]='1';
}
}
int ans[] = new int[n];
for(int i = 0;i<n && k>0;i++){
if(arr[i]=='0'){
ans[i]++;
k--;
arr[i] = '1';
}
}
if(k>0){
if(k%2!=0){
if(arr[n-1]=='1') arr[n-1] = '0';
else arr[n-1]='1';
}
ans[n-1]+=k;
}
for(int i = 0;i<n;i++){
out.print(arr[i]);
}
out.println();
for(int i = 0;i<n;i++){
out.print(ans[i]+" ");
}
out.println();
}
static void solve3(){
int n= sc.nextInt();
long a = sc.nextLong();
long b = sc.nextLong();
long arr[] = new long[n+1];
for(int i =1 ;i<=n;i++) arr[i] = sc.nextLong();
long dp[][] = new long[arr.length+1][arr.length+1];
for(int i = 0;i<dp.length;i++){
Arrays.fill(dp[i],-1);
}
out.println(fun(dp,arr,0,1,a,b));
}
static long fun(long dp[][],long arr[],int i,int j,long a,long b){
if(j==arr.length) return 0;
if(dp[i][j]!=-1) return dp[i][j];
if(i+1==j){
dp[i][j] = b*(arr[j]-arr[i])+fun(dp,arr,i,j+1,a,b);
}else
dp[i][j] = Math.min(b*(arr[j]-arr[i])+fun(dp,arr,i,j+1,a,b),(a*(arr[i+1]-arr[i]))+fun(dp,arr,i+1,j,a,b));
return dp[i][j];
}
static void solve4(){
}
static void reverse(int arr[]){
int i = 0;int j = arr.length-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
}
static class Pair{
int odd;
int even;
Pair(int o,int e){
odd = o;
even = e;
}
}
static long pow(long a, long b) {
if (b == 0) return 1;
long res = pow(a, b / 2);
res = (res * res) % 1_000_000_007;
if (b % 2 == 1) {
res = (res * a) % 1_000_000_007;
}
return res;
}
static int lis(int arr[],int n){
int lis[] = new int[n];
lis[0] = 1;
for(int i = 1;i<n;i++){
lis[i] = 1;
for(int j = 0;j<i;j++){
if(arr[i]>arr[j]){
lis[i] = Math.max(lis[i],lis[j]+1);
}
}
}
int max = Integer.MIN_VALUE;
for(int i = 0;i<n;i++){
max = Math.max(lis[i],max);
}
return max;
}
static boolean isPali(String str){
int i = 0;
int j = str.length()-1;
while(i<j){
if(str.charAt(i)!=str.charAt(j)){
return false;
}
i++;
j--;
}
return true;
}
static long gcd(long a,long b){
if(b==0) return a;
return gcd(b,a%b);
}
static String reverse(String str){
char arr[] = str.toCharArray();
int i = 0;
int j = arr.length-1;
while(i<j){
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}
String st = new String(arr);
return st;
}
static boolean isprime(int n){
if(n==1) return false;
if(n==3 || n==2) return true;
if(n%2==0 || n%3==0) return false;
for(int i = 5;i*i<=n;i+= 6){
if(n%i== 0 || n%(i+2)==0){
return false;
}
}
return true;
}
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = sc.nextInt();
// int t= 1;
while(t-- >0){
// solve();
solve2();
// solve3();
}
// Stop writing your solution here. -------------------------------------
out.close();
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int[] readIntArray(int n){
int arr[] = new int[n];
for(int i = 0;i<n;i++){
arr[i] = Integer.parseInt(next());
}
return arr;
}
int[] reverse(int arr[]){
int n= arr.length;
int i = 0;
int j = n-1;
while(i<j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
j--;i++;
}
return arr;
}
long[] readLongArray(int n){
long arr[] = new long[n];
for(int i = 0;i<n;i++){
arr[i] = Long.parseLong(next());
}
return arr;
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
private static void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
private static void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for (int i=0; i<arr.length; i++){
list.add(arr[i]);
}
Collections.sort(list); // collections.sort uses nlogn in backend
for (int i = 0; i < arr.length; i++){
arr[i] = list.get(i);
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 7fdf8e9614bd219a90bb77a6a312baa5 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class test {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int k = sc.nextInt();
String s = sc.nextLine();
int tm = k;
int[] vis = new int[s.length()];
for(int i = 0;i<vis.length&&k>0;i++) {
if(s.charAt(i)=='1') {
if(tm%2==1) {
vis[i]++;
k--;
}
}
else {
if(tm%2==0) {
vis[i]++;
k--;
}
}
}
vis[n-1]+=k;
StringBuilder tmp = new StringBuilder();
for (int j = 0; j < n; j++) {
if (s.charAt(j) == '0' && (tm - vis[j]) % 2 == 0) {
tmp.append("0");
}
if (s.charAt(j) == '1' && (tm - vis[j]) % 2 == 1) {
tmp.append("0");
}
if (s.charAt(j) == '0' && (tm - vis[j]) % 2 == 1) {
tmp.append("1");
}
if (s.charAt(j) == '1' && (tm - vis[j]) % 2 == 0) {
tmp.append("1");
}
}
pw.println(tmp);
StringBuilder res = new StringBuilder();
for (int i = 0; i < vis.length; i++) {
res.append(vis[i]);
if (i != vis.length - 1)
res.append(" ");
}
pw.println(res);
}
pw.flush();
}
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 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 boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 65d14e9b0a4617cfa3b1fd9159a0b058 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class B_1659 {
public static void main(String[] args)
{
FastReader sc = new FastReader();
PrintWriter out=new PrintWriter(System.out);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int k=sc.nextInt();
char[] ch=sc.next().toCharArray();
if(k%2==1) {
for(int i=0;i<n;i++) {
if(ch[i]=='1') ch[i]='0';
else ch[i]='1';
}
}
int[] arr=new int[n];
for(int i=0;i<n;i++) {
if(ch[i]=='0' && k>0) {
ch[i]='1';
k--;
arr[i]++;
}
}
arr[n-1]+=k;
if(k%2==1) ch[n-1]='0';
out.println(ch);
for(int 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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | ced2048f62e0dcdbcfbabcd51d6f5958 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | //import java.io.IOException;
import java.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.List;
public class BitFlipping {
static InputReader inputReader=new InputReader(System.in);
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static void solve() throws IOException
{
int n=inputReader.nextInt();
int k=inputReader.nextInt();
int temp=k;
String str=inputReader.readString();
int answer[]=new int[n];
for (int i=0;i<n-1;i++)
{
if (str.charAt(i)=='0')
{
if (temp%2!=0)
{
answer[i]=0;
}
else {
if (k > 0)
{
answer[i] = 1;
k--;
}
}
}
else
{
if (temp%2==0)
{
answer[i]=0;
}
else
{
if (k > 0)
{
answer[i] = 1;
k--;
}
}
}
}
answer[n-1]=k;
StringBuffer fans=new StringBuffer();
for (int i=0;i<n;i++)
{
int oper=(temp-answer[i]);
if (str.charAt(i)=='0')
{
if (oper%2==0)
{
fans.append('0');
}
else
{
fans.append('1');
}
}
else
{
if (oper%2==0)
{
fans.append('1');
}
else
{
fans.append('0');
}
}
}
out.println(new String(fans));
for (int ele:answer)
{
out.print(ele+" ");
}
out.println();
}
static PrintWriter out=new PrintWriter((System.out));
static void SortDec(long arr[])
{
List<Long>list=new ArrayList<>();
for(long ele:arr) {
list.add(ele);
}
Collections.sort(list,Collections.reverseOrder());
for (int i=0;i<list.size();i++)
{
arr[i]=list.get(i);
}
}
static void Sort(long arr[])
{
List<Long>list=new ArrayList<>();
for(long ele:arr) {
list.add(ele);
}
Collections.sort(list);
for (int i=0;i<list.size();i++)
{
arr[i]=list.get(i);
}
}
public static void main(String args[])throws IOException
{
int t=inputReader.nextInt();
while (t-->0) {
solve();
}
long s = System.currentTimeMillis();
// out.println(System.currentTimeMillis()-s+"ms");
out.close();
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c))
c = snext();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c))
c = snext();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 7bc977f4ae590e3ed1dd286c211030c6 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B_Bit_Flipping {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int testCases, n;
static long k;
static char a[];
static void solve(int t) {
int s[] = new int[n];
for(int i = 0; i < n; ++i) {
s[i] = a[i] - '0';
}
long copyK = k;
long dp[] = new long[n];
int num[] = {0, 1};
int j = 0;
for(int i = 0; i < n && k > 0; ++i) {
a[i] ^= num[j];
if(k % 2 == a[i] - '0' && i != n - 1) {
dp[i]++;
--k;
j = (j + 1) % 2;
}
}
dp[n - 1] += k;
for(int i = 0; i < n; ++i) {
if((copyK - dp[i]) % 2 == 1 ) {
s[i] ^= 1;
}
}
for(int i : s) {
ans.append(i);
}
ans.append("\n");
for(long i : dp) {
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();
k = in.nextLong();
a = in.next().toCharArray();
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 5a91b9b91c2dc9057d52b1698662dfe5 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String arggs[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
StringBuilder sb = new StringBuilder();
while(t-- > 0) {
int n = sc.nextInt(), k = sc.nextInt(), cnt = 0, arr[] = new int[n];
String s = sc.next();
for(int i=0; i<n-1; i++) {
if(s.charAt(i)-'0'== k%2) {
if(cnt < k) {
cnt++;
arr[i]++;
sb.append(1);
} else sb.append(0);
}else sb.append(1);
}
arr[n-1] += k-cnt;
sb.append((s.charAt(n-1)-'0' == cnt%2 ? 0 : 1) + "\n");
for(int a : arr)
sb.append(a + " ");
sb.append("\n");
}
System.out.println(sb);
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 7bbe8426301961b9345cd8dd7062d7f7 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.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 B {
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(),k=sc.nextInt();
String s=sc.next();
int[]ans=new int[n];
int i=0;
boolean flipped=false;
int kk=k;
boolean allO=false,oEx=false;
while(kk-->0) {
int st=kk+1;
if(st%2==0) {
char trgt=flipped?'1':'0';
while(i<n&&s.charAt(i)!=trgt)i++;
if(i==n) {
allO=true;
break;
}
ans[i]++;
}else {
char trgt=flipped?'0':'1';
while(i<n&&s.charAt(i)!=trgt)i++;
if(i==n) {
oEx=true;
break;
}
ans[i]++;
}
i++;
flipped=!flipped;
}
ans[n-1]+=kk+1;
if(allO||oEx) {
StringBuilder sb=new StringBuilder();
for(int j=0;j<n-1;j++)sb.append("1");
sb.append(allO?"1":"0");
out.println(sb.toString());
}else {
int mx1=-1;
for(int j=n-1;j>=0;j--) {
if(ans[j]==1) {mx1=j;
break;
}
}
for(int j=0;j<=mx1;j++) {
out.print("1");
}
for(int j=mx1+1;j<n;j++) {
int c='1'-s.charAt(j);
out.print(k%2==0?1-c:c);
}
out.println();
}
for(int j=0;j<ans.length;j++)out.print(ans[j]+" ");
out.println();
}
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f7895505f8b85dacfd7fd641d5adc99d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Random;
import java.util.StringTokenizer;
/*
1
6 3
100001
*/
public class B {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int T=fs.nextInt();
for (int tt=0; tt<T; tt++) {
int n=fs.nextInt();
int k=fs.nextInt();
char[] a=fs.next().toCharArray();
if (k%2==1) {
for (int i=0; i<n; i++) {
if (a[i]=='0') a[i]='1';
else a[i]='0';
}
}
int[] choice=new int[n];
for (int i=0; i<a.length; i++) {
if (a[i]=='0' && k>0) {
a[i]='1';
k--;
choice[i]++;
}
}
choice[n-1]+=k;
if (k%2==1) {
a[n-1]='0';
}
out.println(a);
for (int i=0; i<n; i++) {
out.print(choice[i]+" ");
}
out.println();
}
out.close();
}
static final Random random=new Random();
static final int mod=1_000_000_007;
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static long add(long a, long b) {
return (a+b)%mod;
}
static long sub(long a, long b) {
return ((a-b)%mod+mod)%mod;
}
static long mul(long a, long b) {
return (a*b)%mod;
}
static long exp(long base, long exp) {
if (exp==0) return 1;
long half=exp(base, exp/2);
if (exp%2==0) return mul(half, half);
return mul(half, mul(half, base));
}
static long[] factorials=new long[2_000_001];
static long[] invFactorials=new long[2_000_001];
static void precompFacts() {
factorials[0]=invFactorials[0]=1;
for (int i=1; i<factorials.length; i++) factorials[i]=mul(factorials[i-1], i);
invFactorials[factorials.length-1]=exp(factorials[factorials.length-1], mod-2);
for (int i=invFactorials.length-2; i>=0; i--)
invFactorials[i]=mul(invFactorials[i+1], i+1);
}
static long nCk(int n, int k) {
return mul(factorials[n], mul(invFactorials[k], invFactorials[n-k]));
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 10884e0a1d1464375e20f9d20b0264db | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import javax.management.Query;
import java.io.*;
public class practice {
// static int n,k;
//// static String t,s;
// static long[]memo;
// static int n;// int k=sc.nextInt();
static String s;
static HashMap<Long,Integer>hm;
public static void main(String[] args) throws Exception {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int k=sc.nextInt();
String s=sc.next();
int[] a=new int[n];
int[] b=new int[n];
int y=0;
for (int i = 0; i < n-1; i++) {
if(k>0) {
if (k % 2 == ((s.charAt(i) - '0')^y)) {
a[i]=1;
b[i]=1;
k--;
y^=1;
}else{
a[i]=1;
}
}
else{
a[i]=((s.charAt(i) - '0')^y);
}
}
a[n-1]=((s.charAt(n-1) - '0')^y);
b[n-1]=k;
for (int i = 0; i <n ; i++) {
pw.print(a[i]);
}
pw.println();
for (int i = 0; i <n ; i++) {
pw.print(b[i]+" ");
}
pw.println();
}
pw.close();
}
public static void sort(int[] in) {
shuffle(in);
Arrays.sort(in);
}
public static void shuffle(int[] in) {
for (int i = 0; i < in.length; i++) {
int idx = (int) (Math.random() * in.length);
int tmp = in[i];
in[i] = in[idx];
in[idx] = tmp;
}
}
static LinkedList getfact(int n){
LinkedList<Integer>ll=new LinkedList<>();
LinkedList<Integer>ll2=new LinkedList<>();
for (int i = 1; i <= Math.sqrt(n); i++) {
if(n%i==0) {
ll.add(i);
if(i!=n/i)
ll2.addLast(n/i);
}
}
while (!ll2.isEmpty()){
ll.add(ll2.removeLast());
}
return ll;
}
static void rev(int n){
String s1=s.substring(0,n);
s=s.substring(n);
for (int i = 0; i <n ; i++) {
s=s1.charAt(i)+s;
}
}
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, sTree;
Long[]lazy;
SegmentTree(long[] in)
{
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 - 1;
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,j,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*1l;
}
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)
{
if(lazy[node]!=null) {
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] = null;
}
long query(int i, int j)
{
return query(1,1,N,i,j);
}
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;
}
}
// public static long dp(int idx) {
// if (idx >= n)
// return Long.MAX_VALUE/2;
// return Math.min(dp(idx+1),memo[idx]+dp(idx+k));
// }
// if(num==k)
// return dp(0,idx+1);
// if(memo[num][idx]!=-1)
// return memo[num][idx];
// long ret=0;
// if(num==0) {
// if(s.charAt(idx)=='a')
// ret= dp(1,idx+1);
// else if(s.charAt(idx)=='?') {
// ret=Math.max(1+dp(1,idx+1),dp(0,idx+1) );
// }
// }
// else {
// if(num%2==0) {
// if(s.charAt(idx)=='a')
// ret=dp(num+1,idx+1);
// else if(s.charAt(idx)=='?')
// ret=Math.max(1+dp(num+1,idx+1),dp(0,idx+1));
// }
// else {
// if(s.charAt(idx)=='b')
// ret=dp(num+1,idx+1);
// else if(s.charAt(idx)=='?')
// ret=Math.max(1+dp(num+1,idx+1),dp(0,idx+1));
// }
// }
// }
static void putorrem(long x){
if(hm.getOrDefault(x,0)==1){
hm.remove(x);
}
else
hm.put(x,hm.getOrDefault(x,0)-1);
}
public static int max4(int a,int b, int c,int d) {
int [] s= {a,b,c,d};
Arrays.sort(s);
return s[3];
}
public static double logbase2(int k) {
return( (Math.log(k)+0.0)/Math.log(2));
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
long x;
long y;
public pair(long x, long y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair) o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Long(x).hashCode() * 31 + new Long(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return Long.compare(this.y, other.y);
}
return Long.compare(this.x, other.x);
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if (this.y == other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
public tuble add(tuble t){
return new tuble(this.x+t.x,this.y+t.y,this.z+t.z);
}
}
static long mod = 1000000007;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 514bc5102ef74d01e035c5f5757496d6 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.*;
/*
101001
*/
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(), k=fs.nextInt();
int f=k;
char arr[]=fs.next().toCharArray();
int ans[]=new int[n];
if(k%2==0) {
for(int i=0;i<n-1;i++) {
if(arr[i]=='0'&&k>0) {
k--;
ans[i]=1;
}
}
ans[n-1]=k;
}
else {
for(int i=0;i<n-1;i++) {
if(arr[i]=='1'&&k>0) {
k--;
ans[i]=1;
}
}
ans[n-1]=k;
}
for(int i=0;i<n;i++) {
int a= (f-ans[i]);
a%=2;
int b;
if(arr[i]=='1') b=1;
else b=0;
out.print((a^b));
}
out.println();
for(int ele:ans) out.print(ele+" ");
out.println();
}
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 5f2da4ccae2531f8bee2537b2d1c258d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
// int a = 1;
int t;
t = in.nextInt();
//t = 1;
while (t > 0) {
// out.print("Case #"+(a++)+": ");
solver.call(in,out);
t--;
}
out.close();
}
static class TaskA {
public void call(InputReader in, PrintWriter out) {
int n, k, a = 0;
n = in.nextInt();
k = in.nextInt();
char[] arr = in.next().toCharArray();
int[] ans = new int[n];
for (int i = 0; i < n; i++) {
if(arr[i] == '1'){
if(k%2!=0) {
ans[i]++;
a++;
}
}
if(a==k){
break;
}
if(arr[i] == '0'){
if(k%2==0){
ans[i]++;
a++;
}
}
if(a==k){
break;
}
}
if(a!=k){
ans[n - 1] += (k - a);
}
StringBuilder s = new StringBuilder("");
for (int i = 0; i < n; i++) {
if(arr[i]=='1' && (k - ans[i]) %2==0){
s.append("1");
}
else if(arr[i]=='0' && (k - ans[i]) %2!=0){
s.append("1");
}
else{
s.append("0");
}
}
out.println(s);
for(Integer i : ans){
out.print(i+" ");
}
out.println();
}
}
static int gcd(int a, int b )
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static class answer implements Comparable<answer>{
int a, b;
long c;
public answer(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}
public answer(int a, int b) {
this.a = a;
this.b = b;
}
@Override
public int compareTo(answer o) {
return Integer.compare(this.a, o.a);
}
}
static class answer1 implements Comparable<answer1>{
int a, b, c;
public answer1(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}
public answer1(int a, int b) {
this.a = a;
this.b = b;
}
@Override
public int compareTo(answer1 o) {
if(this.b == o.b){
return Integer.compare(this.a, o.a);
}
return Integer.compare(this.b, o.b);
}
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>();
for (Long i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static final Random random=new Random();
static void shuffleSort(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);
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong(){
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | ffc3639ac16715c0154b2b1855d69647 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B1659 {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
solver.solve(in, out);
out.close();
}
private static class Solver {
void solve(InputReader in, PrintWriter out) {
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int k = in.nextInt();
String s = in.next();
char[] chars = s.toCharArray();
int[] ans = new int[chars.length];
if (k % 2 == 0) {
for (int i = 0; i < n; i++) {
if (chars[i] == '0' && k > 0) {
k--;
ans[i]++;
}
}
ans[n - 1] += k;
for (int i = 0; i < n; i++) {
if (ans[i] % 2 != 0) {
chars[i] = (char)('0' + ('1' - chars[i]));
}
}
} else {
for (int i = 0; i < n; i++) {
if (chars[i] == '1' && k > 0) {
k--;
ans[i]++;
}
}
ans[n - 1] += k;
for (int i = 0; i < n; i++) {
if (ans[i] % 2 == 0) {
chars[i] = (char)('0' + ('1' - chars[i]));
}
}
}
out.println(chars);
for (int an : ans) {
out.print(an + " ");
}
out.println();
}
}
}
private static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | fefdeab69d7e668a7c62a1ab5b501f76 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B_Bit_Flipping {
static Scanner in = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
static StringBuilder ans = new StringBuilder();
static int testCases, n;
static long k;
static char a[];
static void solve(int t) {
int num[] = {0, 1};
long dp[] = new long[n];
int s[] = new int[n];
for (int i = 0; i < n; ++i) {
s[i] = a[i] - '0';
}
int j = 0;
long copyK = k;
for (int i = 0; i < n && k > 0; ++i) {
a[i] ^= num[j];
if (k % 2 == a[i] - '0' && i != n - 1) {
++dp[i];
j = (j + 1) % 2;
--k;
}
}
dp[n - 1] += k;
for (int i = 0; i < n; ++i) {
if ((copyK - dp[i]) % 2 == 1) {
s[i] ^= 1;
}
}
for (int i : s) {
ans.append(i);
}
ans.append("\n");
for (long i : dp) {
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();
k = in.nextLong();
a = in.next().toCharArray();
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 5d8240f61e841edf03ec83ffb0e8240d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | /*
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 NewTimeB
{
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 K = Integer.parseInt(st.nextToken());
String line = infile.readLine().trim();
char[] input = line.toCharArray();
int[] arr = new int[N];
for(int i=0; i < N; i++)
arr[i] = input[i]-'0';
int[] res = new int[N];
if(K == 0)
{
sb.append(line).append("\n");
for(int x: res)
sb.append(x+" ");
sb.append("\n");
continue;
}
if((K&1) == 1)
for(int i=0; i < N; i++)
arr[i] ^= 1;
for(int i=0; i < N; i++)
if(K > 0 && arr[i] == 0)
{
res[i] = 1;
arr[i] = 1;
K--;
}
res[N-1] += K;
arr[N-1] ^= (K&1);
for(int x: arr)
sb.append(x);
sb.append('\n');
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;
}
}
/*
K=2
011
000
101
[1,1,0]
011
000
110
[1,0,1]
*/ | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | ab08662a740facdf93fe65903c397daa | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class CodeForces {
/*-------------------------------------------EDITING CODE STARTS HERE-------------------------------------------*/
public static void solve(int tCase) throws IOException {
int n = sc.nextInt();
int k = sc.nextInt();
char[] arr = sc.next().toCharArray();
Pair<String,int[]> ans;
if(k%2==0)ans = find(arr,'0',n,k);
else ans = find(arr,'1',n,k);
out.println(ans.ff);
for(int x : ans.ss)out.print(x+" ");
out.println();
}
private static Pair<String,int[]> find(char[] arr, char a,int n,int k){
int[] ans = new int[n];
int used = 0;
for(int i=0;i<n && k>0;i++)
if(arr[i] == a){
ans[i] = 1;
k--;
used++;
}
ans[0] += 2*(k/2);
used+= 2*(k/2);
k = k%2;
char[] bbr = new char[n];
for(int i=0;i<n;i++){
int rem = used - ans[i];
if(rem % 2 ==1)bbr[i] = arr[i]=='1'?'0':'1';
else bbr[i] = arr[i];
}
if(k==1){
// out.println(Arrays.toString(bbr)+" "+Arrays.toString(ans));
Pair<String ,int[] > pp = null;
for(int i=0;i<n;i++){
if(bbr[i]=='1'){
char[] x = new char[n];
int[] a1 = new int[n];
x[i] = bbr[i];
a1[i]++;
for(int j=0;j<n;j++){
a1[j] += ans[j];
if(i==j)continue;
x[j] = bbr[j]=='1'?'0':'1';
}
pp = new Pair<>(String.valueOf(x),a1);
break;
}
}
for(int i=0;i<n;i++) {
if (bbr[i] == '0') {
char[] x = new char[n];
int[] a1 = new int[n];
x[i] = bbr[i];
a1[i]++;
for (int j = 0; j < n; j++) {
a1[j] += ans[j];
if (i == j) continue;
x[j] = bbr[j] == '1' ? '0' : '1';
}
String y = String.valueOf(x);
if (pp==null || y.compareTo(pp.ff) > 0) pp = new Pair<>(y, a1);
break;
}
}
for(int i=n-1;i>=0;i--){
if(bbr[i]=='1'){
char[] x = new char[n];
int[] a1 = new int[n];
x[i] = bbr[i];
a1[i]++;
for(int j=0;j<n;j++){
a1[j] += ans[j];
if(i==j)continue;
x[j] = bbr[j]=='1'?'0':'1';
}
String y = String.valueOf(x);
if (pp==null || y.compareTo(pp.ff) > 0) pp = new Pair<>(y, a1);
break;
}
}
for(int i=n-1;i>=0;i--) {
if (bbr[i] == '0') {
char[] x = new char[n];
int[] a1 = new int[n];
x[i] = bbr[i];
a1[i]++;
for (int j = 0; j < n; j++) {
a1[j] += ans[j];
if (i == j) continue;
x[j] = bbr[j] == '1' ? '0' : '1';
}
String y = String.valueOf(x);
if (pp==null || y.compareTo(pp.ff) > 0) pp = new Pair<>(y, a1);
break;
}
}
return pp;
}
return new Pair<>(String.valueOf(bbr),ans);
}
public static void main(String[] args) throws IOException {
openIO();
int testCase = 1;
testCase = sc.nextInt();
for (int i = 1; i <= testCase; i++) solve(i);
closeIO();
}
/*-------------------------------------------EDITING CODE ENDS HERE-------------------------------------------*/
/*--------------------------------------HELPER FUNCTIONS STARTS HERE-----------------------------------------*/
// public static int mod = (int) 1e9 + 7;
public static int mod = 998244353;
public static int inf_int = (int) 2e9+10;
public static long inf_long = (long) 2e18;
public static void _sort(int[] arr, boolean isAscending) {
int n = arr.length;
List<Integer> list = new ArrayList<>();
for (int ele : arr) list.add(ele);
Collections.sort(list);
if (!isAscending) Collections.reverse(list);
for (int i = 0; i < n; i++) arr[i] = list.get(i);
}
public static void _sort(long[] arr, boolean isAscending) {
int n = arr.length;
List<Long> list = new ArrayList<>();
for (long ele : arr) list.add(ele);
Collections.sort(list);
if (!isAscending) Collections.reverse(list);
for (int i = 0; i < n; i++) arr[i] = list.get(i);
}
// time : O(1), space : O(1)
public static int _digitCount(long num,int base){
// this will give the # of digits needed for a number num in format : base
return (int)(1 + Math.log(num)/Math.log(base));
}
// time : O(n), space: O(n)
public static long _fact(int n){
// simple factorial calculator
long ans = 1;
for(int i=2;i<=n;i++)
ans = ans * i % mod;
return ans;
}
// time for pre-computation of factorial and inverse-factorial table : O(nlog(mod))
public static long[] factorial , inverseFact;
public static void _ncr_precompute(int n){
factorial = new long[n+1];
inverseFact = new long[n+1];
factorial[0] = inverseFact[0] = 1;
for (int i = 1; i <=n; i++) {
factorial[i] = (factorial[i - 1] * i) % mod;
inverseFact[i] = _modExpo(factorial[i], mod - 2);
}
}
// time of factorial calculation after pre-computation is O(1)
public static int _ncr(int n,int r){
if(r > n)return 0;
return (int)(factorial[n] * inverseFact[r] % mod * inverseFact[n - r] % mod);
}
public static int _npr(int n,int r){
if(r > n)return 0;
return (int)(factorial[n] * inverseFact[n - r] % mod);
}
// euclidean algorithm time O(max (loga ,logb))
public static long _gcd(long a, long b) {
while (a>0){
long x = a;
a = b % a;
b = x;
}
return b;
// if (a == 0)
// return b;
// return _gcd(b % a, a);
}
// lcm(a,b) * gcd(a,b) = a * b
public static long _lcm(long a, long b) {
return (a / _gcd(a, b)) * b;
}
// binary exponentiation time O(logn)
public static long _modExpo(long x, long n) {
long ans = 1;
while (n > 0) {
if ((n & 1) == 1) {
ans *= x;
ans %= mod;
n--;
} else {
x *= x;
x %= mod;
n >>= 1;
}
}
return ans;
}
// function to find a/b under modulo mod. time : O(logn)
public static long _modInv(long a,long b){
return (a * _modExpo(b,mod-2)) % mod;
}
//sieve or first divisor time : O(mx * log ( log (mx) ) )
public static int[] _seive(int mx){
int[] firstDivisor = new int[mx+1];
for(int i=0;i<=mx;i++)firstDivisor[i] = i;
for(int i=2;i*i<=mx;i++)
if(firstDivisor[i] == i)
for(int j = i*i;j<=mx;j+=i)
if(firstDivisor[j]==j)firstDivisor[j] = i;
return firstDivisor;
}
// check if x is a prime # of not. time : O( n ^ 1/2 )
private static boolean _isPrime(long x){
for(long i=2;i*i<=x;i++)
if(x%i==0)return false;
return true;
}
static class Pair<K, V>{
K ff;
V ss;
public Pair(K ff, V ss) {
this.ff = ff;
this.ss = ss;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || this.getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return ff.equals(pair.ff) && ss.equals(pair.ss);
}
@Override
public int hashCode() {
return Objects.hash(ff, ss);
}
@Override
public String toString(){
return ff.toString()+" "+ss.toString();
}
}
/*--------------------------------------HELPER FUNCTIONS ENDS HERE-----------------------------------------*/
/*-------------------------------------------FAST INPUT STARTS HERE---------------------------------------------*/
static FastestReader sc;
static PrintWriter out;
private static void openIO() throws IOException {
sc = new FastestReader();
out = new PrintWriter(System.out);
}
public static void closeIO() throws IOException {
out.flush();
out.close();
sc.close();
}
private static final class FastestReader {
private static final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public FastestReader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastestReader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
private static boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private int skip() throws IOException {
int b;
//noinspection StatementWithEmptyBody
while ((b = read()) != -1 && isSpaceChar(b)) {}
return b;
}
public String next() throws IOException {
int b = skip();
final StringBuilder sb = new StringBuilder();
while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = read();
}
return sb.toString();
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') c = read();
final boolean neg = c == '-';
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
return neg?-ret:ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') c = read();
final boolean neg = c == '-';
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
return neg?-ret:ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') c = read();
final boolean neg = c == '-';
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.')
