G - A/B Matrix CodeForces

You are given four positive integers n, m, a, b (1≤b≤n≤50; 1≤a≤m≤50). Find any such rectangular matrix of size n×m that satisfies all of the following conditions:

each row of the matrix contains exactly a ones;
each column of the matrix contains exactly b ones;
all other elements are zeros.

If the desired matrix does not exist, indicate this.

For example, for n=3, m=6, a=2, b=1, there exists a matrix satisfying the conditions above:

|010001100100001010|

Input

The first line contains an integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.

Each test case is described by four positive integers n, m, a, b (1≤b≤n≤50; 1≤a≤m≤50), where n and m are the sizes of the matrix, and a and b are the number of ones for rows and columns, respectively.

Output

For each test case print:

"YES" (without quotes) and the required matrix (if there are several answers, print any) if it exists, or
"NO" (without quotes) if it does not exist.

To print the matrix n×m, print n rows, each of which consists of m numbers 0 or 1 describing a row of the matrix. Numbers must be printed without spaces.

Example

Input

Output

YES
010001
100100
001010
NO
YES
11
11
YES
1100
1100
0011
0011
YES
1
1

题意：多组输入，每组输入n,m,a,b，要求输出YES和一个n*m的数组，其中每行有a个1，每列有b个1，其余为0.若没有符合条件的数组，输出NO。

贴一个我自己的wrong的代码，暂时还没完全弄明白。

#include<iostream>
#include<algorithm>
using namespace std;int t, n, m, a, b;
int main(){scanf("%d", &t);while(t--){scanf("%d%d%d%d", &n, &m, &a, &b);if(n*a != m*b){    printf("NO\n");continue;}int line[51], list[51];//行、列for(int i=0; i<n; i++)line[i] = a;for(int i=0; i<m; i++)list[i] = b;printf("YSE\n");for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(line[i] && list[j]){printf("1");line[i]--;list[j]--;}else printf("0");}printf("\n");}}
}

吸收各方大佬以后按照他们的方法写的ac的代码

#include<iostream>
using namespace std;
int s[52][52];
int t, n, m, a, b;
int main(){scanf("%d", &t);while(t--){scanf("%d%d%d%d", &n, &m, &a, &b);if(n*a != m*b){    printf("NO\n");continue;}for(int i=0; i<n; i++)for(int j=0; j<m; j++)s[i][j] = 0;for(int i=0, j=0; i<n; i++, j=(j+a)%m)for(int k=0; k<a; k++)s[i][(j+k)%m] = 1;printf("YES\n");for(int i=0; i<n; i++){for(int j=0; j<m; j++)printf("%d", s[i][j]);printf("\n");}}return 0;
}

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