Codeforces Round #644 (Div. 3) G.A/B Matrix

题目链接
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]\left[ \begin{matrix} 0 &1&0&0&0&1\\ 1 &0&0&1&0&0\\ 0 &0&1&0&1&0\\ \end{matrix} \right] ⎣⎡010100001010001100⎦⎤

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

简单构造~
首先判断 NONONO 的情况，a∗n!=b∗ma*n!=b*ma∗n!=b∗m 即可
下面开始构造矩阵，我们按行来，一行连续 aaa 个 111,然后每行都错开 aaa 个即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,t,a,b,cnt=0,g[55][55];
main(){cin>>t;while(t--){memset(g,0,sizeof(g));cin>>n>>m>>a>>b;if(n*a!=m*b) {puts("NO");continue;}for(int i=0;i<n;i++){for(int j=0;j<a;j++)g[i][cnt++]=1,cnt%=m;}puts("YES");for(int i=0;i<n;i++){for(int j=0;j<m;j++) cout<<g[i][j];puts("");}}
}