while ((c = read()) >= '0' && c <= '9')
ret += (c - '0') / (div *= 10);
return neg?-ret:ret;
}
public String nextLine() throws IOException {
final byte[] buf = new byte[(1<<10)]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
break;
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
din.close();
}
}
/*---------------------------------------------FAST INPUT ENDS HERE ---------------------------------------------*/
}
/** Some points to keep in mind :
* 1. don't use Arrays.sort(primitive data type array)
* 2. try to make the parameters of a recursive function as less as possible,
* more use static variables.
*
**/ | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | e44d7b89673acb5c87b3942fd5348da5 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
public static long factorial(int number) {
long f = 1;
int j = 1;
while (j <= number) {
f = (f * j) % 998244353;
j++;
}
return f;
}
public static long combination(int n, int r) {
return factorial(n) / (factorial(n - r) * factorial(r));
}
public static void main(String[] args) throws Exception {
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int k = sc.nextInt();
int tmp=k;
char[] s = sc.next().toCharArray();
int o=0;
int[]res = new int[n];
char[] helper = new char[n];
for(int i=0;i<s.length;i++) {
helper[i]=s[i];
}
for(int i=0;i<n&&k>0;i++) {
s[i]^=o;
if(k%2!=s[i]-'0' || i==n-1) continue;
res[i]++;
k--;
o^=1;
}
res[n-1]+=k;
for(int i=0;i<n;i++) {
if((tmp-res[i])%2==1) {
helper[i]^=1;
}
}
String ans = new String(helper);
pw.println(helper);
for(int i=0;i<n;i++) {
pw.print(res[i]+" ");
}
pw.println();
}
pw.flush();
}
static class SegmentTree {
int[] arr, sTree, lazy;
int N;
public SegmentTree(int[] in) {
arr = in;
N = in.length - 1;
sTree = new int[2 * N];
lazy = new int[2 * N];
build(1, 1, N);
}
public void build(int Node, int left, int right) { // O(n)
if (left == right)
sTree[Node] = arr[left];
else {
int r = 2 * Node + 1;
int l = 2 * Node;
int mid = (left + right) / 2;
build(l, left, mid);
build(r, mid + 1, right);
sTree[Node] = sTree[l] + sTree[r];
}
}
public int query(int i, int j) {
return query(1, 1, N, i, j);
}
public int query(int Node, int left, int right, int i, int j) {
if (right < i || left > j)
return 0;
if (right <= j && i <= left) {
return sTree[Node];
}
int mid = (left + right) / 2;
int l = 2 * Node;
int r = 2 * Node + 1;
return query(l, left, mid, i, j) + query(r, mid + 1, right, i, j);
}
public void updatePoint(int idx, int value) {
int node = idx + N - 1;
arr[idx] = value;
sTree[node] = value;
while (node > 1) {
node /= 2;
int l = 2 * node;
int r = 2 * node + 1;
sTree[node] = sTree[l] + sTree[r];
}
}
public void updateRange(int i, int j, int val) {
updateRange(1, 1, N, i, j, val);
}
public void updateRange(int Node, int left, int right, int i, int j, int val) {
if (i > right || left > j)
return;
if (right <= j && i <= left) {
sTree[Node] += val * (right - left + 1);
lazy[Node] += val;
} else {
int l = 2 * Node;
int r = 2 * Node + 1;
int mid = (left + right) / 2;
propagate(Node, left, right);
updateRange(l, left, mid, i, j, val);
updateRange(r, mid + 1, right, mid, j, val);
sTree[Node] = sTree[l] + sTree[r];
}
}
public void propagate(int node, int left, int right) {
int l = 2 * node;
int r = 2 * node + 1;
int mid = (left + right) / 2;
lazy[l] = lazy[node];
lazy[r] = lazy[node];
sTree[l] += lazy[node] * (mid - left + 1);
sTree[r] += lazy[node] * (right - mid);
lazy[node] = 0;
}
}
public static void sort(int[] in) {
shuffle(in);
Arrays.sort(in);
}
public static void shuffle(int[] in) {
for (int i = 0; i < in.length; i++) {
int idx = (int) (Math.random() * in.length);
int tmp = in[i];
in[i] = in[idx];
in[idx] = tmp;
}
}
static class Pair implements Comparable<Pair> {
int x;
int y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o) {
return x - o.x;
}
public String toString() {
return x + " " + y;
}
}
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
public static void display(char[][] a) {
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++) {
System.out.print(a[i][j]);
}
System.out.println();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | bb41a22096757b27e12a341ebfa23c49 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String arggs[]) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
StringBuilder sb = new StringBuilder();
while(t-- > 0) {
int n = sc.nextInt(), k = sc.nextInt(), cnt = 0, arr[] = new int[n];
String s = sc.next();
for(int i=0; i<n-1; i++) {
if(s.charAt(i)-'0'== k%2) {
if(cnt < k) {
cnt++;
arr[i]++;
sb.append(1);
} else sb.append(0);
}else sb.append(1);
}
arr[n-1] += k-cnt;
sb.append((s.charAt(n-1)-'0' == cnt%2 ? 0 : 1) + "\n");
for(int a : arr)
sb.append(a + " ");
sb.append("\n");
}
System.out.println(sb);
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | e9ba14e19d71cf7bdfd2a86eb47e4f16 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | // "static void main" must be defined in a public class.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t=s.nextInt();
for(int i=0;i<t;i++)
{
int n = s.nextInt();
int k = s.nextInt();
String str = s.next();
int value = k;
int[] arr = new int[str.length()];
char find = '0';
int idx = -1;
if(k%2!=0)
{
for(int j=0;j<str.length();j++)
{
if(str.charAt(j)=='1')
{
idx = j;
break;
}
}
if(idx==-1)
idx = str.length()-1;
arr[idx]=1;
k--;
find = '1';
}
// System.out.println("????????"+" "+idx+" "+find+" "+k);
for(int j=idx+1;j<str.length();j++)
{
if(k==0)
break;
if(str.charAt(j)==find)
{
arr[j]+=1;
k--;
}
}
arr[arr.length-1]+=(k);
// System.out.println("----------->");
// for(int j=0;j<arr.length;j++)
// System.out.println(arr[j]+" (***)");
StringBuilder sb = new StringBuilder("");
for(int j=0;j<arr.length;j++)
{
int val = value - arr[j];
if(val%2==0)
sb.append(str.charAt(j));
else
{
char c = '0';
if(str.charAt(j)=='0')
c='1';
sb.append(c);
}
}
System.out.println(sb);
for(int j=0;j<arr.length;j++)
System.out.print(arr[j]+" ");
System.out.println();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 7f32783a5a0604e52b50240b22cc7e6b | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static FastReader sc=new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=1000000007;
// static long mod=998244353;
static int MAX=Integer.MAX_VALUE;
static int MIN=Integer.MIN_VALUE;
static long MAXL=Long.MAX_VALUE;
static long MINL=Long.MIN_VALUE;
static ArrayList<Integer> graph[];
// static ArrayList<ArrayList<Integer>> graph;
static long fact[];
static int seg[];
static int dp[][];
// static long dp[][];
public static void main (String[] args) throws java.lang.Exception
{
// code goes here
int t=I();
outer:while(t-->0)
{
int n=I(),k=I();
char c[]=S().toCharArray();
int o=0,z=0;
for(int i=0;i<c.length;i++){
if(c[i]=='0')z++;
else o++;
}
int ans[]=new int[c.length];
if(k%2==0){
if(k<=z){
for(int i=0;i<c.length;i++){
if(c[i]=='0' && k-->0){
c[i]='1';
ans[i]=1;
}
}
}else{
if(z%2==0){
for(int i=0;i<c.length;i++){
if(c[i]=='0'){
c[i]='1';
ans[i]=1;
}
}
ans[0]+=k-z;
}else{
for(int i=0;i<c.length;i++){
if(c[i]=='0'){
c[i]='1';
ans[i]=1;
}
}
c[c.length-1]='0';
// ans[0]+=k-z-1;
ans[c.length-1]+=k-z;
}
}
}else{
if(k<=o){
for(int i=0;i<c.length;i++){
if(c[i]=='1' && k-->0){
c[i]='1';
ans[i]=1;
}else if(c[i]=='0'){
c[i]='1';
}else{
c[i]='0';
}
}
}else{
if(o%2==0){
for(int i=0;i<c.length;i++){
if(c[i]=='1'){
ans[i]=1;
}
c[i]='1';
}
c[c.length-1]='0';
ans[c.length-1]+=k-o;
}else{
for(int i=0;i<c.length;i++){
if(c[i]=='1'){
ans[i]=1;
}
c[i]='1';
}
ans[c.length-1]+=k-o;
}
}
}
for(int i=0;i<c.length;i++){
out.print(c[i]);
}out.println();
for(int i=0;i<c.length;i++){
out.print(ans[i]+" ");
}out.println();
}
out.close();
}
public static class pair
{
long a;
long b;
public pair(long aa,long bb)
{
a=aa;
b=bb;
}
}
public static class myComp implements Comparator<pair>
{
//sort in ascending order.
public int compare(pair p1,pair p2)
{
if(p1.a==p2.a)
return 0;
else if(p1.a<p2.a)
return -1;
else
return 1;
}
// sort in descending order.
// public int compare(pair p1,pair p2)
// {
// if(p1.a==p2.a)
// return 0;
// else if(p1.a<p2.a)
// return 1;
// else
// return -1;
// }
}
public static void DFS(int s,boolean visited[])
{
visited[s]=true;
for(int i:graph[s]){
if(!visited[i]){
DFS(i,visited);
}
}
}
public static void setGraph(int n,int m)
{
graph=new ArrayList[n+1];
for(int i=0;i<=n;i++){
graph[i]=new ArrayList<>();
}
for(int i=0;i<m;i++){
int u=I(),v=I();
graph[u].add(v);
graph[v].add(u);
}
}
//LOWER_BOUND and UPPER_BOUND functions
//It returns answer according to zero based indexing.
public static int lower_bound(long arr[],long X,int start, int end) //start=0,end=n-1
{
if(start>end)return -1;
if(arr[arr.length-1]<X)return end;
if(arr[0]>X)return -1;
int left=start,right=end;
while(left<right){
int mid=(left+right)/2;
// if(arr[mid]==X){ //Returns last index of lower bound value.
// if(mid<end && arr[mid+1]==X){
// left=mid+1;
// }else{
// return mid;
// }
// }
if(arr[mid]==X){ //Returns first index of lower bound value.
if(mid>start && arr[mid-1]==X){
right=mid-1;
}else{
return mid;
}
}
else if(arr[mid]>X){
if(mid>start && arr[mid-1]<X){
return mid-1;
}else{
right=mid-1;
}
}else{
if(mid<end && arr[mid+1]>X){
return mid;
}else{
left=mid+1;
}
}
}
return left;
}
//It returns answer according to zero based indexing.
public static int upper_bound(long arr[],long X,int start,int end) //start=0,end=n-1
{
if(arr[0]>=X)return start;
if(arr[arr.length-1]<X)return -1;
int left=start,right=end;
while(left<right){
int mid=(left+right)/2;
if(arr[mid]==X){ //returns first index of upper bound value.
if(mid>start && arr[mid-1]==X){
right=mid-1;
}else{
return mid;
}
}
// if(arr[mid]==X){ //returns last index of upper bound value.
// if(mid<end && arr[mid+1]==X){
// left=mid+1;
// }else{
// return mid;
// }
// }
else if(arr[mid]>X){
if(mid>start && arr[mid-1]<X){
return mid;
}else{
right=mid-1;
}
}else{
if(mid<end && arr[mid+1]>X){
return mid+1;
}else{
left=mid+1;
}
}
}
return left;
}
//END
//Segment Tree Code
public static void buildTree(int a[],int si,int ss,int se)
{
if(ss==se){
// seg[si]=ss;
if(a[ss]%2==0){
seg[si]=1;
}else{
seg[si]=0;
}
return;
}
int mid=(ss+se)/2;
buildTree(a,2*si+1,ss,mid);
buildTree(a,2*si+2,mid+1,se);
// seg[si]=max(seg[2*si+1],seg[2*si+2]);
seg[si]=(seg[2*si+1]&seg[2*si+2]);
}
public static void merge(ArrayList<Integer> f,ArrayList<Integer> a,ArrayList<Integer> b)
{
int i=0,j=0;
while(i<a.size() && j<b.size()){
if(a.get(i)<=b.get(j)){
f.add(a.get(i));
i++;
}else{
f.add(b.get(j));
j++;
}
}
while(i<a.size()){
f.add(a.get(i));
i++;
}
while(j<b.size()){
f.add(b.get(j));
j++;
}
}
public static void update(int si,int ss,int se,int pos,int val)
{
if(ss==se){
seg[si]=(int)add(seg[si],val);
return;
}
int mid=(ss+se)/2;
if(pos<=mid){
update(2*si+1,ss,mid,pos,val);
}else{
update(2*si+2,mid+1,se,pos,val);
}
seg[si]=(int)add(seg[2*si+1],seg[2*si+2]);
}
public static int query(int a[],int si,int ss,int se,int qs,int qe)
{
if(qs>se || qe<ss)return -1;
if(ss>=qs && se<=qe)return seg[si];
int mid=(ss+se)/2;
// return max(query(2*si+1,ss,mid,qs,qe),query(2*si+2,mid+1,se,qs,qe));
int p1=query(a,2*si+1,ss,mid,qs,qe);
int p2=query(a,2*si+2,mid+1,se,qs,qe);
if(p1==-1 && p2==-1)return -1;
else if(p1==-1)return p2;
else if(p2==-1)return p1;
else return (p1&p2);
}
//Segment Tree Code end
//Prefix Function of KMP Algorithm
public static int[] KMP(char c[],int n)
{
int pi[]=new int[n];
for(int i=1;i<n;i++){
int j=pi[i-1];
while(j>0 && c[i]!=c[j]){
j=pi[j-1];
}
if(c[i]==c[j])j++;
pi[i]=j;
}
return pi;
}
public static long kadane(long a[],int n) //largest sum subarray
{
long max_sum=Long.MIN_VALUE,max_end=0;
for(int i=0;i<n;i++){
max_end+=a[i];
if(max_sum<max_end){max_sum=max_end;}
if(max_end<0){max_end=0;}
}
return max_sum;
}
public static long nPr(int n,int r)
{
long ans=divide(fact(n),fact(n-r),mod);
return ans;
}
public static long nCr(int n,int r)
{
long ans=divide(fact[n],mul(fact[n-r],fact[r]),mod);
return ans;
}
public static boolean isSorted(int a[])
{
int n=a.length;
for(int i=0;i<n-1;i++){
if(a[i]>a[i+1])return false;
}
return true;
}
public static boolean isSorted(long a[])
{
int n=a.length;
for(int i=0;i<n-1;i++){
if(a[i]>a[i+1])return false;
}
return true;
}
public static int computeXOR(int n) //compute XOR of all numbers between 1 to n.
{
if (n % 4 == 0)
return n;
if (n % 4 == 1)
return 1;
if (n % 4 == 2)
return n + 1;
return 0;
}
public static int np2(int x)
{
x--;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x++;
return x;
}
public static int hp2(int x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
public static long hp2(long x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
public static ArrayList<Integer> primeSieve(int n)
{
ArrayList<Integer> arr=new ArrayList<>();
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
// Fenwick / BinaryIndexed Tree USE IT - FenwickTree ft1=new FenwickTree(n);
public static class FenwickTree
{
int farr[];
int n;
public FenwickTree(int c)
{
n=c+1;
farr=new int[n];
}
// public void update_range(int l,int r,long p)
// {
// update(l,p);
// update(r+1,(-1)*p);
// }
public void update(int x,int p)
{
for(;x<n;x+=x&(-x))
{
farr[x]+=p;
}
}
public int get(int x)
{
int ans=0;
for(;x>0;x-=x&(-x))
{
ans=ans+farr[x];
}
return ans;
}
}
//Disjoint Set Union
public static class DSU
{
int par[],rank[];
public DSU(int c)
{
par=new int[c+1];
rank=new int[c+1];
for(int i=0;i<=c;i++)
{
par[i]=i;
rank[i]=0;
}
}
public int find(int a)
{
if(a==par[a])
return a;
return par[a]=find(par[a]);
}
public void union(int a,int b)
{
int a_rep=find(a),b_rep=find(b);
if(a_rep==b_rep)
return;
if(rank[a_rep]<rank[b_rep])
par[a_rep]=b_rep;
else if(rank[a_rep]>rank[b_rep])
par[b_rep]=a_rep;
else
{
par[b_rep]=a_rep;
rank[a_rep]++;
}
}
}
public static HashMap<Integer,Integer> primeFact(int a)
{
// HashSet<Long> arr=new HashSet<>();
HashMap<Integer,Integer> hm=new HashMap<>();
int p=0;
while(a%2==0){
// arr.add(2L);
p++;
a=a/2;
}
hm.put(2,hm.getOrDefault(2,0)+p);
for(int i=3;i*i<=a;i+=2){
p=0;
while(a%i==0){
// arr.add(i);
p++;
a=a/i;
}
hm.put(i,hm.getOrDefault(i,0)+p);
}
if(a>2){
// arr.add(a);
hm.put(a,hm.getOrDefault(a,0)+1);
}
// return arr;
return hm;
}
public static boolean isInteger(double N)
{
int X = (int)N;
double temp2 = N - X;
if (temp2 > 0)
{
return false;
}
return true;
}
public static boolean isPalindrome(String s)
{
int n=s.length();
for(int i=0;i<=n/2;i++){
if(s.charAt(i)!=s.charAt(n-i-1)){
return false;
}
}
return true;
}
public static int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long gcd(long a,long b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static long fact(long n)
{
long fact=1;
for(long i=2;i<=n;i++){
fact=((fact%mod)*(i%mod))%mod;
}
return fact;
}
public static long fact(int n)
{
long fact=1;
for(int i=2;i<=n;i++){
fact=((fact%mod)*(i%mod))%mod;
}
return fact;
}
public static boolean isPrime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static boolean isPrime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static void printArray(long a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(int a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(char a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]);
}
out.println();
}
public static void printArray(String a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(boolean a[])
{
for(int i=0;i<a.length;i++){
out.print(a[i]+" ");
}
out.println();
}
public static void printArray(pair a[])
{
for(pair p:a){
out.println(p.a+"->"+p.b);
}
}
public static void printArray(int a[][])
{
for(int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.print(a[i][j]+" ");
}out.println();
}
}
public static void printArray(long a[][])
{
for(int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.print(a[i][j]+" ");
}out.println();
}
}
public static void printArray(char a[][])
{
for(int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.print(a[i][j]+" ");
}out.println();
}
}
public static void printArray(ArrayList<?> arr)
{
for(int i=0;i<arr.size();i++){
out.print(arr.get(i)+" ");
}
out.println();
}
public static void printMapI(HashMap<?,?> hm){
for(Map.Entry<?,?> e:hm.entrySet()){
out.println(e.getKey()+"->"+e.getValue());
}out.println();
}
public static void printMap(HashMap<Long,ArrayList<Integer>> hm){
for(Map.Entry<Long,ArrayList<Integer>> e:hm.entrySet()){
out.print(e.getKey()+"->");
ArrayList<Integer> arr=e.getValue();
for(int i=0;i<arr.size();i++){
out.print(arr.get(i)+" ");
}out.println();
}
}
//Modular Arithmetic
public static long add(long a,long b)
{
a+=b;
if(a>=mod)a-=mod;
return a;
}
public static long sub(long a,long b)
{
a-=b;
if(a<0)a+=mod;
return a;
}
public static long mul(long a,long b)
{
return ((a%mod)*(b%mod))%mod;
}
public static long divide(long a,long b,long m)
{
a=mul(a,modInverse(b,m));
return a;
}
public static long modInverse(long a,long m)
{
int x=0,y=0;
own p=new own(x,y);
long g=gcdExt(a,m,p);
if(g!=1){
out.println("inverse does not exists");
return -1;
}else{
long res=((p.a%m)+m)%m;
return res;
}
}
public static long gcdExt(long a,long b,own p)
{
if(b==0){
p.a=1;
p.b=0;
return a;
}
int x1=0,y1=0;
own p1=new own(x1,y1);
long gcd=gcdExt(b,a%b,p1);
p.b=p1.a - (a/b) * p1.b;
p.a=p1.b;
return gcd;
}
public static long pwr(long m,long n)
{
long res=1;
if(m==0)
return 0;
while(n>0)
{
if((n&1)!=0)
{
res=(res*m);
}
n=n>>1;
m=(m*m);
}
return res;
}
public static long modpwr(long m,long n)
{
long res=1;
m=m%mod;
if(m==0)
return 0;
while(n>0)
{
if((n&1)!=0)
{
res=(res*m)%mod;
}
n=n>>1;
m=(m*m)%mod;
}
return res;
}
public static class own
{
long a;
long b;
public own(long val,long index)
{
a=val;
b=index;
}
}
//Modular Airthmetic
public static void sort(int[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(char[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
char tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(long[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
//max & min
public static int max(int a,int b){return Math.max(a,b);}
public static int min(int a,int b){return Math.min(a,b);}
public static int max(int a,int b,int c){return Math.max(a,Math.max(b,c));}
public static int min(int a,int b,int c){return Math.min(a,Math.min(b,c));}
public static long max(long a,long b){return Math.max(a,b);}
public static long min(long a,long b){return Math.min(a,b);}
public static long max(long a,long b,long c){return Math.max(a,Math.max(b,c));}
public static long min(long a,long b,long c){return Math.min(a,Math.min(b,c));}
public static int maxinA(int a[]){int n=a.length;int mx=a[0];for(int i=1;i<n;i++){mx=max(mx,a[i]);}return mx;}
public static long maxinA(long a[]){int n=a.length;long mx=a[0];for(int i=1;i<n;i++){mx=max(mx,a[i]);}return mx;}
public static int mininA(int a[]){int n=a.length;int mn=a[0];for(int i=1;i<n;i++){mn=min(mn,a[i]);}return mn;}
public static long mininA(long a[]){int n=a.length;long mn=a[0];for(int i=1;i<n;i++){mn=min(mn,a[i]);}return mn;}
public static long suminA(int a[]){int n=a.length;long sum=0;for(int i=0;i<n;i++){sum+=a[i];}return sum;}
public static long suminA(long a[]){int n=a.length;long sum=0;for(int i=0;i<n;i++){sum+=a[i];}return sum;}
//end
public static int[] I(int n){int a[]=new int[n];for(int i=0;i<n;i++){a[i]=I();}return a;}
public static long[] IL(int n){long a[]=new long[n];for(int i=0;i<n;i++){a[i]=L();}return a;}
public static long[] prefix(int a[]){int n=a.length;long pre[]=new long[n];pre[0]=a[0];for(int i=1;i<n;i++){pre[i]=pre[i-1]+a[i];}return pre;}
public static long[] prefix(long a[]){int n=a.length;long pre[]=new long[n];pre[0]=a[0];for(int i=1;i<n;i++){pre[i]=pre[i-1]+a[i];}return pre;}
public static int I(){return sc.I();}
public static long L(){return sc.L();}
public static String S(){return sc.S();}
public static double D(){return sc.D();}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int I(){ return Integer.parseInt(next());}
long L(){ return Long.parseLong(next());}
double D(){return Double.parseDouble(next());}
String S(){
String str = "";
try {
str = br.readLine();
}
catch (IOException e){
e.printStackTrace();
}
return str;
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | d2195f28eb08751701fce67ddd1698e0 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static final int INF = 0x3f3f3f3f;
static final long LNF = 0x3f3f3f3f3f3f3f3fL;
public static void main(String[] args) throws IOException {
initReader();
int t=nextInt();
while (t--!=0){
int n=nextInt();
int k=nextInt();
String s=next();
int[]arr=new int[n];
int p=0;
if(k%2!=0){
p=1;
}
int m=k;
for(int i=0;i<s.length();i++){
if(m==0)break;
if(s.charAt(i)-'0'==p){
arr[i]=1;
m--;
}
}
if(m!=0)arr[n-1]+=m;
for(int i=0;i<n;i++){
if(s.charAt(i)=='1'&&(k-arr[i])%2!=0)pw.print("0");
else if(s.charAt(i)=='1'&&(k-arr[i])%2==0)pw.print("1");
else if(s.charAt(i)=='0'&&(k-arr[i])%2==0)pw.print("0");
else if(s.charAt(i)=='0'&&(k-arr[i])%2!=0)pw.print("1");
}
pw.println();
for(int i=0;i<n;i++){
pw.print(arr[i]+" ");
}
pw.println();
}
pw.close();
}
/***************************************************************************************/
static BufferedReader reader;
static StringTokenizer tokenizer;
static PrintWriter pw;
public static void initReader() throws IOException {
reader = new BufferedReader(new InputStreamReader(System.in));
tokenizer = new StringTokenizer("");
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
// 从文件读写
// reader = new BufferedReader(new FileReader("test.in"));
// tokenizer = new StringTokenizer("");
// pw = new PrintWriter(new BufferedWriter(new FileWriter("test1.out")));
}
public static boolean hasNext() {
try {
while (!tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(reader.readLine());
}
} catch (Exception e) {
return false;
}
return true;
}
public static String next() throws IOException {
while (!tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
public static String nextLine() {
try {
return reader.readLine();
} catch (Exception e) {
return null;
}
}
public static int nextInt() throws IOException {
return Integer.parseInt(next());
}
public static long nextLong() throws IOException {
return Long.parseLong(next());
}
public static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public static char nextChar() throws IOException {
return next().charAt(0);
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | f8e3d308f67943c26120617022aa65c8 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.lang.*;
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();
long k=sc.nextLong();
sc.nextLine();
String s=sc.nextLine();
StringBuilder str=new StringBuilder();
long arr[]=new long[n];
long count=k;
for(int i=0;i<n-1;i++){
if(s.charAt(i)=='1'){
if(count==0){
char c=s.charAt(i);
if(k%2==1){
c=(c=='1')?'0':'1';
}
str.append(c);
}
else if(k%2==1){
arr[i]=1;
str.append('1');
count--;
}
else{
str.append('1');
}
}
else{
if(count==0){
char c=s.charAt(i);
if(k%2==1){
c=(c=='1')?'0':'1';
}
str.append(c);
}
else if(k%2==0){
count--;
arr[i]=1;
str.append('1');
}
else{
str.append('1');
}
}
}
char c=s.charAt(n-1);
if((k-count)%2==1){
c=(c=='1')?'0':'1';
}
str.append(c);
arr[n-1]=count;
System.out.println(str.toString());
for(int i=0;i<n;i++){
System.out.print(arr[i]+" ");
}
System.out.println();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 0e57719a1bfa547433b43d3d5000a5fa | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class BitFlipping{
private static BufferedReader br;
private static BufferedWriter bw;
private static String[] buffer;
private static int index;
private static void before() throws Exception {
try{
new BufferedReader(new FileReader("local.txt"));
local();
}catch(Exception e){
online();
}
buffer = new String[0];
index = 0;
}
private static void after() throws Exception {
br.close();
bw.close();
}
private static void online() throws Exception {
br = new BufferedReader(new InputStreamReader(System.in));
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
private static void local() throws Exception {
br = new BufferedReader(new FileReader("input.txt"));
bw = new BufferedWriter(new FileWriter("output.txt"));
}
private static String next() throws Exception {
if(buffer.length == index)
{
buffer = br.readLine().split(" ");
index = 0;
}
return buffer[index++];
}
private static int nextInt() throws Exception {
return Integer.parseInt(next());
}
private static long nextLong() throws Exception {
return Long.parseLong(next());
}
private static double nextDouble() throws Exception {
return Double.parseDouble(next());
}
private static float nextFloat() throws Exception {
return Float.parseFloat(next());
}
private static boolean nextBoolean() throws Exception {
return Boolean.parseBoolean(next());
}
private static void print(int x) throws Exception {
bw.write(x + "");
}
private static void println(int x) throws Exception {
bw.write(x + "\n");
}
private static void print(long x) throws Exception {
bw.write(x + "");
}
private static void println(long x) throws Exception {
bw.write(x + "\n");
}
private static void print(boolean x) throws Exception {
bw.write(x + "");
}
private static void println(boolean x) throws Exception {
bw.write(x + "\n");
}
private static void print(char x) throws Exception {
bw.write(x);
}
private static void println(char x) throws Exception {
bw.write(x + "\n");
}
private static void print(float x) throws Exception {
bw.write(x + "");
}
private static void println(float x) throws Exception {
bw.write(x + "\n");
}
private static void print(double x) throws Exception {
bw.write(x + "");
}
private static void println(double x) throws Exception {
bw.write(x + "\n");
}
private static void print(String x) throws Exception {
bw.write(x);
}
private static void println(String x) throws Exception {
bw.write(x + "\n");
}
private static void println() throws Exception {
bw.write("\n");
}
public static void main(String[] args) throws Exception {
before();
solve();
after();
}
private static void solve() throws Exception {
int t = nextInt();
while(t-- > 0)
{
int n = nextInt();
int k = nextInt();
String s = next();
char[] ch = s.toCharArray();
int[] arr = new int[n];
if(k % 2 == 0)
{
for(int i = 0; i < n && k > 0; i++)
{
if(ch[i] == '0')
{
ch[i] = '1';
arr[i]++;
k--;
}
}
arr[n - 1] += k;
if(k > 0 && k % 2 == 1)
ch[n-1] = '0';
}
else
{
for(int i = 0; i < n; i++)
{
if(ch[i] == '0')
ch[i] = '1';
else
{
if(k > 0)
{
arr[i]++;
k--;
}
else
ch[i] = '0';
}
}
arr[n - 1] += k;
if(k > 0 && k % 2 == 1)
ch[n - 1] = '0';
}
println(new String(ch));
for(int x : arr)
print(x + " ");
println();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 42cd5e9cb39005e8f7563bd1f610b22d | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes |
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
public class Practice2 {
//
// static class Pair{
// int val;
// int data;
// Pair(int val,int data){
// this.val=val;
// this.data=data;
// }
// }
//
//
public static void printarr(int[] arr) {
int n=arr.length;
for(int i=0;i<n;i++) {
System.out.print(arr[i]+" ");
}
System.out.println();
}
// /** Code for Dijkstra's algorithm **/
public static class ListNode {
int vertex, weight;
ListNode(int v,int w) {
vertex = v;
weight = w;
}
int getVertex() { return vertex; }
int getWeight() { return weight; }
}
public static int[] dijkstra(
int V, HashMap<Integer,ArrayList<ListNode> > graph,
int source) {
int[] distance = new int[V];
for (int i = 0; i < V; i++)
distance[i] = Integer.MAX_VALUE;
distance[0] = 0;
PriorityQueue<ListNode> pq = new PriorityQueue<>(
(v1, v2) -> v1.getWeight() - v2.getWeight());
pq.add(new ListNode(source, 0));
while (pq.size() > 0) {
ListNode current = pq.poll();
for (ListNode n :
graph.get(current.getVertex())) {
if (distance[current.getVertex()]
+ n.getWeight()
< distance[n.getVertex()]) {
distance[n.getVertex()]
= n.getWeight()
+ distance[current.getVertex()];
pq.add(new ListNode(
n.getVertex(),
distance[n.getVertex()]));
}
}
}
// If you want to calculate distance from source to
// a particular target, you can return
// distance[target]
return distance;
}
//Methos to return all divisor of a number
static ArrayList<Long> allDivisors(long n) {
ArrayList<Long> al=new ArrayList<>();
long i=2;
while(i*i<=n) {
if(n%i==0) al.add(i);
if(n%i==0&&i*i!=n) al.add(n/i);
i++;
}
return al;
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static int find(boolean[] vis,int x,int y,int n,int m) {
if(x<0||y<0||x>=n||y>=m) return 0;
// if(vis[i][j]==true) return
return 0;
}
static long power(long N,long R)
{
long x=998244353;
if(R==0) return 1;
if(R==1) return N;
long temp= power(N,R/2)%x; // (a*b)%p = (a%p*b%p)*p
temp=(temp*temp)%x;
if(R%2==0){
return temp%x;
}else{
return (N*temp)%x;
}
}
static int[] sort(int[] arr) {
int n=arr.length;
ArrayList<Integer> al=new ArrayList<>();
for(int i=0;i<n;i++) {
al.add(arr[i]);
}
Collections.sort(al);
for(int i=0;i<n;i++) {
arr[i]=al.get(i);
}
return arr;
}
static long[] sort(long[] arr) {
long n=arr.length;
ArrayList<Long> al=new ArrayList<>();
for(int i=0;i<n;i++) {
al.add(arr[i]);
}
Collections.sort(al);
for(int i=0;i<n;i++) {
arr[i]=al.get(i);
}
return arr;
}
static class Pair{
int val;
int res;
Pair(int val,int res){
this.val=val;
this.res=res;
}
}
public static void main (String[] args) {
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
// out.print();
//out.println();
FastReader sc=new FastReader();
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int k=sc.nextInt();
int a=k;
String str=sc.nextLine();
int[] arr=new int[n];
StringBuffer ans=new StringBuffer();
if(k%2==0) {
for(int i=0;i<n;i++) {
if(i==n-1) {
arr[i]=k;
a -=k;
if(a%2==0) {
if(str.charAt(i)=='0') {
ans.append('0');
}else {
ans.append('1');
}
}else {
if(str.charAt(i)=='0') {
ans.append('1');
}else {
ans.append('0');
}
}
k=0;
continue;
}
if(str.charAt(i)=='0'&&k>0) {
ans.append('1');
arr[i]++;
k--;
}else {
ans.append(str.charAt(i));
}
}
}else {
for(int i=0;i<n;i++) {
if(i==n-1) {
arr[i]=k;
a -=k;
if(a%2==0) {
if(str.charAt(i)=='0') {
ans.append('0');
}else {
ans.append('1');
}
}else {
if(str.charAt(i)=='0') {
ans.append('1');
}else {
ans.append('0');
}
}
k=0;
continue;
}
if(str.charAt(i)=='1'&&k>0) {
ans.append('1');
arr[i]++;
k--;
}else {
if(str.charAt(i)=='0') {
ans.append('1');
}else {
ans.append('0');
}
}
}
}
ans.append("\n");
for(int i=0;i<n;i++) {
ans.append(arr[i]+" ");
}
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;
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 844f34dde313c49255550b16d8295c50 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
// BEFORE 31ST MARCH 2022 !!
//MAX RATING EVER ACHIEVED-1622(LETS SEE WHEN WILL I GET TO CHANGE THIS)
////***************************************************************************
/* public class E_Gardener_and_Tree implements Runnable{
public static void main(String[] args) throws Exception {
new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
}
public void run(){
WRITE YOUR CODE HERE!!!!
JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
}
}
*/
/////**************************************************************************
public class B_Bit_Flipping{
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 k=s.nextLong();
String str=s.nextToken();
if(k==0){
res.append(str+" \n");
for(int i=0;i<n;i++){
res.append("0 ");
}
res.append(" \n");
p++;
continue;
}
long array[]= new long[n];
char array2[]= new char[n];
if(k%2!=0){
//give it to 1
long count=0;
for(int i=0;i<n;i++){
if(count>=k){
break;
}
char ch=str.charAt(i);
if(ch=='1'){
array[i]=1;
count++;
}
if(count>=k){
break;
}
}
long rem=k-count;
array[n-1]+=rem;
for(int i=0;i<n;i++){
long yo=k-array[i];
char ch=str.charAt(i);
if(ch=='1'){
if(yo%2==0){
array2[i]='1';
}
else{
array2[i]='0';
}
}
else{
if(yo%2==0){
array2[i]='0';
}
else{
array2[i]='1';
}
}
}
}
else{
// give it to 0
long count=0;
for(int i=0;i<n;i++){
if(count>=k){
break;
}
char ch=str.charAt(i);
if(ch=='0'){
array[i]=1;
count++;
}
if(count>=k){
break;
}
}
long rem=k-count;
array[n-1]+=rem;
for(int i=0;i<n;i++){
long yo=k-array[i];
char ch=str.charAt(i);
if(ch=='1'){
if(yo%2==0){
array2[i]='1';
}
else{
array2[i]='0';
}
}
else{
if(yo%2==0){
array2[i]='0';
}
else{
array2[i]='1';
}
}
}
}
for(int i=0;i<n;i++){
res.append(array2[i]);
}
res.append(" \n");
for(int i=0;i<n;i++){
res.append(array[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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | be66a3397b2849417580e6e1703d1a97 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//--------------------------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 k = sc.ni();
char[] strr = sc.ns().toCharArray();
if(k == 0) {
for(int i = 0; i < n; i++) {
w.pr(strr[i]+"");
}
w.pl();
for(int i = 0; i < n; i++) {
w.pr(0+" ");
}
w.pl();
return;
}
if(k%2==0 && strr[0]=='0' || k%2==1 && strr[0]=='1') {
k--;
int done = 0;
int[] steps = new int[n];
steps[0] = 1;
for(int i = 1; i < n; i++) {
if(strr[i] == strr[0]) {
if(done < k) {
done++;
steps[i] = 1;
strr[i] = '1';
} else {
strr[i] = '0';
}
} else {
strr[i] = '1';
}
}
strr[0] = '1';
int rem = k-done;
steps[n-1] += rem;
if(rem%2==1) {
strr[n-1] = (strr[n-1]=='1')?'0':'1';
}
for(int i = 0; i < n; i++) {
w.pr(strr[i]+"");
}
w.pl();
for(int i = 0; i < n; i++) {
w.pr(steps[i]+" ");
}
w.pl();
} else {
int done = 0;
int[] steps = new int[n];
for(int i = 1; i < n; i++) {
if(strr[i] != strr[0]) {
if(done < k) {
done++;
steps[i] = 1;
strr[i] = '1';
} else {
strr[i] = '0';
}
} else {
strr[i] = '1';
}
}
strr[0] = '1';
int rem = k-done;
steps[n-1] += rem;
if(rem%2==1) {
strr[n-1] = (strr[n-1]=='1')?'0':'1';
}
for(int i = 0; i < n; i++) {
w.pr(strr[i]+"");
}
w.pl();
for(int i = 0; i < n; i++) {
w.pr(steps[i]+" ");
}
w.pl();
}
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | cb973a4e1071d70ebe59a1bf9eeb302a | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
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 k=input.nextInt();
int k1=k;
String s=input.next();
int arr[]=new int[n];
if(k%2==0)
{
for(int i=0;i<n;i++)
{
if(s.charAt(i)=='0' && k>0)
{
arr[i]=1;
k--;
}
}
arr[n-1]+=k;
}
else
{
for(int i=0;i<n;i++)
{
if(s.charAt(i)=='1' && k>0)
{
arr[i]=1;
k--;
}
}
arr[n-1]+=k;
}
for(int i=0;i<n;i++)
{
int r=k1-arr[i];
if(r%2==0)
{
out.print(s.charAt(i));
}
else
{
if(s.charAt(i)=='1')
{
out.print('0');
}
else
{
out.print('1');
}
}
}
out.println();
for(int 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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 568f94f0bbaca534df2340c45838f1e3 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class B {
FastScanner in;
PrintWriter out;
boolean systemIO = true;
int mod = 1000000007;
public int mult(int x, int y) {
return (int) (x * 1L * y % mod);
}
public int sum(int x, int y) {
if (x + y >= mod) {
return x + y - mod;
}
return x + y;
}
public int diff(int x, int y) {
if (x >= y) {
return x - y;
}
return x - y + mod;
}
public int div(int x, int y) {
return mult(x, modInv(y));
}
public int[][] mult(int[][] a, int[][] b) {
int[][] c = new int[a.length][b[0].length];
for (int i = 0; i < c.length; i++) {
for (int j = 0; j < c[0].length; j++) {
for (int k = 0; k < b.length; k++) {
c[i][j] = sum(c[i][j], mult(a[i][k], b[k][j]));
}
}
}
return c;
}
public int pow(int x, long p) {
if (p == 0) {
return 1;
}
int ans = pow(x, p / 2);
ans = mult(ans, ans);
if ((p & 1) > 0) {
ans = mult(ans, x);
}
return ans;
}
public int[][] pow(int[][] x, long p) {
if (p == 0) {
int[][] ans = { { 1, 0 }, { 0, 1 } };
return ans;
}
int[][] ans = pow(x, p / 2);
ans = mult(ans, ans);
if ((p & 1) > 0) {
ans = mult(ans, x);
}
return ans;
}
public int modInv(int x) {
return pow(x, mod - 2);
}
int[] dx = { -1, 0, 1, 0 };
int[] dy = { 0, -1, 0, 1 };
int sz = 0;
int[] ans;
boolean[][] a;
boolean[][] b;
int n;
boolean[][] used = new boolean[n][n];
public boolean check(int x, int y, int dir) {
if (x + dx[dir] < 0 || x + dx[dir] >= n) {
return false;
}
if (y + dy[dir] < 0 || y + dy[dir] >= n) {
return false;
}
boolean flag = true;
if (dir == 0) {
flag = b[y][x - 1];
b[y][x - 1] = false;
}
if (dir == 2) {
flag = b[y][x];
b[y][x] = false;
}
if (dir == 1) {
flag = a[y - 1][x];
a[y - 1][x] = false;
}
if (dir == 3) {
flag = a[y][x];
a[y][x] = false;
}
return flag;
}
public void dfsRec() {
Stack<Integer> xst = new Stack<>();
Stack<Integer> yst = new Stack<>();
Stack<Integer> prevst = new Stack<>();
Stack<Integer> dirst = new Stack<>();
Stack<Boolean> toAns = new Stack<>();
xst.add(0);
yst.add(0);
prevst.add(-1);
dirst.add(-1);
toAns.add(false);
while (!xst.isEmpty()) {
int x = xst.pop();
int y = yst.pop();
int prev = prevst.pop();
int dir = dirst.pop();
// System.out.println(x + " " + y + " " + prev + " " + dir + " " + toAns.peek());
if (toAns.pop()) {
ans[sz++] = dir ^ 2;
}
used[x][y] = true;
for (dir = dir + 1; dir < 4; dir++) {
if (dir == (prev ^ 2)) {
continue;
}
if (check(x, y, dir)) {
ans[sz++] = dir;
xst.add(x);
yst.add(y);
prevst.add(prev);
dirst.add(dir);
toAns.add(true);
if (!used[x + dx[dir]][y + dy[dir]]) {
xst.add(x + dx[dir]);
yst.add(y + dy[dir]);
prevst.add(dir);
dirst.add(-1);
toAns.add(false);
}
break;
}
}
}
}
public void dfs(int x, int y, int prev) {
System.out.println(x + " " + y);
if (used[x][y]) {
return;
}
used[x][y] = true;
for (int dir = 0; dir < 4; dir++) {
if (dir == (prev ^ 2)) {
continue;
}
if (check(x, y, dir)) {
ans[sz++] = dir;
dfs(x + dx[dir], y + dy[dir], dir);
ans[sz++] = dir ^ 2;
}
}
}
public void solve() {
for (int qwerty = in.nextInt(); qwerty > 0; --qwerty) {
int n = in.nextInt();
int k = in.nextInt();
boolean[] b = new boolean[n];
int[] ans = new int[n];
int left = k;
String s = in.next();
for (int i = 0; i < s.length(); i++) {
b[i] = s.charAt(i) == '1' ^ (k % 2 == 1);
if (!b[i] && left > 0) {
left--;
++ans[i];
b[i] = !b[i];
}
}
if (left > 0) {
ans[n - 1] += left;
if (left % 2 == 1) {
b[n - 1] ^= true;
}
}
for (int i = 0; i < b.length; i++) {
if (b[i]) {
out.print(1);
} else {
out.print(0);
}
}
out.println();
for (int i = 0; i < ans.length; i++) {
out.print(ans[i] + " ");
}
out.println();
}
}
public void add(HashMap<Integer, Integer> map, int x) {
if (map.containsKey(x)) {
map.put(x, map.get(x) + 1);
} else {
map.put(x, 1);
}
}
public void run() {
try {
if (systemIO) {
in = new FastScanner(System.in);
out = new PrintWriter(System.out);
} else {
in = new FastScanner(new File("input.txt"));
out = new PrintWriter(new File("output.txt"));
}
solve();
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String nextLine() {
try {
return br.readLine();
} catch (IOException e) {
return null;
}
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
// AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
public static void main(String[] arg) {
new B().run();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 899896f5c27ff4f26aa3d860fa8b7b5b | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.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(), k = sc.nextInt();
char[] s = sc.next().toCharArray();
int x = (int)(2e5);
long[] no = new long[x], color = new long[x];
for(int i = 0;i<n;i++) no[i] = (s[i]-'0');
int currentBit = k & 1;
for(int i = 0;i<n;i++)
{
if(no[i] == currentBit && k > 0)
{
no[i] ^= 1;
k--;
color[i] = 1;
}
else
{
color[i] = 0;
}
}
if((k & 1) != 0) no[n-1] ^= 1;
color[n-1] += k;
for(int i = 0;i<n;i++) fout.print((no[i] ^ currentBit));
fout.println();
for(int i = 0;i<n;i++) fout.print(color[i] + " ");
fout.println();
}
fout.close();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 8313dcf667b15d316bb0975c55f15389 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.lang.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(), k = ri();
String s = line();
boolean same = k % 2 == 0;
int change[] = new int[n];
for (int i = 0; i < n; i++) {
int t = s.charAt(i) - '0';
if (same) {
change[i] = t;
} else {
change[i] = 1-t;
}
}
int ans[] = new int[n];
// ol(Arrays.toString(change));
int ii = 0;
while (ii < n && k > 0) {
if (change[ii] == 0) {
ans[ii] += 1;
change[ii] = 1;
k -= 1;
}
ii += 1;
}
// ol(Arrays.toString(change));
// ol(Arrays.toString(ans));
// ol(k);
if (k > 0) {
if (k % 2 == 0) {
ans[n-1] += k;
} else {
ans[n-1] += k;
change[n-1] = 0;
}
}
for (int i : change) {
o(i);
}
ol();
for (int an : ans) {
os(an);
}
ol();
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 62372858bf34c65fa86a3833f2231c66 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main {
static long startTime = System.currentTimeMillis();
// for global initializations and methods starts here
// global initialisations and methods end here
static void run() {
boolean tc = true;
//AdityaFastIO r = new AdityaFastIO();
FastReader r = new FastReader();
try (OutputStream out = new BufferedOutputStream(System.out)) {
//long startTime = System.currentTimeMillis();
int testcases = tc ? r.ni() : 1;
int tcCounter = 1;
// Hold Here Sparky------------------->>>
// Solution Starts Here
start:
while (testcases-- > 0) {
int n = r.ni();
int k = r.ni();
char[] s = r.word().toCharArray();
int[] arr = new int[n + 1];
int[] dp = new int[n + 1];
for (int i = 1; i < n + 1; i++) {
arr[i] = s[i - 1] - '0';
}
int check = k & 1;
int shift = 1;
{
int i = 1;
while (i <= n) {
if (arr[i] == check && k > 0) {
arr[i] ^= shift;
k--;
dp[i] = shift;
}
i++;
}
}
if ((k & 1) == shift) arr[n] ^= shift;
dp[n] += k;
for (int i = 1; i < n + 1; i++) {
int ele = (int) (arr[i] ^ check);
out.write((ele + "").getBytes());
}
out.write(("\n").getBytes());
for (int i = 1; i < n + 1; i++) {
int ele = dp[i];
out.write((ele + " ").getBytes());
}
out.write(("\n").getBytes());
}
// Solution Ends Here
} catch (IOException e) {
e.printStackTrace();
}
}
static class AdityaFastIO {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public BufferedReader br;
public StringTokenizer st;
public AdityaFastIO() {
br = new BufferedReader(new InputStreamReader(System.in));
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public AdityaFastIO(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
public String readLine() throws IOException {
byte[] buf = new byte[100000001]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int ni() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg) return -ret;
return ret;
}
public long nl() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg) return -ret;
return ret;
}
public double nd() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') c = read();
boolean neg = (c == '-');
if (neg) c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') while ((c = read()) >= '0' && c <= '9') ret += (c - '0') / (div *= 10);
if (neg) return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null) return;
din.close();
}
}
public static void main(String[] args) throws Exception {
run();
}
static int[] readIntArr(int n, AdityaFastIO r) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = r.ni();
return arr;
}
static long[] readLongArr(int n, AdityaFastIO r) throws IOException {
long[] arr = new long[n];
for (int i = 0; i < n; i++) arr[i] = r.nl();
return arr;
}
static List<Integer> readIntList(int n, AdityaFastIO r) throws IOException {
List<Integer> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.ni());
return al;
}
static List<Long> readLongList(int n, AdityaFastIO r) throws IOException {
List<Long> al = new ArrayList<>();
for (int i = 0; i < n; i++) al.add(r.nl());
return al;
}
static long mod = 998244353;
static long modInv(long base, long e) {
long result = 1;
base %= mod;
while (e > 0) {
if ((e & 1) > 0) result = result * base % mod;
base = base * base % mod;
e >>= 1;
}
return result;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String word() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String line() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int ni() {
return Integer.parseInt(word());
}
long nl() {
return Long.parseLong(word());
}
double nd() {
return Double.parseDouble(word());
}
}
static int MOD = (int) (1e9 + 7);
static long powerLL(long x, long n) {
long result = 1;
while (n > 0) {
if (n % 2 == 1) result = result * x % MOD;
n = n / 2;
x = x * x % MOD;
}
return result;
}
static long powerStrings(int i1, int i2) {
String sa = String.valueOf(i1);
String sb = String.valueOf(i2);
long a = 0, b = 0;
for (int i = 0; i < sa.length(); i++) a = (a * 10 + (sa.charAt(i) - '0')) % MOD;
for (int i = 0; i < sb.length(); i++) b = (b * 10 + (sb.charAt(i) - '0')) % (MOD - 1);
return powerLL(a, b);
}
static long gcd(long a, long b) {
if (a == 0) return b;
else return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
static long lower_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] >= x) r = m;
else l = m;
}
return r;
}
static int upper_bound(int[] arr, int x) {
int l = -1, r = arr.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (arr[m] <= x) l = m;
else r = m;
}
return l + 1;
}
static void addEdge(ArrayList<ArrayList<Integer>> graph, int edge1, int edge2) {
graph.get(edge1).add(edge2);
graph.get(edge2).add(edge1);
}
public static class Pair implements Comparable<Pair> {
int first;
int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(Pair o) {
// TODO Auto-generated method stub
if (this.first != o.first)
return (int) (this.first - o.first);
else return (int) (this.second - o.second);
}
}
public static class PairC<X, Y> implements Comparable<PairC> {
X first;
Y second;
public PairC(X first, Y second) {
this.first = first;
this.second = second;
}
public String toString() {
return "(" + first + "," + second + ")";
}
public int compareTo(PairC o) {
// TODO Auto-generated method stub
return o.compareTo((PairC) first);
}
}
static boolean isCollectionsSorted(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) <= list.get(i - 1)) return false;
return true;
}
static boolean isCollectionsSortedReverseOrder(List<Long> list) {
if (list.size() == 0 || list.size() == 1) return true;
for (int i = 1; i < list.size(); i++) if (list.get(i) >= list.get(i - 1)) return false;
return true;
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 46df3ebbf57df75abd5e1933528e710c | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.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 gl = sc.nextInt();
int k = sc.nextInt();
int temp=k;
String s = sc.next();
StringBuilder sb= new StringBuilder();
int ans []= new int [gl];
for (int i=0; i<gl; i++){
int n = gl-i-1;
if (s.charAt(i)=='1'){
if (k%2==0){
if (temp>n){
ans[i]=((temp-n)/2)*2;
// System.out.println(ans[i]);
}
sb.append('1');
}
else {
if (temp>0){
sb.append('1');
if (temp>n){
int y=temp-n;
if (y%2==0){
ans[i]=y-1;
}
else ans[i]=y;
}
else ans[i]=1;
}
else sb.append('0');
}
}
else {
if (k%2==1){
if (temp>n){
ans[i]=((temp-n)/2)*2;
// System.out.println(ans[i]);
}
sb.append('1');
}
else {
if (temp>0){
sb.append('1');
if (temp>n){
int y=temp-n;
if (y%2==0){
ans[i]=y-1;
}
else ans[i]=y;
}
else ans[i]=1;
}
else sb.append('0');
}
}
temp-=ans[i];
if (i==gl-1 && temp>0){
ans[i]++;
sb.deleteCharAt(i);
sb.append('0');
}
}
StringBuilder x= new StringBuilder();
for (int i=0; i<gl; i++){
x.append(ans[i]+" ");
}
System.out.println(sb.toString());
System.out.println(x.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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 4186d48e3de68219e6dbece060d5061a | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
try{
FastReader read=new FastReader();
StringBuilder sb = new StringBuilder();
int t=read.nextInt();
while(t>0)
{
int n = read.nextInt();
int k = read.nextInt();
String s = read.nextLine();
int p = 0;
int ans[] = new int[n];
for(int i=0;i<n-1;i++)
{
if(p%2==0)
{
if(s.charAt(i)=='1')
{
if(k<1)
{
sb.append('1');
ans[i]=0;
}
else if(k%2!=0)
{
sb.append('1');
ans[i]=1;
p++;
k--;
}
else
{
sb.append('1');
ans[i]=0;
}
}
else
{
if(k<1)
{
sb.append('0');
ans[i]=0;
}
else if(k%2==0)
{
sb.append('1');
ans[i]=1;
p++;
k--;
}
else
{
sb.append('1');
ans[i]=0;
}
}
}
else
{
if(s.charAt(i)=='0')
{
if(k<1)
{
sb.append('1');
ans[i]=0;
}
else if(k%2!=0)
{
sb.append('1');
ans[i]=1;
p++;
k--;
}
else
{
sb.append('1');
ans[i]=0;
}
}
else
{
if(k<1)
{
sb.append('0');
ans[i]=0;
}
else if(k%2==0)
{
sb.append('1');
ans[i]=1;
p++;
k--;
}
else
{
sb.append('1');
ans[i]=0;
}
}
}
}
if(p%2==0)
{
if(s.charAt(n-1)=='1')
{
if(k<1)
{
sb.append('1');
ans[n-1]=0;
}
else
{
sb.append('1');
ans[n-1]=k;
p++;
k--;
}
}
else
{
if(k<1)
{
sb.append('0');
}
else
{
sb.append('0');
ans[n-1]=k;
p++;
k--;
}
}
}
else
{
if(s.charAt(n-1)=='0')
{
if(k<1)
{
sb.append('1');
ans[n-1]=0;
}
else
{
sb.append('1');
ans[n-1]=k;
p++;
k--;
}
}
else
{
if(k<1)
{
sb.append('0');
ans[n-1]=0;
}
else
{
sb.append('0');
ans[n-1]=k;
p++;
k--;
}
}
}
sb.append('\n');
for(int i=0;i<n;i++)
{
sb.append(ans[i]+" ");
}
sb.append('\n');
t--;
}
System.out.println(sb);
}
catch(Exception e)
{return;
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static void sort(int[] A) {
int n = A.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
int tmp = A[i];
int randomPos = i + rnd.nextInt(n - i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static void sort(long[] A) {
int n = A.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
long tmp = A[i];
int randomPos = i + rnd.nextInt(n - i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static void sort(ArrayList<Long> A,char ch) {
int n = A.size();
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
long tmp = A.get(i);
int randomPos = i + rnd.nextInt(n - i);
A.set(i,A.get(randomPos));
A.set(randomPos,tmp);
}
Collections.sort(A);
}
static void sort(ArrayList<Integer> A) {
int n = A.size();
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
int tmp = A.get(i);
int randomPos = i + rnd.nextInt(n - i);
A.set(i,A.get(randomPos));
A.set(randomPos,tmp);
}
Collections.sort(A);
}
static String sort(String s) {
Character ch[] = new Character[s.length()];
for (int i = 0; i < s.length(); i++) {
ch[i] = s.charAt(i);
}
Arrays.sort(ch);
StringBuffer st = new StringBuffer("");
for (int i = 0; i < s.length(); i++) {
st.append(ch[i]);
}
return st.toString();
}
}
class pair implements Comparable<pair>
{
int X,Y,C;
pair(int s,int e,int c)
{
X=s;
Y=e;
C=c;
}
public int compareTo(pair p )
{
return this.C-p.C;
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | cd2b6f04ed7ca0385c8b1429c4f05923 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import java.math.*;
/**
* @author Naitik
*
*/
public class Main
{
static FastReader sc=new FastReader();
static long dp[][];
// static boolean v[][][];
static int mod=998244353;;
// static int mod=1000000007;
static int max;
static int bit[];
//static long fact[];
static HashMap<Long,Integer> map;
//static StringBuffer sb=new StringBuffer("");
//static HashMap<Integer,Integer> map;
static PrintWriter out=new PrintWriter(System.out);
public static void main(String[] args)
{
// StringBuffer sb=new StringBuffer("");
int ttt=1;
ttt =i();
outer :while (ttt-- > 0)
{
int n=i();
int k=i();
char A[]=s().toCharArray();
ArrayList<Integer> l=new ArrayList<Integer>();
int f=0;
int j=0;
for(int i=0;i<n && k>0;i++) {
if(f==1) {
if(A[i]=='1')
A[i]='0';
else
A[i]='1';
}
if(i==n-1) {
j++;
continue;
}
if(A[i]=='1') {
if(k%2!=0) {
l.add(1);
k--;
f^=1;
}
else {
l.add(0);
}
}
else {
if(k%2==0) {
l.add(1);
k--;
f^=1;
A[i]='1';
}
else {
l.add(0);
A[i]='1';
}
}
j++;
}
if(f==1) {
for(int i=j;i<n;i++) {
if(A[i]=='1')
A[i]='0';
else
A[i]='1';
}
}
while(l.size()<n) {
l.add(0);
}
int y=l.get(l.size()-1);
y+=k;
l.remove(l.size()-1);
l.add(y);
for(char ch : A) {
out.print(ch);
}
out.println();
for(int i : l) {
out.print(i+" ");
}
out.println();
}
//System.out.println(sb.toString());
out.close();
//CHECK FOR N=1 //CHECK FOR M=0
//CHECK FOR N=1 //CHECK FOR M=0
//CHECK FOR N=1
}
static class Pair implements Comparable<Pair>
{
int x;
int y;
// int z;
Pair(int x,int y){
this.x=x;
this.y=y;
// this.z=z;
}
@Override
public int compareTo(Pair o) {
if(this.x>o.x)
return 1;
else if(this.x<o.x)
return -1;
else {
if(this.y>o.y)
return 1;
else if(this.y<o.y)
return -1;
else
return 0;
}
}
public int hashCode()
{
final int temp = 14;
int ans = 1;
ans =x*31+y*13;
return ans;
}
@Override
public boolean equals(Object o)
{
if (this == o) {
return true;
}
if (o == null) {
return false;
}
if (this.getClass() != o.getClass()) {
return false;
}
Pair other = (Pair)o;
if (this.x != other.x || this.y!=other.y) {
return false;
}
return true;
}
//
/* FOR TREE MAP PAIR USE */
// public int compareTo(Pair o) {
// if (x > o.x) {
// return 1;
// }
// if (x < o.x) {
// return -1;
// }
// if (y > o.y) {
// return 1;
// }
// if (y < o.y) {
// return -1;
// }
// return 0;
// }
}
static int find(int A[],int a) {
if(A[a]==a)
return a;
return A[a]=find(A, A[a]);
}
//static int find(int A[],int a) {
// if(A[a]==a)
// return a;
// return find(A, A[a]);
//}
//FENWICK TREE
static void update(int i, int x){
for(; i < bit.length; i += (i&-i))
bit[i] += x;
}
static int sum(int i){
int ans = 0;
for(; i > 0; i -= (i&-i))
ans += bit[i];
return ans;
}
//END
static void add(long v) {
if(!map.containsKey(v)) {
map.put(v, 1);
}
else {
map.put(v, map.get(v)+1);
}
}
static void remove(long v) {
if(map.containsKey(v)) {
map.put(v, map.get(v)-1);
if(map.get(v)==0)
map.remove(v);
}
}
public static int upper(int A[],int k,int si,int ei)
{
int l=si;
int u=ei;
int ans=-1;
while(l<=u) {
int mid=(l+u)/2;
if(A[mid]<=k) {
ans=mid;
l=mid+1;
}
else {
u=mid-1;
}
}
return ans;
}
public static int lower(int A[],int k,int si,int ei)
{
int l=si;
int u=ei;
int ans=-1;
while(l<=u) {
int mid=(l+u)/2;
if(A[mid]<=k) {
l=mid+1;
}
else {
ans=mid;
u=mid-1;
}
}
return ans;
}
static int[] copy(int A[]) {
int B[]=new int[A.length];
for(int i=0;i<A.length;i++) {
B[i]=A[i];
}
return B;
}
static long[] copy(long A[]) {
long B[]=new long[A.length];
for(int i=0;i<A.length;i++) {
B[i]=A[i];
}
return B;
}
static int[] input(int n) {
int A[]=new int[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextInt();
}
return A;
}
static long[] inputL(int n) {
long A[]=new long[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextLong();
}
return A;
}
static String[] inputS(int n) {
String A[]=new String[n];
for(int i=0;i<n;i++) {
A[i]=sc.next();
}
return A;
}
static long sum(int A[]) {
long sum=0;
for(int i : A) {
sum+=i;
}
return sum;
}
static long sum(long A[]) {
long sum=0;
for(long i : A) {
sum+=i;
}
return sum;
}
static void reverse(long A[]) {
int n=A.length;
long B[]=new long[n];
for(int i=0;i<n;i++) {
B[i]=A[n-i-1];
}
for(int i=0;i<n;i++)
A[i]=B[i];
}
static void reverse(int A[]) {
int n=A.length;
int B[]=new int[n];
for(int i=0;i<n;i++) {
B[i]=A[n-i-1];
}
for(int i=0;i<n;i++)
A[i]=B[i];
}
static void input(int A[],int B[]) {
for(int i=0;i<A.length;i++) {
A[i]=sc.nextInt();
B[i]=sc.nextInt();
}
}
static int[][] input(int n,int m){
int A[][]=new int[n][m];
for(int i=0;i<n;i++) {
for(int j=0;j<m;j++) {
A[i][j]=i();
}
}
return A;
}
static char[][] charinput(int n,int m){
char A[][]=new char[n][m];
for(int i=0;i<n;i++) {
String s=s();
for(int j=0;j<m;j++) {
A[i][j]=s.charAt(j);
}
}
return A;
}
static int nextPowerOf2(int n)
{
if(n==0)
return 1;
n--;
n |= n >> 1;
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;
return n;
}
static int highestPowerof2(int x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
static long highestPowerof2(long x)
{
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
return x ^ (x >> 1);
}
static int max(int A[]) {
int max=Integer.MIN_VALUE;
for(int i=0;i<A.length;i++) {
max=Math.max(max, A[i]);
}
return max;
}
static int min(int A[]) {
int min=Integer.MAX_VALUE;
for(int i=0;i<A.length;i++) {
min=Math.min(min, A[i]);
}
return min;
}
static long max(long A[]) {
long max=Long.MIN_VALUE;
for(int i=0;i<A.length;i++) {
max=Math.max(max, A[i]);
}
return max;
}
static long min(long A[]) {
long min=Long.MAX_VALUE;
for(int i=0;i<A.length;i++) {
min=Math.min(min, A[i]);
}
return min;
}
static long [] prefix(long A[]) {
long p[]=new long[A.length];
p[0]=A[0];
for(int i=1;i<A.length;i++)
p[i]=p[i-1]+A[i];
return p;
}
static long [] prefix(int A[]) {
long p[]=new long[A.length];
p[0]=A[0];
for(int i=1;i<A.length;i++)
p[i]=p[i-1]+A[i];
return p;
}
static long [] suffix(long A[]) {
long p[]=new long[A.length];
p[A.length-1]=A[A.length-1];
for(int i=A.length-2;i>=0;i--)
p[i]=p[i+1]+A[i];
return p;
}
static long [] suffix(int A[]) {
long p[]=new long[A.length];
p[A.length-1]=A[A.length-1];
for(int i=A.length-2;i>=0;i--)
p[i]=p[i+1]+A[i];
return p;
}
static void fill(int dp[]) {
Arrays.fill(dp, -1);
}
static void fill(int dp[][]) {
for(int i=0;i<dp.length;i++)
Arrays.fill(dp[i], -1);
}
static void fill(int dp[][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
Arrays.fill(dp[i][j],-1);
}
}
}
static void fill(int dp[][][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
for(int k=0;k<dp[0][0].length;k++) {
Arrays.fill(dp[i][j][k],-1);
}
}
}
}
static void fill(long dp[]) {
Arrays.fill(dp, -1);
}
static void fill(long dp[][]) {
for(int i=0;i<dp.length;i++)
Arrays.fill(dp[i], -1);
}
static void fill(long dp[][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
Arrays.fill(dp[i][j],-1);
}
}
}
static void fill(long dp[][][][]) {
for(int i=0;i<dp.length;i++) {
for(int j=0;j<dp[0].length;j++) {
for(int k=0;k<dp[0][0].length;k++) {
Arrays.fill(dp[i][j][k],-1);
}
}
}
}
static int min(int a,int b) {
return Math.min(a, b);
}
static int min(int a,int b,int c) {
return Math.min(a, Math.min(b, c));
}
static int min(int a,int b,int c,int d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static int max(int a,int b) {
return Math.max(a, b);
}
static int max(int a,int b,int c) {
return Math.max(a, Math.max(b, c));
}
static int max(int a,int b,int c,int d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long min(long a,long b) {
return Math.min(a, b);
}
static long min(long a,long b,long c) {
return Math.min(a, Math.min(b, c));
}
static long min(long a,long b,long c,long d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static long max(long a,long b) {
return Math.max(a, b);
}
static long max(long a,long b,long c) {
return Math.max(a, Math.max(b, c));
}
static long max(long a,long b,long c,long d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long power(long x, long y, long p)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static long power(long x, long y)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x);
y = y >> 1;
x = (x * x);
}
return res;
}
static void print(int A[]) {
for(int i : A) {
out.print(i+" ");
}
out.println();
}
static void print(long A[]) {
for(long i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static long mod(long x) {
return ((x%mod + mod)%mod);
}
static String reverse(String s) {
StringBuffer p=new StringBuffer(s);
p.reverse();
return p.toString();
}
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
static void sort(int[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static void sort(long[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
static String sort(String s) {
Character ch[]=new Character[s.length()];
for(int i=0;i<s.length();i++) {
ch[i]=s.charAt(i);
}
Arrays.sort(ch);
StringBuffer st=new StringBuffer("");
for(int i=0;i<s.length();i++) {
st.append(ch[i]);
}
return st.toString();
}
static HashMap<Integer,Integer> hash(int A[]){
HashMap<Integer,Integer> map=new HashMap<Integer, Integer>();
for(int i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static HashMap<Long,Integer> hash(long A[]){
HashMap<Long,Integer> map=new HashMap<Long, Integer>();
for(long i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static TreeMap<Integer,Integer> tree(int A[]){
TreeMap<Integer,Integer> map=new TreeMap<Integer, Integer>();
for(int i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static TreeMap<Long,Integer> tree(long A[]){
TreeMap<Long,Integer> map=new TreeMap<Long, Integer>();
for(long i : A) {
if(map.containsKey(i)) {
map.put(i, map.get(i)+1);
}
else {
map.put(i, 1);
}
}
return map;
}
static boolean prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean prime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 1e01b7495da723afa5ed678ffd6ca6c1 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static FastReader s = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
private static int[] rai(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = s.nextInt();
}
return arr;
}
private static int[][] rai(int n, int m) {
int[][] arr = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr[i][j] = s.nextInt();
}
}
return arr;
}
private static long[] ral(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = s.nextLong();
}
return arr;
}
private static long[][] ral(int n, int m) {
long[][] arr = new long[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr[i][j] = s.nextLong();
}
}
return arr;
}
private static int ri() {
return s.nextInt();
}
private static long rl() {
return s.nextLong();
}
private static String rs() {
return s.next();
}
static int gcd(int a,int b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
static long gcd(long a,long b)
{
if(b==0)
{
return a;
}
return gcd(b,a%b);
}
static boolean isPrime(int n) {
//check if n is a multiple of 2
if(n==1)
{
return false;
}
if(n==2)
{
return true;
}
if (n % 2 == 0) return false;
//if not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2) {
if (n % i == 0)
return false;
}
return true;
}
static boolean[] sieveOfEratosthenes(int n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
boolean prime[] = new boolean[n+1];
for(int i=0;i<n;i++)
prime[i] = true;
for(int p = 2; p*p <=n; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p] == true)
{
// Update all multiples of p
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int MOD = 998244353;
static boolean calc(long a, long b, long c)
{
long diff = c-b;
long req = b-diff;
if(req >= a && req%a==0)
{
return true;
}
diff = b-a;
req = b+diff;
if(req >= c && req%c == 0)
{
return true;
}
diff = (c-a);
if(Math.abs(diff)%2 ==1)
{
return false;
}
diff /=2;
req = c-diff;
return req >= b && req % b == 0;
}
public static void main(String[] args) {
StringBuilder ans = new StringBuilder();
int t = ri();
// int t=1;
while (t-- > 0)
{
int n=ri();
int k=ri();
char[] arr=rs().toCharArray();
int[] count = new int[n];
int used = 0;
for(int i=0;i<n && used<k;i++)
{
if(k%2==1)
{
if(arr[i]=='1')
{
count[i] = 1;
used++;
}
}
else
{
if(arr[i]=='0')
{
count[i] = 1;
used++;
}
}
}
int val =k - used;
if(val%2==0)
{
count[0]+=val;
}
else
{
count[0]+=(val-1);
count[n-1]+=1;
}
for(int i=0;i<n;i++)
{
int changed = k-count[i];
if(changed%2==1)
{
if(arr[i]=='0')
{
arr[i]='1';
}
else
{
arr[i]='0';
}
}
ans.append(arr[i]);
}
ans.append("\n");
for(int i:count)
{
ans.append(i).append(" ");
}
ans.append("\n");
}
out.print(ans.toString());
out.flush();
}
/*
abdeba
abedba
*/
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 257464eb157f5dc1e08984ab8cb43d8e | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.*;
import java.util.*;
import java.math.*;
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(),K=i();
char X[]=in.next().toCharArray();
int f[]=new int[N];
if(K%2==0)
{
int c=0;
for(int i=0; i<N; i++)
{
if(K==c)
{
break;
}
if(X[i]=='0') {
c++;
f[i]=1;
}
}
f[N-1]+=(K-c);
for(int i=0; i<N; i++)
{
if(f[i]==0)ans.append(X[i]);
else
{
int x=X[i]-'0';
x+=(f[i]%2);
x%=2;
ans.append(x);
}
}
ans.append("\n");
}
else
{
int c=0;
for(int i=0; i<N; i++)
{
if(K==c)
{
break;
}
if(X[i]=='1') {
c++;
f[i]=1;
}
}
f[N-1]+=(K-c);
for(int i=0; i<N; i++)
{
int x=X[i]-'0';
x+=1;
x%=2;
if(f[i]%2==0)ans.append(x); //flip it
else
{
x+=1;
x%=2;
ans.append(x);
}
}
ans.append("\n");
}
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | c7cbee4c339cd40e651aa9fc767808c1 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | //make sure to make new file!
import java.io.*;
import java.util.*;
public class B782{
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 m = Integer.parseInt(st.nextToken());
char[] input = f.readLine().toCharArray();
int[] array = new int[n];
for(int k = 0; k < n; k++){
array[k] = Character.getNumericValue(input[k]);
}
int[] answer = new int[n];
int[] result = new int[n];
int added = 0;
if(m%2 == 0){
//maximize
for(int k = 0; k < n; k++){
result[k] = array[k];
if(added < m){
if(array[k] == 0){
answer[k]++;
added++;
result[k] = 1;
}
}
}
} else {
//minimize but add 1-array to result
for(int k = 0; k < n; k++){
result[k] = 1-array[k];
if(added < m){
if(array[k] == 1){
answer[k]++;
added++;
result[k] = 1;
}
}
}
}
if((m-added)%2 == 1){
result[n-1] = 1-result[n-1];
}
answer[n-1] += (m-added);
StringJoiner sj = new StringJoiner("");
StringJoiner sj2 = new StringJoiner(" ");
for(int k = 0; k < n; k++){
sj.add("" + result[k]);
sj2.add("" + answer[k]);
}
out.println(sj.toString());
out.println(sj2.toString());
}
out.close();
}
} | Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 22f7d81ea4116d0a10ac735b8f9e7c8b | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class BitFlipping {
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(), K = in.nextInt();
String s = in.next();
boolean[] arr = new boolean[N];
for (int i = 0; i < N; i++) {
arr[i] = s.charAt(i) == '1';
}
if (K % 2 == 1) {
for (int i = 0; i < N; i++) {
arr[i] = !arr[i];
}
}
int[] res = new int[N];
for (int i = 0; i < N; i++) {
if (!arr[i] && K > 0) {
arr[i] = true;
res[i]++;
K--;
}
}
res[N - 1] += K;
if (K > 0 && K % 2 == 1) {
arr[N - 1] = !arr[N - 1];
}
StringBuilder sb = new StringBuilder("");
for (boolean b : arr) {
sb.append(b ? "1" : "0");
}
out.println(sb.toString());
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 | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 6345aefe1cc1d38f069468ec8bc8f457 | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Random;
import java.util.StringTokenizer;
/*
1
6 3
100001
*/
public class B {
public static void main(String[] args) {
FastScanner fs=new FastScanner();
PrintWriter out=new PrintWriter(System.out);
int T=fs.nextInt();
for (int tt=0; tt<T; tt++) {
int n=fs.nextInt();
int k=fs.nextInt();
char[] a=fs.next().toCharArray();
if (k%2==1) {
for (int i=0; i<n; i++) {
if (a[i]=='0') a[i]='1';
else a[i]='0';
}
}
int[] choice=new int[n];
for (int i=0; i<a.length; i++) {
if (a[i]=='0' && k>0) {
a[i]='1';
k--;
choice[i]++;
}
}
choice[n-1]+=k;
if (k%2==1) {
a[n-1]='0';
}
out.println(a);
for (int i=0; i<n; i++) {
out.print(choice[i]+" ");
}
out.println();
}
out.close();
}
static final Random random=new Random();
static final int mod=1_000_000_007;
static void ruffleSort(int[] a) {
int n=a.length;//shuffle, then sort
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
a[oi]=a[i]; a[i]=temp;
}
Arrays.sort(a);
}
static long add(long a, long b) {
return (a+b)%mod;
}
static long sub(long a, long b) {
return ((a-b)%mod+mod)%mod;
}
static long mul(long a, long b) {
return (a*b)%mod;
}
static long exp(long base, long exp) {
if (exp==0) return 1;
long half=exp(base, exp/2);
if (exp%2==0) return mul(half, half);
return mul(half, mul(half, base));
}
static long[] factorials=new long[2_000_001];
static long[] invFactorials=new long[2_000_001];
static void precompFacts() {
factorials[0]=invFactorials[0]=1;
for (int i=1; i<factorials.length; i++) factorials[i]=mul(factorials[i-1], i);
invFactorials[factorials.length-1]=exp(factorials[factorials.length-1], mod-2);
for (int i=invFactorials.length-2; i>=0; i--)
invFactorials[i]=mul(invFactorials[i+1], i+1);
}
static long nCk(int n, int k) {
return mul(factorials[n], mul(invFactorials[k], invFactorials[n-k]));
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 37cdb416b011c14c0a7543180cecbf7f | train_110.jsonl | 1650206100 | You are given a binary string of length $$$n$$$. You have exactly $$$k$$$ moves. In one move, you must select a single bit. The state of all bits except that bit will get flipped ($$$0$$$ becomes $$$1$$$, $$$1$$$ becomes $$$0$$$). You need to output the lexicographically largest string that you can get after using all $$$k$$$ moves. Also, output the number of times you will select each bit. If there are multiple ways to do this, you may output any of them.A binary string $$$a$$$ is lexicographically larger than a binary string $$$b$$$ of the same length, if and only if the following holds: in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ contains a $$$1$$$, and the string $$$b$$$ contains a $$$0$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class B {
public static void main(String[] args) throws NumberFormatException, IOException {
MyReader reader = new MyReader();
int t = reader.nextInt();
for(int testcase=0; testcase<t; testcase++) {
int n = reader.nextInt();
int k = reader.nextInt();
String s = reader.nextLine();
int[] arr = new int[n];
for(int i=0; i<n; i++) {
if(s.charAt(i) == '1') arr[i] = 1;
else arr[i] = 0;
}
if(k % 2 == 1) {
for(int i=0; i<n; i++) {
arr[i]++;
arr[i] %= 2;
}
}
int count0 = 0;
int[] f = new int[n];
for(int i=0; i<n-1; i++) {
if(arr[i] == 0) count0++;
}
if(k>count0) {
//make everything other than last 1
for(int i=0; i<n-1; i++) {
if(arr[i] == 0) {
arr[i] = 1;
f[i]++;
}
}
k -= count0;
f[n-1] = k;
if(k%2 == 1) {
arr[n-1]++;
arr[n-1] %= 2;
}
}
else {
for(int i=0; i<n; i++) {
if(k == 0) break;
if(arr[i] == 0) {
arr[i] = 1;
f[i]++;
k--;
}
}
}
StringBuilder sb1 = new StringBuilder();
for(int i=0; i<n; i++) {
sb1.append(arr[i]);
}
StringBuilder sb2 = new StringBuilder();
for(int i=0; i<n; i++) {
sb2.append(f[i] + " ");
}
System.out.println(sb1.toString());
System.out.println(sb2.toString());
}
}
static class MyReader {
BufferedReader br;
StringTokenizer st;
public MyReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() throws IOException {
while (st == null || !st.hasMoreElements()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(next());
}
long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(next());
}
double nextDouble() throws NumberFormatException, IOException {
return Double.parseDouble(next());
}
String nextLine() throws IOException {
return br.readLine();
}
}
}
| Java | ["6\n\n6 3\n\n100001\n\n6 4\n\n100011\n\n6 0\n\n000000\n\n6 1\n\n111001\n\n6 11\n\n101100\n\n6 12\n\n001110"] | 1 second | ["111110\n1 0 0 2 0 0 \n111110\n0 1 1 1 0 1 \n000000\n0 0 0 0 0 0 \n100110\n1 0 0 0 0 0 \n111111\n1 2 1 3 0 4 \n111110\n1 1 4 2 0 4"] | NoteHere is the explanation for the first testcase. Each step shows how the binary string changes in a move. Choose bit $$$1$$$: $$$\color{red}{\underline{1}00001} \rightarrow \color{red}{\underline{1}}\color{blue}{11110}$$$. Choose bit $$$4$$$: $$$\color{red}{111\underline{1}10} \rightarrow \color{blue}{000}\color{red}{\underline{1}}\color{blue}{01}$$$. Choose bit $$$4$$$: $$$\color{red}{000\underline{1}01} \rightarrow \color{blue}{111}\color{red}{\underline{1}}\color{blue}{10}$$$. The final string is $$$111110$$$ and this is the lexicographically largest string we can get. | Java 8 | standard input | [
"bitmasks",
"constructive algorithms",
"greedy",
"strings"
] | 38375efafa4861f3443126f20cacf3ae | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. Each test case has two lines. The first line has two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$; $$$0 \leq k \leq 10^9$$$). The second line has a binary string of length $$$n$$$, each character is either $$$0$$$ or $$$1$$$. The sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,300 | For each test case, output two lines. The first line should contain the lexicographically largest string you can obtain. The second line should contain $$$n$$$ integers $$$f_1, f_2, \ldots, f_n$$$, where $$$f_i$$$ is the number of times the $$$i$$$-th bit is selected. The sum of all the integers must be equal to $$$k$$$. | standard output | |
PASSED | 006fd57fb26d328ea9e05f9fd002ffa2 | 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 codeforces {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
long t = scan.nextLong();
for (long i = 0; i < t; i++) {
long games = scan.nextLong();
long r = scan.nextLong();
long b = scan.nextLong();
System.out.println(compute(games, r, b));
}
}
public static String compute(long games, long r, long b) {
long rest = r%(b+1);
long ratio = r/(b+1);
String sol = "";
long j = 0;
while(j <= b) {
for (int i = 0; i < ratio; i++) {
sol += "R";
}
if (j < rest) sol += "R";
if (j < b) sol = sol + "B";
j++;
}
return sol;
}
public static long findMax(long games, long r, long b) {
long i = 1;
while(true) {
if ((b+1)*i + b >= games) return i;
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 11 | 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 | f83f9996276c444491388bdcc073d707 | 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.lang.*;
import java.util.*;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner a=new Scanner(System.in);
int y=a.nextInt();
while (y>0){
int xx=a.nextInt();
int m=a.nextInt();
int n=a.nextInt();
if (n==1){
for (int i=0;i<m/2;i++){
System.out.print("R");
}
System.out.print("B");
for (int i=m/2;i<m;i++){
System.out.print("R");
}
}else{
int x=m%n;
int d=m/n;
int u=d;
int hh=-1;
if (d<x){
u=d+1;
int zz=m-(n*d);
int zz1=n-zz;
for (int i=0;i<zz;i++){
for (int j=0;j<u;j++){
System.out.print("R");
}
System.out.print("B");
}
for (int i=0;i<zz1;i++){
for (int j=0;j<d;j++){
System.out.print("R");
}
System.out.print("B");
}
}
else{
while (m-u*n<u){
u--;
if (m-u*n<=u){
}else{
u++;
break;
}
}
for (int i=0;i<n;i++){
for (int j=0;j<u;j++){
System.out.print("R");
}
System.out.print("B");
}
for (int i=0;i<(m-n*u);i++){
System.out.print("R");
}
}
}
System.out.println();
y--;
}
}
} | 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 11 | 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 | a2c5ec3e2ef04f3e41bca968049a32b7 | 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 ProblemB {
static int MOD = (int)1e9+7;
static int MAX = (int)1e6+1;
static LinkedHashSet<Integer>temp;
public static void main(String[] args) throws Exception {
Scanner in = new Scanner();
StringBuilder sb = new StringBuilder();
int test = in.readInt();
while(test-->0){
int n = in.readInt();
int r = in.readInt();
int b = in.readInt();
char res[] = new char[n+1];
for(int i = 2;i<=100;i++){
int mx = i,tb= b;
Arrays.fill(res, 'R');
for(int j =i;j<=n;j+=i){
if(tb>0){res[j] = 'B';tb--;mx =n-j;}
}
if(mx<=i){
for(int j =1;j<=n;j++){
if(tb>0&& j+1<n && res[j] == 'R' && res[j+1] == res[j]){
res[j] = 'B';tb--;
}
}
break;
}
}
for(int i = 1;i<=n;i++)sb.append(res[i]);
sb.append("\n");
}
System.out.println(sb);
}
public static ArrayList<Integer> subtract(int a[],int b[]){
int n = a.length-1;
ArrayList<Integer>l = new ArrayList<>();
boolean carry = false;
for(int i =n;i>0;i--){
if(carry)a[i]--;
if(a[i]>=b[i-1]){
l.add(a[i]-b[i-1]);carry = false;
}
else {
a[i]+=10;
l.add(a[i]-b[i-1]);
carry = true;
}
}
Collections.reverse(l);
return l;
}
public static int[] fill(String s){
int a[] = new int[s.length()];
for(int i = 0;i<s.length();i++)a[i] = s.charAt(i)-'0';
return a;
}
public static long factorial(int n){
long fac = 1;
for(int i = 1;i<=n;i++)fac*=i;
return fac;
}
public static ArrayList<Integer> factors(int n){
ArrayList<Integer>l = new ArrayList<>();
for(int i = 1;i*i<=n;i++){
if(n%i == 0){
l.add(i);
if(n/i != i)l.add(n/i);
}
}
return l;
}
public static void build(int seg[],int ind,int l,int r,int a[]){
if(l == r){
seg[ind] = a[l];
return;
}
int mid = (l+r)/2;
build(seg,(2*ind)+1,l,mid,a);
build(seg,(2*ind)+2,mid+1,r,a);
seg[ind] = Math.min(seg[(2*ind)+1],seg[(2*ind)+2]);
}
public static long gcd(long a, long b){
return b ==0 ?a:gcd(b,a%b);
}
public static long nearest(TreeSet<Long> list,long target){
long nearest = -1,sofar = Integer.MAX_VALUE;
for(long it:list){
if(Math.abs(it - target)<sofar){
sofar = Math.abs(it - target);
nearest = it;
}
}
return nearest;
}
public static ArrayList<Long> factors(long n){
ArrayList<Long>l = new ArrayList<>();
for(long i = 1;i*i<=n;i++){
if(n%i == 0){
l.add(i);
if(n/i != i)l.add(n/i);
}
}
Collections.sort(l);
return l;
}
public static void swap(char a[],int i, int j){
char t = a[i];
a[i] = a[j];
a[j] = t;
}
public static boolean isSorted(long a[]){
for(int i =0;i<a.length;i++){
if(i+1<a.length && a[i]>a[i+1])return false;
}
return true;
}
public static int first(ArrayList<Long>list,long target){
int index = -1;
int l = 0,r = list.size()-1;
while(l<=r){
int mid = (l+r)/2;
if(list.get(mid) == target){
index = mid;
r = mid-1;
}else if(list.get(mid)>target){
r = mid-1;
}else l = mid+1;
}
return index;
}
public static int last(ArrayList<Long>list,long target){
int index = -1;
int l = 0,r = list.size()-1;
while(l<=r){
int mid = (l+r)/2;
if(list.get(mid) == target){
index= mid;
l = mid+1;
}else if(list.get(mid)<target){
l = mid+1;
}else r = mid-1;
}
return index;
}
public static void sort(int arr[]){
ArrayList<Integer>list = new ArrayList<>();
for(int it:arr)list.add(it);
Collections.sort(list);
for(int i = 0;i<arr.length;i++)arr[i] = list.get(i);
}
static class Pair{
int x,y,z;
public Pair(int x,int y){
this.x = x;
this.y = y;
}
}
static class Scanner{
BufferedReader br;
StringTokenizer st;
public Scanner(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public String read()
{
while (st == null || !st.hasMoreElements()) {
try { st = new StringTokenizer(br.readLine()); }
catch (Exception e) { e.printStackTrace(); }
}
return st.nextToken();
}
public int readInt() { return Integer.parseInt(read()); }
public long readLong() { return Long.parseLong(read()); }
public double readDouble(){return Double.parseDouble(read());}
public String readLine() throws Exception { 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 11 | 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 | 9ca1da8e428d273c48d364ad49d30af4 | 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 ProblemB {
static int MOD = (int)1e9+7;
static int MAX = (int)1e6+1;
static LinkedHashSet<Integer>temp;
public static void main(String[] args) throws Exception {
Scanner in = new Scanner();
StringBuilder sb = new StringBuilder();
int test = in.readInt();
while(test-->0){
int n = in.readInt();
int r = in.readInt();
int b = in.readInt();
char res[] = new char[n+1];
for(int i = 2;i<=100;i++){
int mx = i,tb= b;
Arrays.fill(res, 'R');
for(int j =i;j<=n;j+=i){
if(tb>0){res[j] = 'B';tb--;mx =Math.max(i-1, n-j);}
}
if(mx<=i){
for(int j =1;j<=n;j++){
if(tb>0&& j+1<n && res[j] == 'R' && res[j+1] == res[j]){
res[j] = 'B';tb--;
}
}
break;
}
}
for(int i = 1;i<=n;i++)sb.append(res[i]);
sb.append("\n");
}
System.out.println(sb);
}
public static ArrayList<Integer> subtract(int a[],int b[]){
int n = a.length-1;
ArrayList<Integer>l = new ArrayList<>();
boolean carry = false;
for(int i =n;i>0;i--){
if(carry)a[i]--;
if(a[i]>=b[i-1]){
l.add(a[i]-b[i-1]);carry = false;
}
else {
a[i]+=10;
l.add(a[i]-b[i-1]);
carry = true;
}
}
Collections.reverse(l);
return l;
}
public static int[] fill(String s){
int a[] = new int[s.length()];
for(int i = 0;i<s.length();i++)a[i] = s.charAt(i)-'0';
return a;
}
public static long factorial(int n){
long fac = 1;
for(int i = 1;i<=n;i++)fac*=i;
return fac;
}
public static ArrayList<Integer> factors(int n){
ArrayList<Integer>l = new ArrayList<>();
for(int i = 1;i*i<=n;i++){
if(n%i == 0){
l.add(i);
if(n/i != i)l.add(n/i);
}
}
return l;
}
public static void build(int seg[],int ind,int l,int r,int a[]){
if(l == r){
seg[ind] = a[l];
return;
}
int mid = (l+r)/2;
build(seg,(2*ind)+1,l,mid,a);
build(seg,(2*ind)+2,mid+1,r,a);
seg[ind] = Math.min(seg[(2*ind)+1],seg[(2*ind)+2]);
}
public static long gcd(long a, long b){
return b ==0 ?a:gcd(b,a%b);
}
public static long nearest(TreeSet<Long> list,long target){
long nearest = -1,sofar = Integer.MAX_VALUE;
for(long it:list){
if(Math.abs(it - target)<sofar){
sofar = Math.abs(it - target);
nearest = it;
}
}
return nearest;
}
public static ArrayList<Long> factors(long n){
ArrayList<Long>l = new ArrayList<>();
for(long i = 1;i*i<=n;i++){
if(n%i == 0){
l.add(i);
if(n/i != i)l.add(n/i);
}
}
Collections.sort(l);
return l;
}
public static void swap(char a[],int i, int j){
char t = a[i];
a[i] = a[j];
a[j] = t;
}
public static boolean isSorted(long a[]){
for(int i =0;i<a.length;i++){
if(i+1<a.length && a[i]>a[i+1])return false;
}
return true;
}
public static int first(ArrayList<Long>list,long target){
int index = -1;
int l = 0,r = list.size()-1;
while(l<=r){
int mid = (l+r)/2;
if(list.get(mid) == target){
index = mid;
r = mid-1;
}else if(list.get(mid)>target){
r = mid-1;
}else l = mid+1;
}
return index;
}
public static int last(ArrayList<Long>list,long target){
int index = -1;
int l = 0,r = list.size()-1;
while(l<=r){
int mid = (l+r)/2;
if(list.get(mid) == target){
index= mid;
l = mid+1;
}else if(list.get(mid)<target){
l = mid+1;
}else r = mid-1;
}
return index;
}
public static void sort(int arr[]){
ArrayList<Integer>list = new ArrayList<>();
for(int it:arr)list.add(it);
Collections.sort(list);
for(int i = 0;i<arr.length;i++)arr[i] = list.get(i);
}
static class Pair{
int x,y,z;
public Pair(int x,int y){
this.x = x;
this.y = y;
}
}
static class Scanner{
BufferedReader br;
StringTokenizer st;
public Scanner(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public String read()
{
while (st == null || !st.hasMoreElements()) {
try { st = new StringTokenizer(br.readLine()); }
catch (Exception e) { e.printStackTrace(); }
}
return st.nextToken();
}
public int readInt() { return Integer.parseInt(read()); }
public long readLong() { return Long.parseLong(read()); }
public double readDouble(){return Double.parseDouble(read());}
public String readLine() throws Exception { 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 11 | 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 | 9a88e981b80ca655aaa9175a4351f08c | 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 javax.sound.midi.Soundbank;
import java.awt.*;
import java.io.*;
import java.util.*;
import java.util.zip.InflaterInputStream;
public class Main {
static int N = 1001;
// Array to store inverse of 1 to N
static long[] factorialNumInverse = new long[N + 1];
// Array to precompute inverse of 1! to N!
static long[] naturalNumInverse = new long[N + 1];
// Array to store factorial of first N numbers
static long[] fact = new long[N + 1];
// Function to precompute inverse of numbers
public static void InverseofNumber(int p)
{
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for(int i = 2; i <= N; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] * (long)(p - p / i) % p;
}
public static void InverseofFactorial(int p)
{
factorialNumInverse[0] = factorialNumInverse[1] = 1;
// Precompute inverse of natural numbers
for(int i = 2; i <= N; i++)
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p;
}
// Function to calculate factorial of 1 to N
public static void factorial(int p)
{
fact[0] = 1;
// Precompute factorials
for(int i = 1; i <= N; i++)
{
fact[i] = (fact[i - 1] * (long)i) % p;
}
}
public static long ncp(int N, int R, int p)
{
// n C r = n!*inverse(r!)*inverse((n-r)!)
long ans = ((fact[N] * factorialNumInverse[R]) %
p * factorialNumInverse[N - R]) % p;
return ans;
}
public static void main(String[] args) throws IOException {
Reader.init(System.in);
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
int t = Reader.nextInt();
while (t > 0){
int n = Reader.nextInt();
int r = Reader.nextInt();
int b = Reader.nextInt();
int l = (r)%(b+1);
int v = r/(b+1);
while ( r > 0){
int i = 0;
while ( i < v ){
output.write("R");
i++;
r--;
}
if ( l > 0){
output.write("R");
l--;
r--;
}
if ( b > 0){
output.write("B");
b--;
}
}
output.write("\n");
t--;
}
output.flush();
}
public static boolean help(int m,int r,int b){
boolean ans = true;
while ( r > 0 ){
if ( b == 0 && r > m){
ans =false;
break;
}
else{
r-=(int)min(m,r);
b--;
}
}
return ans;
}
public static void primeFactorisation(long n,HashMap<Long,Integer> map){
long i = 2;
while ( n%i == 0){
if ( map.containsKey(i)){
map.put(i,map.get(i)+1);
}
else{
map.put(i,1);
}
n/=i;
}
i = 3;
long last = (long)Math.pow(n,0.5);
while ( i <= last){
while ( n%i == 0){
if ( map.containsKey(i)){
map.put(i,map.get(i)+1);
}
else{
map.put(i,1);
}
n/=i;
}
i++;
}
if ( n > 2){
map.put(n,1);
}
}
public static boolean prime(int n){
int i = 2;
int r = (int)Math.pow(n,0.5);
boolean ans = true;
while ( i<=r){
if ( n%i == 0){
ans = false;
break;
}
i++;
}
return ans;
}
public static long log2(long N)
{
long result = (long)(Math.log(N) / Math.log(2));
return result;
}
private static int bs(int low,int high,int [] array,long find){
if ( low <= high ){
int mid = low + (high-low)/2;
if ( array[mid] > find){
high = mid -1;
return bs(low,high,array,find);
}
else if ( array[mid] < find){
low = mid+1;
return bs(low,high,array,find);
}
return mid;
}
return -1;
}
private static long max(long a, long b) {
return Math.max(a,b);
}
private static long min(long a,long b){
return Math.min(a,b);
}
public static long modularExponentiation(long a,long b,long mod){
if ( b == 1){
return a;
}
else{
long ans = modularExponentiation(a,b/2,mod)%mod;
if ( b%2 == 1){
return (a*((ans*ans)%mod))%mod;
}
return ((ans*ans)%mod);
}
}
public static long sum(long n){
return (n*(n+1))/2;
}
public static long abs(long a){
return a < 0 ? (-1*a) : a;
}
public static long gcd(long a,long b){
if ( a == 0){
return b;
}
else{
return gcd(b%a,a);
}
}
}
class Reader {
static BufferedReader reader;
static StringTokenizer tokenizer;
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input));
tokenizer = new StringTokenizer("");
}
static String next() throws IOException {
while ( ! tokenizer.hasMoreTokens() ) {
tokenizer = new StringTokenizer(
reader.readLine() );
}
return tokenizer.nextToken();
}
static int nextInt() throws IOException {
return Integer.parseInt( next() );
}
static double nextDouble() throws IOException {
return Double.parseDouble( next() );
}
static long nextLong() throws IOException{
return Long.parseLong(next());
}
}
class NComparator implements Comparator<Node>{
@Override
public int compare(Node o1, Node o2) {
if ( o2.infected == 1 && o1.infected == 0){
return -1;
}
else if ( o2.infected == 0 && o1.infected == 1){
return 1;
}
else{
if ( o2.value > o1.value){
return 1;
}
else if ( o2.value < o1.value){
return -1;
}
else{
return 0;
}
}
}
}
class DComparator implements Comparator<Integer>{
@Override
public int compare(Integer o1, Integer o2) {
if ( o2 > o1){
return 1;
}
else if ( o2 < o1){
return -1;
}
else{
return 0;
}
}
}
class Node{
int value;
int infected;
Node(int A,int B){
value= A;
infected = 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 11 | 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 | ca01fad21d0a0d2778ff4c133f74a0d1 | 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 {
public static void main(String[] args) throws IOException {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(reader.readLine());
while (t > 0) {
t--;
StringTokenizer st = new StringTokenizer(reader.readLine());
int n = Integer.parseInt(st.nextToken());
int r = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
StringBuilder sb = new StringBuilder();
int segments = b + 1;
for (int i = 0; i < segments; i++) {
int segmentSize = r / (b + 1);
sb.append("R".repeat(segmentSize));
r -= segmentSize;
if (b > 0) {
sb.append("B");
b--;
}
}
sb.append("R".repeat(r));
out.println(sb);
}
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 11 | 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 | a4970b687f0445dfa11b97128af52a02 | 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.Objects;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws InterruptedException {
Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 29);
thread.start();
thread.join();
}
static class TaskAdapter implements Runnable {
@Override
public void run() {
FastReader in = new FastReader();
//FastestNumberReader in = new FastestNumberReader();
OutputWriter out = new OutputWriter();
Solver solver = new Solver();
int testCount = 1;
testCount = in.nextInt();
for (int i = 1; i <= testCount; i++) {
try {
solver.solve(i, in, out);
} catch (IOException e) {
e.printStackTrace();
}
}
out.close();
}
}
static class Solver {
private void solve(int testNumber, FastReader reader, OutputWriter writer) throws IOException {
int n = reader.nextInt();
int r = reader.nextInt();
int parts = reader.nextInt() + 1;
StringBuilder ans = new StringBuilder();
int minLength = r / parts;
int longerPartsAmount = r % parts;
int i = 0;
int sequenceLength = 0;
int sequenceCnt = 0;
for (; sequenceCnt < longerPartsAmount; i++) {
if (sequenceLength == minLength + 1) {
sequenceLength = 0;
ans.append("B");
sequenceCnt++;
} else {
ans.append("R");
sequenceLength++;
}
}
for (; i < n; i++) {
if (sequenceLength == minLength) {
sequenceLength = 0;
ans.append("B");
} else {
ans.append("R");
sequenceLength++;
}
}
writer.println(ans);
}
}
static class Pair<F extends Comparable<? super F>, S extends Comparable<? super S>>
implements Comparable<Pair<F, S>> {
F first;
S second;
public Pair(F first, S second) {
this.first = first;
this.second = second;
}
@Override
public int compareTo(Pair<F, S> p) {
int result = this.first.compareTo(p.first);
if (result == 0) {
return this.second.compareTo(p.second);
}
return result;
}
@Override
public int hashCode() {
return Objects.hash(first, second);
}
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return Objects.equals(first, pair.first) && Objects.equals(second, pair.second);
}
@Override
public String toString() {
return "Pair{" + "first=" + first + ", second=" + second + '}';
}
}
static class FastestNumberReader {
private final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer;
private int bytesRead;
public FastestNumberReader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public FastestNumberReader(String fileName) throws IOException {
din = new DataInputStream(
new FileInputStream(fileName));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
} else {
continue;
}
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
public String readLine(int length) throws IOException {
byte[] buf = new byte[length + 1]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
} else {
continue;
}
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
public void close() throws IOException {
din.close();
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
int nextInt() {
return Integer.parseInt(next());
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter() {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void flush() {
writer.flush();
}
public void print(Object obj) {
writer.print(obj);
}
public void println(Object obj) {
writer.println(obj);
}
public void printf(String format, Object... objects) {
writer.printf(format, objects);
}
public void printfln(String format, Object... objects) {
writer.printf(format, objects);
writer.println();
}
public void close() {
writer.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 11 | 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 | b361c51b690b9c1f2f4d619789c7f7fa | 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 java.lang.Exception {
int testCase = sc.nextInt();
while(testCase-->0){
int n = sc.nextInt(), red = sc.nextInt(), blue = sc.nextInt();
int remain = red%(blue+1);
for(int i=0; i<blue+1; i++){
int breakPoint = red/(blue+1);
if(remain>0){
breakPoint++; remain--;
}
while(breakPoint-->0){
System.out.print("R");
}
if(i!=blue)
System.out.print("B");
}
System.out.println();
}
}
// Fast Reader Class
public static FastReader sc = new FastReader();
public static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
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 | ["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 11 | 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 | 10411264d3b81b510641d9b64fd87572 | 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 java.lang.Exception {
int testCase = sc.nextInt();
while(testCase-->0){
int n = sc.nextInt();
int red = sc.nextInt();
int blue = sc.nextInt();
int partitions = blue+1;
int [] arr = new int[partitions];
int remain = red%partitions;
for(int i=0; i<partitions; i++){
arr[i] = red/partitions;
if(remain>0){
arr[i]++;
remain--;
}
}
for(int i=0; i<partitions; i++){
while(arr[i]-->0){
System.out.print("R");
}
if(i!=partitions-1)
System.out.print("B");
}
System.out.println();
}
}
// Fast Reader Class
public static FastReader sc = new FastReader();
public static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
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 | ["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 11 | 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 | b2abb36277626df1cd1a83fdbcf9c6c1 | 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 class TreeNode{
int val;
TreeNode left, right;
}
public PriorityQueue<Integer> pq = new PriorityQueue<>();
public int kthSmallest(TreeNode root, int k) {
traverseTree(root, k);
return pq.poll();
}
public void traverseTree(TreeNode node, int k){
if(node == null) return;
if(pq.size()>=k){
pq.offer(node.val);
pq.poll();
}else{
pq.offer(node.val);
}
traverseTree(node.left,k);
traverseTree(node.right,k);
}
public static void main (String[] args) throws java.lang.Exception {
int testCase = sc.nextInt();
while(testCase-->0){
int n = sc.nextInt();
int red = sc.nextInt();
int blue = sc.nextInt();
int partitions = blue+1;
if(red%partitions==0){
int breakPoint = red/partitions;
int redCount = 0;
while(red>0 || blue>0){
if(redCount == breakPoint){
redCount = 0;
System.out.print("B");
blue--;
}else{
System.out.print("R");
redCount++;
red--;
}
}
System.out.println();
}else{
int [] arr = new int[partitions];
int remain = red%partitions;
for(int i=0; i<partitions; i++){
arr[i] = red/partitions;
if(remain>0){
arr[i]++;
remain--;
}
}
for(int i=0; i<partitions; i++){
while(arr[i]-->0){
System.out.print("R");
}
if(i!=partitions-1)
System.out.print("B");
}
System.out.println();
}
}
}
// Fast Reader Class
public static FastReader sc = new FastReader();
public static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
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 | ["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 11 | 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 | 9dd87ba9ff7f4acbb9c2f9b97c5a6915 | 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 Solution {
private static final String SPACE = "\\s+";
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int tt = Integer.parseInt(in.readLine());
for (int t = 1; t <= tt; t++) {
String[] tokens = in.readLine().split(SPACE);
int n = Integer.parseInt(tokens[0]);
int r = Integer.parseInt(tokens[1]);
int b = Integer.parseInt(tokens[2]);
int size = r / (b + 1);
int extras = r % (b + 1);
StringBuilder sb = new StringBuilder();
while (true) {
for (int i = 0; i < size && r > 0; i++) {
sb.append('R');
r--;
}
if (extras > 0) {
sb.append('R');
r--;
extras--;
}
if (b > 0) {
sb.append('B');
b--;
}
if (r == 0 && b == 0) break;
}
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 11 | 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 | b658bbab256806748070950e6591c7de | 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 Main{
public static void main(String[] args) throws Exception{
Scanner cin = new Scanner(System.in);
int t=cin.nextInt();
for(int i=0;i<t;i++) {
int n=cin.nextInt();
int r=cin.nextInt();
int b=cin.nextInt();
int count=0;
int yu=r%(b+1);//有这么多个需要多插一个的
int index=r/(b+1);
for(int j=0;j<yu;j++) {
for(int k=0;k<index+1;k++) {
System.out.print("R");
count++;
}
System.out.print("B");
}
for(int j=yu;j<b;j++) {
for(int k=0;k<index;k++) {
System.out.print("R");
count++;
}
System.out.print("B");
}
for(;count<r;count++) {
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 11 | 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 | a81140e913988f3c7f73a338426c3005 | 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.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class Main {
// Graph
// prefix sums
//inputs
public static void main(String args[])throws Exception{
Input sc=new Input();
precalculates p=new precalculates();
StringBuilder sb=new StringBuilder();
int t=sc.readInt();
for(int f=0;f<t;f++){
int d[]=sc.readArray();
int n=d[0],r=d[1],b=d[2];
int rep=r/(b+1);
int a[]=new int[b+1];
for(int i=0;i<a.length;i++){
a[i]=rep;
r-=rep;
}
for(int i=0;i<a.length;i++){
if(r>0){
a[i]++;
r--;
}
}
for(int i=0;i<b;i++){
for(int j=0;j<a[i];j++){
sb.append("R");
}
sb.append("B");
}
for(int j=0;j<a[b];j++){
sb.append("R");
}
sb.append("\n");
}
System.out.print(sb);
}
}
class Input{
BufferedReader br;
StringTokenizer st;
Input(){
br=new BufferedReader(new InputStreamReader(System.in));
st=new StringTokenizer("");
}
public int[] readArray() throws Exception{
st=new StringTokenizer(br.readLine());
int a[]=new int[st.countTokens()];
for(int i=0;i<a.length;i++){
a[i]=Integer.parseInt(st.nextToken());
}
return a;
}
public long[] readArrayLong() throws Exception{
st=new StringTokenizer(br.readLine());
long a[]=new long[st.countTokens()];
for(int i=0;i<a.length;i++){
a[i]=Long.parseLong(st.nextToken());
}
return a;
}
public int readInt() throws Exception{
st=new StringTokenizer(br.readLine());
return Integer.parseInt(st.nextToken());
}
public long readLong() throws Exception{
st=new StringTokenizer(br.readLine());
return Long.parseLong(st.nextToken());
}
public String readString() throws Exception{
return br.readLine();
}
public int[][] read2dArray(int n,int m)throws Exception{
int a[][]=new int[n][m];
for(int i=0;i<n;i++){
st=new StringTokenizer(br.readLine());
for(int j=0;j<m;j++){
a[i][j]=Integer.parseInt(st.nextToken());
}
}
return a;
}
}
class precalculates{
public long gcd(long p, long q) {
if (q == 0) return p;
else return gcd(q, p % q);
}
public int[] prefixSumOneDimentional(int a[]){
int n=a.length;
int dp[]=new int[n];
for(int i=0;i<n;i++){
if(i==0)
dp[i]=a[i];
else
dp[i]=dp[i-1]+a[i];
}
return dp;
}
public int[] postSumOneDimentional(int a[]) {
int n = a.length;
int dp[] = new int[n];
for (int i = n - 1; i >= 0; i--) {
if (i == n - 1)
dp[i] = a[i];
else
dp[i] = dp[i + 1] + a[i];
}
return dp;
}
public int[][] prefixSum2d(int a[][]){
int n=a.length;int m=a[0].length;
int dp[][]=new int[n+1][m+1];
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
dp[i][j]=a[i-1][j-1]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
}
}
return dp;
}
public long pow(long a,long b){
long mod=998244353;
long ans=1;
if(b<=0)
return 1;
if(b%2==0){
ans=pow(a,b/2)%mod;
return ((ans%mod)*(ans%mod))%mod;
}else{
ans=pow(a,b-1)%mod;
return ((a%mod)*(ans%mod))%mod;
}
}
}
class GraphInteger{
HashMap<Integer,vertex> vtces;
class vertex{
HashMap<Integer,Integer> children;
public vertex(){
children=new HashMap<>();
}
}
public GraphInteger(){
vtces=new HashMap<>();
}
public void addVertex(int a){
vtces.put(a,new vertex());
}
public void addEdge(int a,int b,int cost){
if(!vtces.containsKey(a)){
vtces.put(a,new vertex());
}
if(!vtces.containsKey(b)){
vtces.put(b,new vertex());
}
vtces.get(a).children.put(b,cost);
// vtces.get(b).children.put(a,cost);
}
public boolean isCyclicDirected(){
boolean isdone[]=new boolean[vtces.size()+1];
boolean check[]=new boolean[vtces.size()+1];
for(int i=1;i<=vtces.size();i++) {
if (!isdone[i] && isCyclicDirected(i,isdone, check)) {
return true;
}
}
return false;
}
private boolean isCyclicDirected(int i,boolean isdone[],boolean check[]){
if(check[i])
return true;
if(isdone[i])
return false;
check[i]=true;
isdone[i]=true;
Set<Integer> set=vtces.get(i).children.keySet();
for(Integer ii:set){
if(isCyclicDirected(ii,isdone,check))
return true;
}
check[i]=false;
return false;
}
}
class union_find {
int n;
int[] sz;
int[] par;
union_find(int nval) {
n = nval;
sz = new int[n + 1];
par = new int[n + 1];
for (int i = 0; i <= n; i++) {
par[i] = i;
sz[i] = 1;
}
}
int root(int x) {
if (par[x] == x)
return x;
return par[x] = root(par[x]);
}
boolean find(int a, int b) {
return root(a) == root(b);
}
int union(int a, int b) {
int ra = root(a);
int rb = root(b);
if (ra == rb)
return 0;
if(a==b)
return 0;
if (sz[a] > sz[b]) {
int temp = ra;
ra = rb;
rb = temp;
}
par[ra] = rb;
sz[rb] += sz[ra];
return 1;
}
}
/*
static int mod=998244353;
private static int add(int x, int y) {
x += y;
return x % MOD;
}
private static int mul(int x, int y) {
int res = (int) (((long) x * y) % MOD);
return res;
}
private static int binpow(int x, int y) {
int z = 1;
while (y > 0) {
if (y % 2 != 0)
z = mul(z, x);
x = mul(x, x);
y >>= 1;
}
return z;
}
private static int inv(int x) {
return binpow(x, MOD - 2);
}
private static int devide(int x, int y) {
return mul(x, inv(y));
}
private static int C(int n, int k, int[] fact) {
return devide(fact[n], mul(fact[k], fact[n-k]));
}
*/
| 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 11 | 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 | 7c8050e2a543ed44e656d5a60d321353 | 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 com.sun.source.tree.Tree;
import java.io.*;
import java.sql.Array;
import java.util.*;
public class Main {
static Kattio io;
static long mod = 998244353, inv2 = 499122177;
static {
io = new Kattio();
}
public static void main(String[] args) {
int t = io.nextInt();
for (int i = 0; i < t; i++) {
solve();
}
io.close();
}
private static void solve() {
int n = io.nextInt(),r=io.nextInt(), b = io.nextInt();
int [] buck = new int[b+1];
int ind = 0 ;
while (r!=0){
buck[ind%buck.length]++;
ind++;
r --;
}
// io.println(Arrays.toString(buck));
for (int i = 0; i < buck.length; i++) {
for (int j = 0; j < buck[i]; j++) {
io.print("R");
}
if(i!=buck.length-1)
io.print("B");
}
io.println();
}
static class pair implements Comparable<pair> {
private final int x;
private final int y;
public pair(final int x, final int y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return "pair{" +
"x=" + x +
", y=" + y +
'}';
}
@Override
public boolean equals(final Object o) {
if (this == o) {
return true;
}
if (!(o instanceof pair)) {
return false;
}
final pair pair = (pair) o;
if (x != pair.x) {
return false;
}
return y == pair.y;
}
@Override
public int hashCode() {
int result = x;
result = (int) (31 * result + y);
return result;
}
@Override
public int compareTo(pair pair) {
if(this.y==pair.y)
return Integer.compare(this.x, pair.x);
return Integer.compare(this.y, pair.y);
}
}
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() {
this(System.in, System.out);
}
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(new FileWriter(problemName + ".out"));
r = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) {}
return null;
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["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 11 | 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 | 5a1ffee04bacee0e30b858391103ce77 | 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 {
static int[][] size;
static int[][][] parent;
static char[][] a;
public static void main(String[] args) {
MyScanner in = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int t = in.nextInt();
while(t -- > 0) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
int bucket = r / (b + 1);
int rem = r - bucket * (b+1); // at most < bucket
for(int i = 1; i <= b + 1; i++) {
for(int j = 1; j <= bucket; j++) {
out.print('R');
}
if(rem >0) {
out.print('R');
rem--;
}
if(i != b + 1) {
out.print('B');
}
}
out.println("");
}
out.close();
return;
}
//-----------PrintWriter for faster output---------------------------------
public static PrintWriter out;
//-----------MyScanner class for faster input----------
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
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 11 | 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 | 748e7efbd9bcd2f51aada7f22591db3f | 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 | //Utilities
import java.io.*;
import java.util.*;
public class a {
static int t;
static int n, a, b;
public static void main(String[] args) throws IOException {
t = in.iscan();
while (t-- > 0) {
n = in.iscan(); a = in.iscan(); b = in.iscan();
char[] res = new char[n];
if (a >= b) {
for (int i = 0; i <= n; i++) {
if (a - b * i <= i) {
for (int j = 0, cnt = 0; j < n; j++) {
if (cnt == i) {
res[j] = 'B';
b--;
cnt = 0;
continue;
}
res[j] = 'R';
a--;
cnt++;
}
int idx = 0;
while (b > 0) {
if (res[idx] == 'R' && (idx == n-1 || res[idx+1] == 'R')) {
res[idx] = 'B';
idx++;
b--; a++;
}
idx++;
}
break;
}
}
}
else {
for (int i = 0; i <= n; i++) {
if (b - a * i <= i) {
for (int j = 0, cnt = 0; j < n; j++) {
if (cnt == i) {
res[j] = 'R';
a--;
cnt = 0;
continue;
}
res[j] = 'B';
b--;
cnt++;
}
int idx = 0;
while (a > 0) {
if (res[idx] == 'B' && (idx == n-1 || res[idx+1] == 'B')) {
res[idx] = 'R';
idx++;
a--; b++;
}
idx++;
}
break;
}
}
}
out.println(res);
}
out.close();
}
static INPUT in = new INPUT(System.in);
static PrintWriter out = new PrintWriter(System.out);
private static class INPUT {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar, numChars;
public INPUT (InputStream stream) {
this.stream = stream;
}
public INPUT (String file) throws IOException {
this.stream = new FileInputStream (file);
}
public int cscan () throws IOException {
if (curChar >= numChars) {
curChar = 0;
numChars = stream.read (buf);
}
if (numChars == -1)
return numChars;
return buf[curChar++];
}
public int iscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
int res = 0;
do {
res = (res << 1) + (res << 3);
res += c - '0';
c = cscan ();
}
while (!space (c));
return res * sgn;
}
public String sscan () throws IOException {
int c = cscan ();
while (space (c))
c = cscan ();
StringBuilder res = new StringBuilder ();
do {
res.appendCodePoint (c);
c = cscan ();
}
while (!space (c));
return res.toString ();
}
public double dscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
double res = 0;
while (!space (c) && c != '.') {
if (c == 'e' || c == 'E')
return res * UTILITIES.fast_pow (10, iscan ());
res *= 10;
res += c - '0';
c = cscan ();
}
if (c == '.') {
c = cscan ();
double m = 1;
while (!space (c)) {
if (c == 'e' || c == 'E')
return res * UTILITIES.fast_pow (10, iscan ());
m /= 10;
res += (c - '0') * m;
c = cscan ();
}
}
return res * sgn;
}
public long lscan () throws IOException {
int c = cscan (), sgn = 1;
while (space (c))
c = cscan ();
if (c == '-') {
sgn = -1;
c = cscan ();
}
long res = 0;
do {
res = (res << 1) + (res << 3);
res += c - '0';
c = cscan ();
}
while (!space (c));
return res * sgn;
}
public boolean space (int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
public static class UTILITIES {
static final double EPS = 10e-6;
public static void sort(int[] a, boolean increasing) {
ArrayList<Integer> arr = new ArrayList<Integer>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static void sort(long[] a, boolean increasing) {
ArrayList<Long> arr = new ArrayList<Long>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static void sort(double[] a, boolean increasing) {
ArrayList<Double> arr = new ArrayList<Double>();
int n = a.length;
for (int i = 0; i < n; i++) {
arr.add(a[i]);
}
Collections.sort(arr);
for (int i = 0; i < n; i++) {
if (increasing) {
a[i] = arr.get(i);
}
else {
a[i] = arr.get(n-1-i);
}
}
}
public static int lower_bound (int[] arr, int x) {
int low = 0, high = arr.length, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x)
high = mid;
else
low = mid + 1;
}
return low;
}
public static int upper_bound (int[] arr, int x) {
int low = 0, high = arr.length, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] > x)
high = mid;
else
low = mid + 1;
}
return low;
}
public static void updateMap(HashMap<Integer, Integer> map, int key, int v) {
if (!map.containsKey(key)) {
map.put(key, v);
}
else {
map.put(key, map.get(key) + v);
}
if (map.get(key) == 0) {
map.remove(key);
}
}
public static long gcd (long a, long b) {
return b == 0 ? a : gcd (b, a % b);
}
public static long lcm (long a, long b) {
return a * b / gcd (a, b);
}
public static long fast_pow_mod (long b, long x, int mod) {
if (x == 0) return 1;
if (x == 1) return b;
if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;
return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
}
public static long fast_pow (long b, long x) {
if (x == 0) return 1;
if (x == 1) return b;
if (x % 2 == 0) return fast_pow (b * b, x / 2);
return b * fast_pow (b * b, x / 2);
}
public static long choose (long n, long k) {
k = Math.min (k, n - k);
long val = 1;
for (int i = 0; i < k; ++i)
val = val * (n - i) / (i + 1);
return val;
}
public static long permute (int n, int k) {
if (n < k) return 0;
long val = 1;
for (int i = 0; i < k; ++i)
val = (val * (n - i));
return val;
}
// start of permutation and lower/upper bound template
public static void nextPermutation(int[] nums) {
//find first decreasing digit
int mark = -1;
for (int i = nums.length - 1; i > 0; i--) {
if (nums[i] > nums[i - 1]) {
mark = i - 1;
break;
}
}
if (mark == -1) {
reverse(nums, 0, nums.length - 1);
return;
}
int idx = nums.length-1;
for (int i = nums.length-1; i >= mark+1; i--) {
if (nums[i] > nums[mark]) {
idx = i;
break;
}
}
swap(nums, mark, idx);
reverse(nums, mark + 1, nums.length - 1);
}
public static void swap(int[] nums, int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
public static void reverse(int[] nums, int i, int j) {
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
static int lower_bound (int[] arr, int hi, int cmp) {
int low = 0, high = hi, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= cmp) high = mid;
else low = mid + 1;
}
return low;
}
static int upper_bound (int[] arr, int hi, int cmp) {
int low = 0, high = hi, mid = -1;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] > cmp) high = mid;
else low = mid + 1;
}
return low;
}
// end of permutation and lower/upper bound template
}
} | 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 11 | 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 | d3a03d7df5990dee7e3ba3bef0683e4a | 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.FileReader;
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 cf1659A {
// https://codeforces.com/problemset/problem/1659/A
public static void main(String[] args) {
Kattio io = new Kattio();
int t = io.nextInt(), chunk, r, b, rem;
char divider, fill;
for (int i = 0; i < t; i++) {
io.nextInt();
r = io.nextInt();
b = io.nextInt();
chunk = r / (b + 1);
rem = r % (b + 1);
while (r > 0) {
for (int k = 0; k < chunk; k++) {
if (r > 0) io.print('R');
r--;
}
if (rem > 0) {
io.print('R');
rem--;
r--;
}
if (b > 0) io.print('B');
b--;
}
io.println();
}
io.close();
}
// Kattio
static class Kattio extends PrintWriter {
private BufferedReader r;
private StringTokenizer st;
// standard input
public Kattio() { this(System.in,System.out); }
public Kattio(InputStream i, OutputStream o) {
super(o);
r = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public Kattio(String problemName) throws IOException {
super(problemName+".out");
r = new BufferedReader(new FileReader(problemName+".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(r.readLine());
return st.nextToken();
} catch (Exception e) {}
return null;
}
public String nextLine() {
try {
st = null;
return r.readLine();
} catch (Exception e) {}
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
}
| Java | ["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 11 | 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 | 0884e5bd9a67ab5b5c00fd3e096fad9b | 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 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 div=r/(b+1);
int rem=r%(b+1);
String v="";
StringBuilder sb=new StringBuilder();
StringBuilder nsb=new StringBuilder();
for(int i=0;i<div+1;i++)
{
sb.append('R');
}
for(int i=0;i<rem;i++)
{
nsb.append(sb);
nsb.append('B');
}
for(int i=rem;i<b;i++)
{
for(int j=0;j<div;j++)
nsb.append('R');
nsb.append('B');
}
for(int i=0;i<div;i++)
nsb.append('R');
String s=nsb.toString();
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 11 | 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 | 8b2593cd4f4495f98dfc51a513a7853e | 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 sc = new FastScanner();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int p = r/(b+1);
int q = r % (b+1);
for (int i = 0; i < (b+1); i++) {
for (int j = 0; j < r/(b+1); j++) {
System.out.print("R");
}
if(q > 0) {System.out.print("R");q--;}
if(i < b) System.out.print("B");
}
System.out.println();
}
}
}
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());
}
int[] sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int i : arr) list.add(i);
Collections.sort(list);
int[] res = new int[arr.length];
for (int i = 0; i < arr.length; i++) res[i] = list.get(i);
return res;
}
void debug(int a) {
System.out.println("This is var " + 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 11 | 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 | a40184bb478bdec60b436d6b19763998 | 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 Solution {
static boolean cases = true;
// Solution
static void solve(int t) {
int n = sc.nextInt(), r = sc.nextInt(), b = sc.nextInt();
int k = r / (b + 1);
int add = r % (b + 1);
while (r > 0) {
int need = k + ((add > 0) ? 1 : 0);
add = Math.max(0, add - 1);
r -= need;
while (need-- > 0) System.out.print("R");
if (b > 0) {
System.out.print("B");
--b;
}
}
System.out.println();
}
public static void main(String[] args) {
int c = 1;
for (int t = (cases ? sc.nextInt() : 0); t > 1; t--, c++) solve(c);
solve(c);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens()) try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() { return Long.parseLong(next()); }
}
private static final FastScanner sc = new FastScanner();
private static PrintWriter out = new PrintWriter(System.out);
} | Java | ["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 11 | 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 | efce98ae6a01c6481cd290aaccd60ccb | 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 Solution {
static boolean cases = true;
// Solution
static void solve(int t) {
int n = sc.nextInt(), r = sc.nextInt(), b = sc.nextInt();
int k = r / (b + 1);
int add = r % (b + 1);
while (r > 0) {
int need = k + ((add > 0) ? 1 : 0);
add = Math.max(0, add - 1);
r -= need;
while (need-- > 0) System.out.print("R");
if (b > 0) {
System.out.print("B");
--b;
}
}
System.out.println();
}
public static void main(String[] args) {
int c = 1;
for (int t = (cases ? sc.nextInt() : 0); t > 1; t--, c++) solve(c);
solve(c);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens()) try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
long nextLong() { return Long.parseLong(next()); }
}
private static final FastScanner sc = new FastScanner();
private static PrintWriter out = new PrintWriter(System.out);
} | Java | ["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 11 | 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 | f4bf42fe0160163bedd35cef2debe58f | 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.DataInputStream;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;
import java.util.function.Consumer;
import java.util.stream.Collectors;
import static java.lang.Math.*;
public class Round782_A {
@SuppressWarnings("FieldCanBeLocal")
private static Reader in;
private static PrintWriter out;
private static String YES = "YES";
private static String NO = "NO";
private static void solve() throws IOException{
//Your Code Goes Here;
int n = in.nextInt(), r = in.nextInt(), b = in.nextInt();
int k = r / (b + 1);
int bg = k + 1;
int nBg = r % (b + 1);
int nSm = b + 1 - nBg;
for(int i = 0; i < nSm; i++){
if(i != 0) System.out.print('B');
for(int j = 0; j < k; j++){
System.out.print('R');
}
}
for(int i = 0; i < nBg; i++){
System.out.print('B');
for(int j = 0; j < bg; j++){
System.out.print('R');
}
}
System.out.println();
}
public static void main(String[] args) throws IOException, InterruptedException {
in = new Reader();
out = new PrintWriter(new OutputStreamWriter(System.out));
int T = in.nextInt();
while(T --> 0){
solve();
}
out.flush();
in.close();
out.close();
}
static final Random random = new Random();
static final int mod = 1_000_000_007;
private static boolean check(long[] arr, long n){
long ch = arr[0];
for(int i = 0; i < n; i++){
if(ch != arr[i]){
return false;
}
}
return true;
}
//This function gives the max occuring of any element in an array;(HashMap version)
private static long maxFreqHashMap(long[] arr, long n){
Map<Long, Long> hp = new HashMap<Long, Long>();
for(int i = 0; i < n; i++) {
long key = arr[i];
if(hp.containsKey(key)) {
long freq = hp.get(key);
freq++;
hp.put(key, freq);
}
else {
hp.put(key, 1L);
}
}
long max_count = 0, res = 1, count = 0;
for(Map.Entry<Long, Long> val : hp.entrySet()) {
if (max_count < val.getValue()) {
res = Math.toIntExact(val.getKey());
max_count = Math.toIntExact(val.getValue());
count = max_count;
}
}
return count;
}
//This function gives the max occuring of any element in an array; this also known as the "MOORE's Algorithm"
private static long maxFreq(long []arr, long n) {
// using moore's voting algorithm
long res = 0;
long count = -1;
for(int i = 1; i < n; i++) {
if(arr[i] == arr[(int) res]) {
count++;
}
else {
count--;
}
if(count == 0) {
res = i;
count = 1;
}
}
return arr[(int) res];
/*
you've to add below code in the solve()
long freq = maxFreq(arr, n);
int count = 0;
for(int i = 0; i < n; i++){
if(arr[i] == freq){
count++;
}
}
*/
}
private static int LCSubStr(char[] X, char[] Y, int m, int n) {
int[][] LCStuff = new int[m + 1][n + 1];
int result = 0;
for (int i = 0; i <= m; i++){
for (int j = 0; j <= n; j++) {
if (i == 0 || j == 0) {
LCStuff[i][j] = 0;
}
else if (X[i - 1] == Y[j - 1]) {
LCStuff[i][j] = LCStuff[i - 1][j - 1] + 1;
result = Integer.max(result, LCStuff[i][j]);
}
else {
LCStuff[i][j] = 0;
}
}
}
return result;
}
private static long longCusBsearch(long[] arr, long n, long h){
long ans = h;
long l = 1;
long r = h;
while(l <= r){
long mid = (l + r) / 2;
long ok = 0;
for(long i = 0; i < n; i++){
if(i == n - 1) {
ok += mid;
}
else{
long x = arr[(int) (i + 1)] - arr[(int) i];
if(x >= mid) {
ok += mid;
}
else{
ok += x;
}
}
}
if(ok >= h){
ans = mid;
r = mid - 1;
}
else{
l = mid + 1;
}
}
return ans;
}
public static int intCusBsearch(int[] arr, int n, int h){
int ans = h, l = 1, r = h;
while(l <= r){
int mid = (l + r) / 2;
int ok = 0;
for(int i = 0; i < n; i++){
if(i == n - 1){
ok += mid;
}
else{
int x = arr[i + 1] - arr[i];
if(x >= mid){
ok += mid;
}
else{
ok += x;
}
}
}
if(ok >= h){
ans = mid;
r = mid - 1;
}
else{
l = mid + 1;
}
}
return ans;
}
/* Method to check if x is power of 2*/
private static boolean isPowerOfTwo (int x) {
/* First x in the below expression is
for the case when x is 0 */
return x!=0 && ((x&(x-1)) == 0);
}
//Taken From "Second Thread"
static void ruffleSort(int[] a){
int n = a.length;//shuffles, then sort;
for(int i=0; i<n; i++){
int oi = random.nextInt(n), temp = a[oi];
a[oi] = a[i]; a[i] = temp;
}
sort(a);
}
private static void longRuffleSort(long[] a){
long n = a.length;
for(long i = 0;i<n;i++){
long oi = random.nextInt((int) n), temp = a[(int) oi];
a[(int) oi] = a[(int) i]; a[(int) i] = temp;
}
longSort(a);
}
private static int gcd(int a, int b){
if(a == 0 || b == 0){
return 0;
}
while(b != 0){
int temp;
temp = a % b;
a = b;
b = temp;
}
return a;
}
private static long gcd(long a, long b){
if(a == 0 || b == 0){
return 0;
}
while(b != 0){
long temp;
temp = a % b;
a = b;
b = temp;
}
return a;
}
private static int lowestCommonMultiple(int a, int b){
return (a / gcd(a, b) * b);
}
/*
The Below func: the for loop runs in sqrt times. The second if is used to if the
divisors are equal to print only one of them otherwise we're printing both;
*/
private static void pDivisors(int n){
for(int i=1;i<Math.sqrt(n);i++){
if(n % i == 0){
if(n / i == i){
System.out.println(i + " ");
}
else{
System.out.println(i + " " + (n / i) + " ");
}
}
}
}
private static void returnNothing(){return;}
private static int numOfdigitsinN(int a){return (int) (Math.floor(Math.log10(a)) + 1);}
//prime Num till N: it takes any number and prints all the prime till that
//num;
private static boolean isPrime(int n){
if(n <= 1) return false;
if(n <= 3) return true;
//This is checked so that we can skip middle five number in below loop:
if(n % 2 == 0 || n % 3 == 0) return false;
for(int i = 5; i * i <= n; i = i + 6){
if(n % i == 0 || n % (i + 2) == 0){
return false;
}
}
return true;
}
//below func is to print the isPrime func();
private static void printTheIsPrimeFunc(int n){
for(int i=2;i<=n;i++){
if(isPrime(i)) System.out.println(i + " ");
}
}
public static boolean primeFinder(int n){
int m = 0;
int flag = 0;
m = n / 2;
if(n == 0 ||n == 1){
return false;
}
else{
for(int i=2;i<=m;i++){
if(n % i == 0){
return false;
}
}
return true;
}
}
private static boolean evenOdd(int num){
//System.out.println((num & 1) == 0 ? "EVEN" : "ODD");
return (num & 1) == 0 ? true : false;
}
private static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int)n + 1];
for (int i = 0; i <= n; i++) {
prime[i] = true;
}
for (long p = 2; p * p <= n; p++) {
if (prime[(int)p] == true) {
for (long i = p * p; i <= n; i += p)
prime[(int)i] = false;
}
}
return prime;
}
private static int log2(int n){
int result = (int) (Math.log(n) / Math.log(2));
return result;
}
private static long add(long a, long b){
return (a + b) % mod;
}
private static long lcm(long a, long b){
return (a / gcd((int) a, (int) b) * b);
}
private 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);
}
private static void longSort(long[] a){
ArrayList<Long> l = new ArrayList<Long>();
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 int[][] prefixsum( int n , int m , int arr[][] ){
int prefixsum[][] = new int[n+1][m+1];
for( int i = 1 ;i <= n ;i++) {
for( int j = 1 ; j<= m ; j++) {
int toadd = 0;
if( arr[i-1][j-1] == 1) {
toadd = 1;
}
prefixsum[i][j] = toadd + prefixsum[i][j-1] + prefixsum[i-1][j] - prefixsum[i-1][j-1];
}
}
return prefixsum;
}
//call this method when you want to read an integer array;
private static int[] readArray(int len) throws IOException{
int[] a = new int[len];
for(int i=0;i<len;i++)a[i] = in.nextInt();
return a;
}
//call this method when you want to read an Long array;
private static long[] readLongArray(int len) throws IOException{
long[] a = new long[len];
for(int i=0;i<len;i++) a[i] = in.nextLong();
return a;
}
//call this method to print the integer array;
private static void printArray(int[] array){
for(int now : array) out.print(now);out.print(' ');out.close();
}
//call this method to print the long array;
private static void printLongArray(long[] array){
for(long now:array) out.print(now); out.print(' '); out.close();
}
/*A way of Reading input...collected from a online blog
from here => https://codeforces.com/contest/1625/submission/144334744;*/
private static class Reader {
private static final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
Reader() {//Constructor;
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {//To take user input for String values;
final byte[] buf = new byte[1024]; // 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 nextSign() throws IOException {//For taking the signs like plus or minus ...
byte c = read();
while ('+' != c && '-' != c) {
c = read();
}
return '+' == c ? 0 : 1;
}
private static boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
public int skip() throws IOException {
int b;
// noinspection ALL
while ((b = read()) != -1 && isSpaceChar(b)) {
;
}
return b;
}
public char nc() throws IOException {
return (char) skip();
}
public String next() throws IOException {
int b = skip();
final StringBuilder sb = new StringBuilder();
while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = read();
}
return sb.toString();
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) {
return -ret;
}
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) {
buffer[0] = -1;
}
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) {
fillBuffer();
}
return buffer[bufferPointer++];
}
public void close() throws IOException {
din.close();
}
}
} | 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 11 | 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 | e3753ab2adbdd338b4311049acb25032 | 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.*;
public class second
{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String str = "";
int t = sc.nextInt();
for(int i=0;i<t;i++){
int n = sc.nextInt();
int r = sc.nextInt();
int b = sc.nextInt();
int d = r/(b+1);
int rem = r-(b+1)*d;
for(int z=0;z<d;z++){
str=str+"R"+"";
}
for(int f=0;f<b;f++){
if(rem>0){
str=str+"R";
rem--;
}
str=str+"B";
for(int z=0;z<d;z++){
str=str+"R"+"";
}
}
System.out.println(str);
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 11 | 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 | f4072ec4272ad2975cc7356b9121a0f4 | 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.math.*;
import java.util.*;
public class P03 {
static class FastReader{
BufferedReader br ;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(st==null || !st.hasMoreElements()){try {st = new StringTokenizer(br.readLine());}catch(IOException e) {e.printStackTrace();}}
return st.nextToken();
}
int nextInt(){return Integer.parseInt(next());}
long nextLong(){return Long.parseLong(next());}
double nextDouble(){return Double.parseDouble(next());}
float nextFloat(){return Float.parseFloat(next());}
String nextLine(){
String str = "";
try {
str = br.readLine();
}catch(IOException e) {
e.printStackTrace();
}
return str ;
}
}
static class BPair {
Integer index;boolean value;
public BPair(Integer index, boolean value) {this.value =value;this.index =index;}
int getIndex() {return index;}
boolean getValue() {return value;}
}
static class Pair implements Comparable<Pair> {
Integer index,value;
public Pair(Integer index, Integer value) {this.value = value;this.index = index;}
@Override public int compareTo(Pair o){return value - o.value;}
int getValue(){ return value;}
int getindex(){return index;}
}
private static long gcd(long l, long m){
if(m==0) {return l;}
return gcd(m,l%m);
}
static void swap(long[] arr, int i, int j){long temp = arr[i];arr[i] = arr[j];arr[j] = temp;}
static int partition(long[] arr, int low, int high){
long pivot = arr[high];
int i = (low - 1);
for(int j = low; j <= high - 1; j++){
if (arr[j] < pivot){
i++;
swap(arr, i, j);
}
}
swap(arr, i + 1, high);
return (i + 1);
}
static void quickSort(long[] arr, int low, int high){
if (low < high){
int pi = partition(arr, low, high);
quickSort(arr, low, pi - 1);
quickSort(arr, pi + 1, high);
}
}
static long M = 1000000007;
static long powe(long a,long b) {
long res =1;
while(b>0){
if((b&1)!=0) {
res=(res*a)&M;
}
a=(a*a)%M;
b>>=1;
}
return res;
}
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;
}
static boolean [] seive(int n) {
boolean a[] = new boolean[n+1];
Arrays.fill(a, true);
a[0]=false;
a[1]=false;
for(int i =2;i<=Math.sqrt(n);i++) {
for(int j =2*i;j<=n;j+=i) {
a[j]=false;
}
}
return a;
}
private static void pa(char[] cs){for(int i =0;i<cs.length;i++){System.out.print(cs[i]+" ");}System.out.println();}
private static void pa(long[] b){for(int i =0;i<b.length;i++) {System.out.print(b[i]+" ");}System.out.println();}
private static void pm(Character[][] a){for(int i =0;i<a.length;i++){for(int j=0;j<a[i].length;j++){System.out.print(a[i][j]);}System.out.println();}}
private static void pm(int [][] a) {for(int i =0;i<a.length;i++) {for(int j=0;j<a[0].length;j++) {System.out.print(a[i][j]+" ");}System.out.println();}}
private static void pa(int[] a){for(int i =0;i<a.length;i++){System.out.print(a[i]+" ");}System.out.println();}
private static boolean isprime(int n){if (n <= 1) return false;else if (n == 2) return true;else if (n % 2 == 0) return false;for (int i = 3; i <= Math.sqrt(n); i += 2){if (n % i == 0) return false;}return true;}
static int lb(long[] b, long a){
int l=-1,r=b.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(b[m]>=a) r=m;
else l=m;
}
return r;
}
static int ub(long a[], long 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;
}
static void rec(String digits,ArrayList<String> ans,String temp,int in ,HashMap<Integer,String>hm,int l){
if(temp.length()==l){
ans.add(temp);
return ;
}
String s = hm.get(digits.charAt(in)-'0');
int n = s.length();
for(int i =0;i<n;i++){
rec(digits,ans,temp+s.charAt(i),in+1,hm,l);
}
}
public static int longestCommonSubsequence(String text1, String text2) {
int n = text1.length();
int m = text2.length();
int dp [][]= new int [n+1][m+1];
for(int i =0;i<=n;i++){
for(int j =0;j<=m;j++){
if(i==0||j==0)dp[i][j]=0;
else if(text1.charAt(i-1)==text2.charAt(j-1)){
dp[i][j]=1+dp[i-1][j-1];
}
else dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
}
}
return dp[n][m];
}
static //MAIN FUNCTION ------->
ArrayList<String> alpos = new ArrayList<>();
public static void main(String[] args) throws IOException {
long start2 = System.currentTimeMillis();
// FastReader fs = new FastReader();
Scanner fs = new Scanner (System.in);
int t = fs.nextInt();
// int t =1;
while(t-->0) {
int n = fs.nextInt();
int r = fs.nextInt();
int b = fs.nextInt();
int x = (int)Math.floor((double)r/(double)(b+1));
int k = r%(b+1);
// System.out.println(x+" "+k);
String ans ="";
for(int i =0;i<b+1;i++) {
for(int j=0;j<x;j++) {
ans+='R';
}
if(k>0) {
ans+='R';
k--;
}
if(i!=b)ans+='B';
}
System.out.println(ans);
}
long end2 = System.currentTimeMillis();
// System.out.println("time :"+(end2-start2));
// 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 11 | 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 | 26436ca2e3e509def9206f8d7b86a7a6 | 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.util.Map.Entry;
import java.math.*;
import java.sql.Array;
public class Simple{
public static class Pair implements Comparable<Pair>{
int x;
int y;
public Pair(int x,int y){
this.x = x;
this.y = y;
}
public int compareTo(Pair other){
return (int) (this.y - other.y);
}
public boolean equals(Pair other){
if(this.x == other.x && this.y == other.y)return true;
return false;
}
public int hashCode(){
return 31*x + y;
}
// @Override
// public int compareTo(Simple.Pair o) {
// // TODO Auto-generated method stub
// return 0;
// }
}
static int power(int x, int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0) {
// If y is odd, multiply x
// with result
if (y % 2 == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns n^(-1) mod p
static int modInverse(int n, int p)
{
return power(n, p - 2, p);
}
// Returns nCr % p using Fermat's
// little theorem.
static int nCrModPFermat(int n, int r,
int p)
{
if (n<r)
return 0;
// Base case
if (r == 0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
int[] fac = new int[n + 1];
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i % p;
return (fac[n] * modInverse(fac[r], p)
% p * modInverse(fac[n - r], p)
% p)
% p;
}
static int nCrModp(int n, int r, int p)
{
if (r > n - r)
r = n - r;
// The array C is going to store last
// row of pascal triangle at the end.
// And last entry of last row is nCr
int C[] = new int[r + 1];
C[0] = 1; // Top row of Pascal Triangle
// One by constructs remaining rows of Pascal
// Triangle from top to bottom
for (int i = 1; i <= n; i++) {
// Fill entries of current row using previous
// row values
for (int j = Math.min(i, r); j > 0; j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
C[j] = (C[j] + C[j - 1]) % p;
}
return C[r];
}
static int gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
public static class Node{
int root;
ArrayList<Node> al;
Node par;
public Node(int root,ArrayList<Node> al,Node par){
this.root = root;
this.al = al;
this.par = par;
}
}
// public static int helper(int arr[],int n ,int i,int j,int dp[][]){
// if(i==0 || j==0)return 0;
// if(dp[i][j]!=-1){
// return dp[i][j];
// }
// if(helper(arr, n, i-1, j-1, dp)>=0){
// return dp[i][j]=Math.max(helper(arr, n, i-1, j-1,dp)+arr[i-1], helper(arr, n, i-1, j,dp));
// }
// return dp[i][j] = helper(arr, n, i-1, j,dp);
// }
// public static void dfs(ArrayList<ArrayList<Integer>> al,int levelcount[],int node,int count){
// levelcount[count]++;
// for(Integer x : al.get(node)){
// dfs(al, levelcount, x, count+1);
// }
// }
public static long __gcd(long a, long b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
public static long helper(int arr1[][],int m){
Arrays.sort(arr1, (int a[],int b[])-> a[0]-b[0]);
long ans =0;
for(int i=1;i<m;i++){
long count =0;
for(int j=0;j<i;j++){
if(arr1[i][1] > arr1[j][1])count++;
}
ans+=count;
}
return ans;
}
public static class DSU{
int n;
int par[];
int rank[];
public DSU(int n){
this.n = n;
par = new int[n+1];
rank = new int[n+1];
for(int i=1;i<=n;i++){
par[i] = i ;
rank[i] = 0;
}
}
public int findPar(int node){
if(node==par[node]){
return node;
}
return par[node] = findPar(par[node]);//path compression
}
public void union(int u,int v){
u = findPar(u);
v = findPar(v);
if(rank[u]<rank[v]){
par[u] = v;
}
else if(rank[u]>rank[v]){
par[v] = u;
}
else{
par[v] = u;
rank[u]++;
}
}
}
public static void dfs(ArrayList<ArrayList<Integer>> adj, int l[],int r[],boolean vis[],int node,long dp[][]){
long ansl = 0;
long ansr = 0;
vis[node] = true;
for(Integer x : adj.get(node)){
if(!vis[x]){
vis[x] = true;
dfs(adj, l, r, vis, x, dp);
ansl+=Math.max(Math.abs(l[node]-l[x])+dp[x][0], Math.abs(l[node]-r[x])+dp[x][1]);
ansr+=Math.max(Math.abs(r[node]-l[x])+dp[x][0], Math.abs(r[node]-r[x])+dp[x][1]);
}
}
dp[node][0] = ansl;
dp[node][1] = ansr;
}
public static void main(String args[]){
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int t1 = 1;t1<=t;t1++){
int n = s.nextInt();
int r = s.nextInt();
int l = s.nextInt();
String str ="";
// int count = 0;
// int part1 = r/(l+1);
// if(r%(l+1)!=0)part1++;
ArrayList<Integer> al = new ArrayList<>();
int l1 =l;
for(int i=1;i<=l1;i++){
int num = r/(l+1);
al.add(num);
r-=num;
l--;
}
int k = 0;
// System.out.println(part);
// System.out.println(al.toString());
int count = 0;
for(int i=1;i<=n;i++){
if(k<l1&&al.get(k)==count){
str+='B';
k++;
count =0;
}
else{
str+='R';
count++;
}
}
System.out.println(str);
}
}
}
/*
4 2 2 7
0 2 5
-2 3
5
0*x1 + 1*x2 + 2*x3 + 3*x4
*/ | 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 11 | 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 | cd3a01bacc0553d293ea087876521871 | 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 Firstclass {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-->0){
int n =s.nextInt(),r=s.nextInt(),b=s.nextInt();
int res = r/(b+1);
int num = r%(b+1);
int cont=0;
for (int i = 0; i < n; i++) {
if (cont!=res){
System.out.print("R");
cont++;
}
else {
if(num>0){
System.out.print("R");
num--;
i++;
}
System.out.print("B");
cont=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 11 | 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 | 667ff91ead0356eb7e6fa8cd9b7496d4 | 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_Red_Versus_Blue {
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
long t = sc.nextLong();
while (t-- > 0)
{
int n=sc.nextInt();
long r=sc.nextLong();
long b=sc.nextLong();
int a[]=new int[n];
Arrays.fill(a, 1);
int i=0;
while(i<n){
int gap= (int) Math.ceil((float)r/(b+1));
while(gap>0)
{
a[i]=1;
gap--;
r--;
i++;
}
if (i<n)
{
a[i]=0;
b--;
i++;
}
}
for(i=0;i<n;i++)
{
if(a[i]==1)
{
System.out.print("R");
}
else{
System.out.print("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 11 | 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 | 0f19f36a08fe9e5fe8018593510217ec | 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_Red_Versus_Blue {
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
long t = sc.nextLong();
while (t-- > 0)
{
int n=sc.nextInt();
long r=sc.nextLong();
long b=sc.nextLong();
int a[]=new int[n];
Arrays.fill(a, 1);
int m=(int) Math.ceil((double)n/(b+1));
int p=m-1;
int i=0;
while(i<n){
int gap= (int) Math.ceil((float)r/(b+1));
while(gap>0){
a[i]=1;
gap--;
r--;
i++;
}
if (i<n){
a[i]=0;
b--;
i++;
}
}
// System.out.println("m="+m);
// while(b>0)
// {
// if(p>=n-m && b!=0)
// {
// a[p]=0;
// p++;
// b--;
// }
// else
// {a[p]=0;
// b--;
// p=p+m;}
// }
for(i=0;i<n;i++)
{
if(a[i]==1)
{
System.out.print("R");
}
else{
System.out.print("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 11 | 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 | 738bc9f1b1b3638957a38f5207dc1609 | 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.StringTokenizer;
public class redVsBlue {
public static void main(String[] args) throws IOException {
Reader x = new Reader();
int t = x.nextInt();
String[] answers = new String[t];
int n, r, b, p;
StringBuilder s, y;
for (int i = 0; i < t; i++) {
n = x.nextInt();
r = x.nextInt();
b = x.nextInt();
p = r % (b + 1);
s = new StringBuilder();
y = new StringBuilder();
for (int j = 0; j < r / (b + 1); j++) {
y.append("R");
}
for (int j = 0; j < b + 1; j++) {
if (j > 0) {
s.append("B");
}
s.append(y);
if (p > 0) {
s.append("R");
p--;
}
}
answers[i] = s.toString();
}
for (int i = 0; i < t; i++) {
System.out.println(answers[i]);
}
}
public String addChar(String str, char ch, int position) {
return str.substring(0, position) + ch + str.substring(position);
}
static class Reader {
String filename;
BufferedReader br;
StringTokenizer st;
PrintWriter out;
boolean fileOut = false;
public Reader() {
this.br = new BufferedReader(new InputStreamReader(System.in));
}
public Reader(String name) throws IOException {
this.filename = name;
this.br = new BufferedReader(new FileReader(this.filename + ".in"));
this.out = new PrintWriter(new BufferedWriter(new FileWriter(this.filename + ".out")));
this.fileOut = true;
}
public String next() throws IOException {
while(this.st == null || !this.st.hasMoreTokens()) {
try {
this.st = new StringTokenizer(this.br.readLine());
} catch (Exception var2) {
}
}
return this.st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(this.next());
}
public void println(Object text) {
if (this.fileOut) {
this.out.println(text);
} else {
System.out.println(text);
}
}
public void close() {
if (this.fileOut) {
this.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 11 | 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 | 2a476c97eb2d84360e212c9ab7bfecf3 | 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[] strgs) {
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();
String s="";
int limit=r/(b+1);
for(int i=0;i<limit;i++) {
s+="R";
}
int rem=r%(b+1);
int equal=b+1-rem;
String ans="";
for(int i=1;i<=equal;i++) {
ans+=s+"B";
}
for(int i=1;i<=rem;i++) {
ans+=s+"RB";
}
System.out.println(ans.substring(0,n));
}
}
}
| 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 11 | 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 | afa6b8b50c2ac404e67cf15c9c97d536 | 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 red_vs_blue {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
StringBuffer str = new StringBuffer("");
while(t-- > 0)
{
int N = sc.nextInt();
int R = sc.nextInt();
int B = sc.nextInt();
int rem = 0, que = 0;
que = R/B;
rem = R%B;
int rint = 0, bint = 0;
if(B == 1)
{
que = R/2;
int temp = que;
while(temp-- > 0)
{
str.append('R');
R--;
rint++;
}
if(B-- > 0)
{
str.append('B');
bint++;
}
while(R-- > 0)
{
str.append('R');
rint++;
}
}
else
{
que = (R/(B+1));
rem = (R%(B+1));
int temp = 0;
while(rem-- > 0)
{
temp = que+1;
while(temp-- > 0)
{
str.append('R');
R--;
rint++;
}
if(B-- > 0)
{
str.append('B');
bint++;
}
}
while(str.length() < N)
{
temp = que;
while(temp-- > 0 && R > 0)
{
str.append('R');
R--;
rint++;
}
if(B-- > 0)
{
str.append('B');
bint++;
}
}
}
System.out.println(str);
str.delete(0, str.length());
str.append("");
}
}
}
| 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 11 | 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 | 8e019cb4cd0dcaa8a63d0bfa07cae436 | 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.Collections;
import java.util.List;
import java.util.Scanner;
public class meomeo{
public static void main(String[] args) {
try(final Scanner scanner = new Scanner(System.in)){
int noTests = scanner.nextInt();
while(noTests-- >0){
final Integer n = scanner.nextInt();
final Integer r = scanner.nextInt();
Integer b = scanner.nextInt();
Integer noR = 0;
Integer diffNoR = 0;
StringBuilder sb = new StringBuilder("B");
if(b== 1 && r%2 == 0){
noR = r/2;
String startString = "R".repeat(noR);
sb.insert(0, startString);
sb.append(startString);
System.out.println(sb.toString());
continue;
}
if(b==1&& r%2!=0){
noR = r/2;
diffNoR = r/2 + 1;
String startString = "R".repeat(diffNoR);
String endString = "R".repeat(noR);
sb.insert(0, startString);
sb.append(endString);
System.out.println(sb.toString());
continue;
}
String res = "B".repeat(b);
//res = res + "";
StringBuilder result = new StringBuilder(res);
final Integer oldB = b;
b = b+1;
Integer divStringR = r/b;
Integer modStringR = r%b;
if(modStringR != 0 && modStringR <= oldB && modStringR !=1){
divStringR +=1;
}
if(modStringR ==0 && divStringR <=b){
modStringR +=1;
}
String stringR = "R".repeat(divStringR);
String endString = "R".repeat(modStringR);
Integer count = 0;
Integer countB = -1;
for(int i = 0; i<n;i++){
if(result.toString().charAt(i) == 'B'){
countB++;
if(i==0){
result.insert(i, stringR);
count += stringR.length();
}
if(i != 0 && i!= result.toString().length() -1 ){
if((r-count)/(b-countB) < divStringR -1){
divStringR -=1;
stringR = "R".repeat(divStringR);
}
result.insert(i+1, stringR);
count += stringR.length();
}
if(i == result.length() - 1){
Integer getIndex = n - result.length();
if(getIndex < endString.length()){
String subString = endString.substring(0, getIndex);
result.append(subString);
}
else {
endString = "R".repeat(getIndex);
result.append(endString);
//break;
}
}
}
}
System.out.println(result.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 11 | 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 | fa489e56f770c946f6f1c7be17eaa790 | 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 Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int r = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
int div = r / (b+1);
int mod = r % (b+1);
for(int i = 0; i < (b+1); i++)
{
int temp = div;
if(mod-->0)
temp++;
while(temp-->0)
output.write("R");
if(i!=b)
output.write("B");
}
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// int k = Integer.parseInt(st.nextToken());
// char arr[] = br.readLine().toCharArray();
// output.write();
// int n = Integer.parseInt(st.nextToken());
// HashMap<Character, Integer> map = new HashMap<Character, Integer>();
// if
// output.write("YES\n");
// else
// output.write("NO\n");
// long a = Long.parseLong(st.nextToken());
// long b = Long.parseLong(st.nextToken());
// if(flag == 1)
// output.write("NO\n");
// else
// output.write("YES\n" + x + " " + y + " " + z + "\n");
// output.write(n+ "\n");
}
output.flush();
}
} | Java | ["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 11 | 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 | f850386e0fcb7808425589807f189a87 | 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 Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
StringTokenizer st;
while (t-- > 0) {
// int n = Integer.parseInt(br.readLine());
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int r = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
int r2 = r;
int nchunks = r2/(b+1);
int addr = r % (b+1);
while(r2 > 0)
{
int mkl = nchunks;
if(addr>0)
mkl++;
addr = Math.max(0, addr - 1);
r2-=mkl;
while(mkl-->0)
{
output.write("R");
}
if(b > 0)
{
output.write("B");
b--;
}
}
output.write("\n");
// int arr[] = new int[n];
// for(int i = 0; i < n; i++)
// {
// arr[i] = e;
// }
// int k = Integer.parseInt(st.nextToken());
// char arr[] = br.readLine().toCharArray();
// output.write();
// int n = Integer.parseInt(st.nextToken());
// HashMap<Character, Integer> map = new HashMap<Character, Integer>();
// if
// output.write("YES\n");
// else
// output.write("NO\n");
// long a = Long.parseLong(st.nextToken());
// long b = Long.parseLong(st.nextToken());
// if(flag == 1)
// output.write("NO\n");
// else
// output.write("YES\n" + x + " " + y + " " + z + "\n");
// output.write(n+ "\n");
}
output.flush();
}
} | Java | ["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 11 | 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 | 2e111341f21e0172d4a7e2cbd2fc6e50 | 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 Scanner obj = new Scanner(System.in);
public static PrintWriter out = new PrintWriter(System.out);
public static int i() {
return obj.nextInt();
}
public static void main(String[] args) {
int len = i();
while (len-- != 0) {
int n=obj.nextInt();
int r=obj.nextInt();
int b=obj.nextInt();
int val=r/(b+1);
int in=0;
StringBuffer sb=new StringBuffer("");
while(b>0)
{
int c=1;
while(r>0 && c<=val)
{
sb.append("R");
r--;
c++;
}
sb.append("B");
in=sb.length();
b--;
}
for(int i=0;i<val && r>0;i++,r--) sb.append("R");
StringBuffer ans=new StringBuffer("");
for(int i=0;i<sb.length();i++)
{
if(sb.charAt(i)=='B' && r-1>=0)
{
ans.append("R");
r--;
}
ans.append(sb.charAt(i));
}
out.println(ans.toString());
}
out.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 11 | 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 | 7f2bb9d12515eb029d042919a6da1996 | 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.util.*;
public class Test2 {
static Map<Integer,Integer> mp=new HashMap();
private static int fun(int n)
{
if(n==0)
return 0;
if(mp.containsKey(n))
return mp.get(n);
else {
mp.put(n, 1+fun(n^fun2(n)));
return mp.get(n);
}
}
private static int fun2(int n)
{
StringBuilder s=new StringBuilder(Integer.toBinaryString(n));
s.reverse();
int decimal=Integer.parseInt(s.toString(),2);
return decimal;
}
static int findGcd(int x, int y)
{
if (x == 0)
//returns y is x==0
return y;
//calling function that returns GCD
return findGcd(y % x, x);
}
//function finds the LCM
static int findLcm(int x, int y)
{
//returns the LCM
return (x / findGcd(x, y)) * y;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t--!=0)
{
int n=sc.nextInt();
int r=sc.nextInt();
int b=sc.nextInt();
String s="";
int rt=r/(b+1);
for(int i=b+1;i>=1;i--)
{
for(int j=0;j<r/i;j++)
System.out.print('R');
r-=r/i;
if(i>1)
System.out.print('B');
}
System.out.println();
}
}
}
/*9
3 5 1 4
WBWWW
BBBWB
WWBBB
4 3 2 1
BWW
BBW
WBB
WWB
2 3 2 2
WWW
WWW
2 2 1 1
WW
WB
5 9 5 9
WWWWWWWWW
WBWBWBBBW
WBBBWWBWW
WBWBWBBBW
WWWWWWWWW
1 1 1 1
B
1 1 1 1
W
1 2 1 1
WB
2 1 1 1
W
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 11 | 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 | f636b0649d9f0ea63c23778513bd19f5 | 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 spoj;
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
//FileReader fr = new FileReader(new File("input.txt"));
InputStreamReader sr = new InputStreamReader(System.in);
BufferedReader reader = new BufferedReader(sr);
StringBuilder sb = new StringBuilder();
int t = Integer.parseInt(reader.readLine());
while (t-- > 0) {
String[] s = reader.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int r = Integer.parseInt(s[1]);
int b = Integer.parseInt(s[2]);
StringBuilder res = new StringBuilder();
int x = r / (b + 1);
int rem = r % (b + 1);
for (int i = 0; i < n; i += x) {
for (int j = 0; j < x; j++) {
res.append('R');
}
if (rem > 0) {
res.append('R');
rem--;
}
if (b > 0) {
res.append('B');
b--;
}
}
sb.append(res.subSequence(0, n) + "\n");
}
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 11 | 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 | 7a8487a5dd1056d0a26fde92738626b8 | 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 : Ryan Ranaut
__Hope is a big word, never lose it__
------------- --------------*/
import java.io.*;
import java.util.*;
public class Codeforces1 {
static PrintWriter out = new PrintWriter(System.out);
static final int mod = 1_000_000_007;
static long min = (long) (-1e16);
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
/*--------------------------------------------------------------------------*/
//Try seeing general case
//Minimization Maximization - BS..... Connections - Graphs.....
//Greedy not worthy - Try DP
public static void main(String[] args) {
FastReader s = new FastReader();
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
int r = s.nextInt();
int b = s.nextInt();
out.println(find(n, r, b));
}
out.close();
}
public static String find(int n, int r, int b)
{
StringBuilder sb = new StringBuilder();
int gap = r/(b+1);
int rem = r%(b+1);
for(int i=0;i<b+1;i++)
{
for(int j=0;j<gap;j++)
sb.append("R");
if(rem>0)
{
rem--;
sb.append("R");
}
if(i == b)
break;
sb.append("B");
}
return sb.toString();
}
/*----------------------------------End of the road--------------------------------------*/
static class DSU {
int[] parent;
int[] ranks;
int[] groupSize;
int size;
public DSU(int n) {
size = n;
parent = new int[n];//0 based
ranks = new int[n];
groupSize = new int[n];//Size of each component
for (int i = 0; i < n; i++) {
parent[i] = i;
ranks[i] = 1;
groupSize[i] = 1;
}
}
public int find(int x)//Path Compression
{
if (parent[x] == x)
return x;
else
return parent[x] = find(parent[x]);
}
public void union(int x, int y)//Union by rank
{
int x_rep = find(x);
int y_rep = find(y);
if (x_rep == y_rep)
return;
if (ranks[x_rep] < ranks[y_rep]) {
parent[x_rep] = y_rep;
groupSize[y_rep] += groupSize[x_rep];
} else if (ranks[x_rep] > ranks[y_rep]) {
parent[y_rep] = x_rep;
groupSize[x_rep] += groupSize[y_rep];
} else {
parent[y_rep] = x_rep;
ranks[x_rep]++;
groupSize[x_rep] += groupSize[y_rep];
}
size--;//Total connected components
}
}
public static int gcd(int x, int y) {
return y == 0 ? x : gcd(y, x % y);
}
public static long gcd(long x, long y) {
return y == 0L ? x : gcd(y, x % y);
}
public static int lcm(int a, int b) {
return (a * b) / gcd(a, b);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static long pow(long a, long b) {
if (b == 0L)
return 1L;
long tmp = 1;
while (b > 1L) {
if ((b & 1L) == 1)
tmp *= a;
a *= a;
b >>= 1;
}
return (tmp * a);
}
public static long modPow(long a, long b, long mod) {
if (b == 0L)
return 1L;
long tmp = 1;
while (b > 1L) {
if ((b & 1L) == 1L)
tmp *= a;
a *= a;
a %= mod;
tmp %= mod;
b >>= 1;
}
return (tmp * a) % mod;
}
static long mul(long a, long b) {
return a * b;
}
static long fact(int n) {
long ans = 1;
for (int i = 2; i <= n; i++) ans = mul(ans, i);
return ans;
}
static long fastPow(long base, long exp) {
if (exp == 0) return 1;
long half = fastPow(base, exp / 2);
if (exp % 2 == 0) return mul(half, half);
return mul(half, mul(half, base));
}
static void debug(int... a) {
for (int x : a)
out.print(x + " ");
out.println();
}
static void debug(long... a) {
for (long x : a)
out.print(x + " ");
out.println();
}
static void debugMatrix(int[][] a) {
for (int[] x : a)
out.println(Arrays.toString(x));
}
static void debugMatrix(long[][] a) {
for (long[] x : a)
out.println(Arrays.toString(x));
}
static void reverseArray(int[] a) {
int n = a.length;
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void sort(int[] a) {
ArrayList<Integer> ls = new ArrayList<>();
for (int x : a) ls.add(x);
Collections.sort(ls);
for (int i = 0; i < a.length; i++)
a[i] = ls.get(i);
}
static void sort(long[] a) {
ArrayList<Long> ls = new ArrayList<>();
for (long x : a) ls.add(x);
Collections.sort(ls);
for (int i = 0; i < a.length; i++)
a[i] = ls.get(i);
}
static class Pair {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
} | 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 11 | 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 | d0c56a1a12a07f9b0bb132e5c5c91106 | 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.*;
/**
* A simple template for competitive programming problems.
*/
public class Solution {
//InputReader in = new InputReader("input.txt");
final InputReader in = new InputReader(System.in);
final PrintWriter out = new PrintWriter(System.out);
static final int mod = 1000000007;
void solve() {
int t = in.nextInt();
while(t-->0) {
int n = in.nextInt();
int r = in.nextInt();
int b = in.nextInt();
int div = r/(b+1);
int rem = r%(b+1);
for(int i=0; i<b; i++) {
for(int j=0; j<div; j++) {
out.print("R");
}
if(rem>0) {
out.print("R"); rem--;
}
out.print("B");
}
for(int i=0; i<div; i++) {
out.print("R");
}
out.println();
}
}
public static void main(final String[] args) throws FileNotFoundException {
final Solution s = new Solution();
final Long t1 = System.currentTimeMillis();
s.solve();
System.err.println(System.currentTimeMillis() - t1 + " ms");
s.out.close();
}
public Solution() throws FileNotFoundException {
}
private static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[1024];
private int curChar;
private int numChars;
Random r = new Random();
InputReader(final InputStream stream) {
this.stream = stream;
}
InputReader(final String fileName) {
InputStream stream = null;
try {
stream = new FileInputStream(fileName);
} catch (final FileNotFoundException e) {
e.printStackTrace();
}
this.stream = stream;
}
int[] nextArray(final int n) {
final int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
int[][] nextMatrix(final int n, final int m) {
final int[][] matrix = new int[n][m];
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
matrix[i][j] = nextInt();
return matrix;
}
String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
final StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
final StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
int[] randomArray(int n, int low, int up) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = low + r.nextInt(up - low + 1);
}
return arr;
}
double nextDouble() {
return Double.parseDouble(nextString());
}
private int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (final IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
private boolean isSpaceChar(final int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(final int c) {
return c == '\n' || c == '\r' || c == -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 11 | 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 | f5c8af0bc9d33e472d095e9ceb2ca657 | 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 test {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(System.out);
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
StringTokenizer tokenizer = new StringTokenizer(br.readLine());
int n = Integer.parseInt(tokenizer.nextToken());
int r = Integer.parseInt(tokenizer.nextToken());
int b = Integer.parseInt(tokenizer.nextToken());
pw.println(solve(n, r, b));
}
pw.flush();
pw.close();
br.close();
}
public static String solve(int n, int r, int b) {
StringBuilder result = new StringBuilder();
int parts = r / (b + 1);
int rem = r % (b + 1);
for (int i = 0; i < n; i++) {
int count = 0;
while (count < parts && r > 0) {
result.append("R");
count++;
r--;
}
if (rem > 0) {
result.append("R");
r--;
rem--;
}
if (b > 0) {
result.append("B");
b--;
}
if (r == 0 && b == 0)
break;
}
return result.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 11 | 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 | 2fea392e4d12fef5e1dba7256b0482ae | 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 scanner = new Scanner(System.in);
int t = scanner.nextInt();
while (t-- > 0) {
int n = scanner.nextInt(), r = scanner.nextInt(), b = scanner.nextInt(), cr = 0, cb = 0;
double max = Math.ceil((double)r/(b + 1));
for (int i = 0, j = 0; i < n; i++) {
if (j >= max && b > 0) {
System.out.print('B');
j = 0;
b--;
max = Math.ceil((double)r/(b + 1));
}
else {
System.out.print('R');
j++;
r--;
}
// System.out.print('R');
// cr++;
// r--;
// if (cr > (double)r/(b + 1) && b > 0 && i < n - 1) {
// System.out.print('B');
// cb++;
// b--;
// cr = 0;
// 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 11 | 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 | fb059c13377b7e2ab2e4dad94c0072b6 | 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 Home {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
int t = sc.nextInt();
for (int i = 0; i < t; ++i) {
int n = sc.nextInt(), r = sc.nextInt(), b = sc.nextInt();
int div = r/(b+1);
int[] points = new int[b+1];
Arrays.fill(points, div);
int rem = r % (div*(b+1));
for (int j = 0; j < rem; ++j) {
points[j] += 1;
}
for (int j = 0; j < points.length-1; ++j) {
System.out.print("R".repeat(points[j]));
System.out.print("B");
}
System.out.println("R".repeat(points[points.length-1]));
}
}
}
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());
}
double nextDouble() {
return Double.parseDouble(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 11 | 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 | de16a9c7ba3f542c1dbc11ddf30725fa | 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 spidername {
static Scanner sc = new Scanner(System.in);
static int nxt() {
int x = sc.nextInt();
return x;
}
static void solve() {
int n = nxt();
int r = nxt();
int b = nxt();
String ans = "";
int m = (r) / (b + 1);
int cnt = 0;
int extra = r % (b + 1);
for (int i = 0; i < r; i++) {
ans += 'R';
cnt++;
if (extra > 0) {
if (cnt == m + 1 && b > 0) {
cnt = 0;
extra--;
b--;
ans += 'B';
}
} else {
if (cnt == m && b > 0) {
cnt = 0;
extra--;
b--;
ans += 'B';
}
}
}
while (b > 0) {
ans += 'B';
b--;
}
System.out.println(ans);
}
public static void main(String args[]) {
int t = nxt();
while (t-- > 0) {
solve();
}
}
}
| 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 11 | 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 | 08fe4a135338279ef0228df9d2b52035 | 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.IOException;
import java.io.InputStream;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;
import java.util.*;
import static java.util.Arrays.*;
public class CodeforcesTemp {
public static void main(String[] args) throws IOException {
Reader scan = new Reader();
FastPrinter out = new FastPrinter();
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--;
}
out.println(sb.toString());
}
out.close();
}
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(int length) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = this.nextLong();
return array;
}
public long[] nextLongArray(int length, java.util.function.LongUnaryOperator map) {
long[] array = new long[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsLong(this.nextLong());
return array;
}
public int[] nextIntArray(int length) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = this.nextInt();
return array;
}
public int[][] nextIntArrayMulti(int length, int width) {
int[][] arrays = new int[width][length];
for (int i = 0; i < length; i++) {
for (int j = 0; j < width; j++)
arrays[j][i] = this.nextInt();
}
return arrays;
}
public int[] nextIntArray(int length, java.util.function.IntUnaryOperator map) {
int[] array = new int[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsInt(this.nextInt());
return array;
}
public double[] nextDoubleArray(int length) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = this.nextDouble();
return array;
}
public double[] nextDoubleArray(int length, java.util.function.DoubleUnaryOperator map) {
double[] array = new double[length];
for (int i = 0; i < length; i++)
array[i] = map.applyAsDouble(this.nextDouble());
return array;
}
public long[][] nextLongMatrix(int height, int width) {
long[][] mat = new long[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextLong();
}
return mat;
}
public int[][] nextIntMatrix(int height, int width) {
int[][] mat = new int[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextInt();
}
return mat;
}
public double[][] nextDoubleMatrix(int height, int width) {
double[][] mat = new double[height][width];
for (int h = 0; h < height; h++)
for (int w = 0; w < width; w++) {
mat[h][w] = this.nextDouble();
}
return mat;
}
public char[][] nextCharMatrix(int height, int width) {
char[][] mat = new char[height][width];
for (int h = 0; h < height; h++) {
String s = this.next();
for (int w = 0; w < width; w++) {
mat[h][w] = s.charAt(w);
}
}
return mat;
}
}
static class 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]);
}
}
}
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 shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void rsort(int[] a) {shuffle(a); sort(a);}
static void rsort(long[] a) {shuffle(a); sort(a);}
static void rsort(double[] a) {shuffle(a); sort(a);}
static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
}
| 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 11 | 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 | 5d559aba4e2b89ef5d3215f13f896370 | 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 | /*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
//package CodeForces782Div2;
import java.util.Scanner;
/**
*
* @author jordy
*/
public class redvblue
{
static String ans;
static int n, r, b;
public static void main(String[] args)
{
Scanner reader = new Scanner(System.in);
int t = reader.nextInt();
for (int i = 0; i < t; i++)
{
//System.out.println("--------------");
n = reader.nextInt();
r = reader.nextInt();
b = reader.nextInt();
// r/(b+1) per region
// r%(b+1) of them will have 1 extra
StringBuilder build = new StringBuilder();
int perregion = (r/(b+1));
int extra = r%(b+1);
for (int j = 0; j < b+1; j++)
{
for (int k = 0; k < perregion; k++)
{
build.append("R");
}
if(extra>0)
{
build.append("R");
extra--;
}
if(j!=b)build.append("B");
}
System.out.println(build.toString());
}
}
}
/*
1
35 30 5
1
50 40 10
1
11 7 4
3
7 4 3
6 5 1
19 13 6
5
98 74 24
83 59 24
89 88 1
100 76 24
82 58 24
1
83 59 24
*/
| 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 11 | 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 | 270dd2406feb8e00bb1f7ee18edbbcc6 | 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 Group{
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 idx=r/(b+1);
int ext=r%(b+1);
while(b>0){
for(int i=0; i<idx; i++){
System.out.print('R');
r--;
}
if(ext>0){
System.out.print('R');
ext--;
r--;
}
System.out.print('B');
b--;
}
while(r-->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 11 | 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 | 96bd017ad974a0f52768150e26dea929 | 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 | /***** ---> :) Vijender Srivastava (: <--- *****/
import java.util.*;
import java.lang.*;
// import java.lang.reflect.Array;
import java.io.*;
public class Main
{
static FastReader sc =new FastReader();
static PrintWriter out=new PrintWriter(System.out);
static long mod=(long)32768;
static StringBuilder sb = new StringBuilder();
/* start */
public static void main(String [] args)
{
// int testcases = 1;
int testcases = i();
// calc();
while(testcases-->0)
{
solve();
}
out.flush();
out.close();
}
static void solve()
{
int n = i();
int a = i();
int b = i();
int cnt = a/(b+1),x=0,y=0;
if(a%(b+1)!=0) cnt++;
int i = 0;
while(i<n)
{
if(x<cnt)
{
p("R");
x++;
} else {
p("B");
x = 0;
a-=cnt;
b--;
if(b+1>0)
cnt = a/(b+1);
if(a%(b+1)!=0) cnt++;
}
i++;
}
pl("");
}
/* end */
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 p(Object o)
{
out.print(o);
}
static void pl(Object o)
{
out.println(o);
}
static int i() {
return sc.nextInt();
}
static String s() {
return sc.next();
}
static long l() {
return sc.nextLong();
}
static char[] inputC()
{
String s = sc.nextLine();
return s.toCharArray();
}
static int[] input(int n) {
int A[]=new int[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextInt();
}
return A;
}
static long[] inputL(int n) {
long A[]=new long[n];
for(int i=0;i<n;i++) {
A[i]=sc.nextLong();
}
return A;
}
static long[] putL(long a[]) {
long A[]=new long[a.length];
for(int i=0;i<a.length;i++) {
A[i]=a[i];
}
return A;
}
static String[] inputS(int n) {
String A[]=new String[n];
for(int i=0;i<n;i++) {
A[i]=sc.next();
}
return A;
}
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static String reverse(String s) {
StringBuffer p=new StringBuffer(s);
p.reverse();
return p.toString();
}
static int min(int a,int b) {
return Math.min(a, b);
}
static int min(int a,int b,int c) {
return Math.min(a, Math.min(b, c));
}
static int min(int a,int b,int c,int d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static int max(int a,int b) {
return Math.max(a, b);
}
static int max(int a,int b,int c) {
return Math.max(a, Math.max(b, c));
}
static int max(int a,int b,int c,int d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long min(long a,long b) {
return Math.min(a, b);
}
static long min(long a,long b,long c) {
return Math.min(a, Math.min(b, c));
}
static long min(long a,long b,long c,long d) {
return Math.min(a, Math.min(b, Math.min(c, d)));
}
static long max(long a,long b) {
return Math.max(a, b);
}
static long max(long a,long b,long c) {
return Math.max(a, Math.max(b, c));
}
static long max(long a,long b,long c,long d) {
return Math.max(a, Math.max(b, Math.max(c, d)));
}
static long sum(int A[]) {
long sum=0;
for(int i : A) {
sum+=i;
}
return sum;
}
static long sum(long A[]) {
long sum=0;
for(long i : A) {
sum+=i;
}
return sum;
}
static void print(int A[]) {
for(int i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static void print(long A[]) {
for(long i : A) {
System.out.print(i+" ");
}
System.out.println();
}
static long mod(long x) {
return ((x%mod + mod)%mod);
}
static long power(long x, long y)
{
if(y==0)
return 1;
if(x==0)
return 0;
long res = 1;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) ;
y = y >> 1;
x = (x * x);
}
return res;
}
static boolean prime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static boolean prime(long n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
double sq=Math.sqrt(n);
for (int i = 5; i <= sq; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
static long[] sort(long a[]) {
ArrayList<Long> arr = new ArrayList<>();
for(long i : a) {
arr.add(i);
}
Collections.sort(arr);
for(int i = 0; i < arr.size(); i++) {
a[i] = arr.get(i);
}
return a;
}
static int[] sort(int a[])
{
ArrayList<Integer> arr = new ArrayList<>();
for(Integer i : a) {
arr.add(i);
}
Collections.sort(arr);
for(int i = 0; i < arr.size(); i++) {
a[i] = arr.get(i);
}
return a;
}
//pair class
private static class Pair implements Comparable<Pair> {
long first, second;
public Pair(long f, long s) {
first = f;
second = s;
}
@Override
public int compareTo(Pair p) {
if (first < p.first)
return 1;
else if (first > p.first)
return -1;
else {
if (second > p.second)
return 1;
else if (second < p.second)
return -1;
else
return 0;
}
}
}
} | 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 11 | 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 | ced66d78456d255fd375c44137791b28 | 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.*;
import java.util.*;
public class solution {
static final Random random = new Random();
static void sort(int arr[]) {
int n = arr.length;
for(int i = 0; i < n; i++) {
int j = random.nextInt(n),temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
Arrays.sort(arr);
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long[] readArrayL(int n) {
long a[]=new long[n];
for(int i=0;i<n;i++) a[i]=nextLong();
return a;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
static long factorial( long n )
{
long res = 1 ;
while(n>1)
{
res = (res*n)%998244353;
n-- ;
}
return res ;
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));//PRINTLN METHOD
static FastScanner sc = new FastScanner();//FASTSCANNER OBJECT
public static void main(String[] args){
int t = sc.nextInt() ;
while(t-- != 0 )
{
long n = sc.nextLong();
long r = sc.nextInt();
long b = sc.nextInt();
long p = (r)/(b+1) ;
long q = r%(b+1) ;
if(b== 1 )
{
for(int i = 1 ; i<= n/2 ; i++)
{
System.out.print("R");
}System.out.print("B");
for(int i = (int)n/2 +1 ; i < n ; i++)
{
System.out.print("R") ;
}System.out.println();
}
else {
while(n>0)
{
for(int i = 0 ; i < p ; i++)
{
if(n>0) {
System.out.print("R"); n-- ; }
}
if(q>0)
{
System.out.print("R") ; q-- ; n--;
}
if(n>0) {
System.out.print("B");n-- ; }
}System.out.println();
}
}
///////////////////////////////////////////////
out.flush();out.close();
}//END OF MAIN METHOD
}//END OF MAIN CLASS | 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 11 | 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 | 3681eb9bc2c72e792af3a7a36a10d81f | 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.DataInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
public class A {
static final Reader in = new Reader();
static final PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
int t = in.nextInt();
for (int tc = 1; tc <= t; tc++) {
solve();
out.println();
}
out.close();
}
static void solve() {
int n = in.nextInt(), r = in.nextInt(), b = in.nextInt();
String ans = "";
boolean poss = true;
int repeats = r;
while (poss && repeats > 0) {
int curR = r, curB = b;
StringBuilder s = new StringBuilder();
s.append("R".repeat(repeats));
curR -= repeats;
while (true) {
s.append("B");
curB -= 1;
int subtract = Math.min(curR, repeats);
s.append("R".repeat(subtract));
curR -= subtract;
if (curR <= 0 || curB <= 0) {
break;
}
}
s.append("B".repeat(Math.min(1, curB)));
curB -= Math.min(1, curB);
s.append("R".repeat(curR));
if (curB > 0) {
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == 'B') {
int subtract = Math.min(repeats - 1, curB);
s.insert(i, "B".repeat(subtract));
curB -= subtract;
i += subtract;
}
if (curB <= 0) {
break;
}
}
}
int lastB = s.lastIndexOf("B");
int trailingR = Math.max(n - lastB - 1, 0);
if (trailingR > repeats) {
poss = false;
}
else {
ans = s.toString();
}
repeats--;
}
out.print(ans);
}
static class Reader {
private final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
Reader(InputStream in) {
try {
din = new DataInputStream(in);
} catch (Exception e) {
throw new RuntimeException();
}
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
String next() {
int c;
while ((c = read()) != -1 && (c == ' ' || c == '\n' || c == '\r')) ;
StringBuilder s = new StringBuilder();
while (c != -1) {
if (c == ' ' || c == '\n' || c == '\r')
break;
s.append((char) c);
c = read();
}
return s.toString();
}
String nextLine() {
int c;
while ((c = read()) != -1 && (c == ' ' || c == '\n' || c == '\r')) ;
StringBuilder s = new StringBuilder();
while (c != -1) {
if (c == '\n' || c == '\r')
break;
s.append((char) c);
c = read();
}
return s.toString();
}
int nextInt() {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
ret = ret * 10 + c - '0';
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
int[] nextInts(int n) {
int[] ar = new int[n];
for (int i = 0; i < n; ++i)
ar[i] = nextInt();
return ar;
}
long nextLong() {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
ret = ret * 10 + c - '0';
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
long[] nextLongs(int n) {
long[] ar = new long[n];
for (int i = 0; i < n; ++i)
ar[i] = nextLong();
return ar;
}
double nextDouble() {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
ret = ret * 10 + c - '0';
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
while ((c = read()) >= '0' && c <= '9')
ret += (c - '0') / (div *= 10);
if (neg)
return -ret;
return ret;
}
void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
byte read() {
try {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
} catch (IOException e) {
throw new RuntimeException();
}
}
void close() throws IOException {
din.close();
}
}
} | Java | ["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 11 | 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 | a4f6c07817eab39e9f2418ecf203b073 | 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 forces {
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);
int y = r % (b + 1);
String s = "";
int c = b + 1;
while (c-- > 0) {
for (int i = 0; i < x; i++) {
s = s + "R";
}
if (y > 0) {
s = s + 'R';
y--;
}
if (b > 0) {
b--;
s = s + '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 11 | 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 | b49ea045b2b4580f5f3510494063af85 | 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.FileNotFoundException;
import java.io.FileReader;
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 RedVBlue {
public static void main(String[] args) {
IO io = new IO();
int numtimes = io.nextInt();
for (int i = 0; i < numtimes; i++) {
int n = io.nextInt();
int r = io.nextInt();
int b = io.nextInt();
int x = 0;
double max = (double) (r) / (b + 1);
String s = "";
double count = max;
while (r > 0) {
while ((int) count > 0 && r > 0) {
s += "R";
count -= 1;
r--;
}
if (b > 0) {
s += "B";
b--;
}
count += max;
}
System.out.println(s);
}
}
static class IO extends PrintWriter {
public BufferedReader reader;
public StringTokenizer tokens;
public IO(InputStream in, OutputStream out) {
super(out);
reader = new BufferedReader(new InputStreamReader(in));
}
public IO() {
super(System.out);
reader = new BufferedReader(new InputStreamReader(System.in));
}
public IO(String inputFileName, String outputFileName) throws FileNotFoundException {
super(outputFileName);
reader = new BufferedReader(new FileReader(inputFileName));
}
public String next() {
while (tokens == null || !tokens.hasMoreTokens()) {
try {
tokens = new StringTokenizer(reader.readLine());
} catch (IOException e) {
System.out.println("BIGPROBLEM");
throw new RuntimeException(e);
}
}
return tokens.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public float nextFloat() {
return Float.parseFloat(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 11 | 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 | c3f20edade2401aaf24c3f38b8c51c80 | 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(),r=sc.nextInt(),b=sc.nextInt();
StringBuilder sb=new StringBuilder();
int s_r=r;
int y=b;
int s_b=b+1;
int x;
if(r%s_b==0){
x=r/s_b;
}else{
x=(r/s_b)+1;
}
StringBuilder st=new StringBuilder();
for (int i = 1; i <=x ; i++) {
st.append('R');
}
String tt=st.toString();
sb.append(tt);
s_r-=x;
s_b--;
while (y-->0){
sb.append('B');
int c;
if(s_r%s_b==0){
c=s_r/s_b;
}else{
c=(s_r/s_b)+1;
}
StringBuilder temp=new StringBuilder();
for (int i = 0; i <c ; i++) {
temp.append('R');
}
sb.append(temp);
s_r-=c;
s_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 11 | 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 | f4fa0435316037bb33fbf449a0f8e81c | 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 {
static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
int T = in.nextInt();
while (T-- > 0) {
solve();
}
}
private static void solve() {
int n, r, b;
n = in.nextInt();
r = in.nextInt();
b = in.nextInt();
if(r > b){
int little = r % (b+1);
int time_cap = r / (b+1);
int cap = 0;
int b_cnt = 0;
for (int i = 0; i < r; i++){
System.out.print("R");
cap++;
int cab_next = time_cap;
if(little > 0){
cab_next++;
}
if(cap >= cab_next && b_cnt < b){
b_cnt++;
System.out.print("B");
cap=0;
little--;
}
}
while (b_cnt++ < b){
System.out.print("B");
}
}else {
int little = b % (r+1);
int time_cap = b / (r+1);
int cap = 0;
int r_cnt = 0;
for (int i = 0; i < b; i++){
System.out.print("B");
cap++;
int cab_next = time_cap;
if(little > 0){
cab_next++;
}
if(cap >= cab_next && r_cnt < r){
r_cnt++;
System.out.print("R");
cap=0;
little--;
}
}
while (r_cnt++ < r){
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 11 | 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 |
Subsets and Splits