G. A/B Matrix

合集 - 构造(16)

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16.G. A/B Matrix2024-07-20

收起

原题链接

题解

每行有 a 个，所以总共有 $n\cdot a$ 个

每列有 b 个，所以总共有 $m\cdot b$ 个

所以要满足 $na=mb$

想象一下这个场景：每一行，每次往当前列中，最左端的一最少的列的开始连续放置1

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

void solve()
{
    int n,m,a,b;
    cin>>n>>m>>a>>b;

    if(n*a!=m*b)
    {
        cout<<"no\n";
        return;
    }
    cout<<"yes\n";
    int start=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(j>=start&&j<=start+a-1||j<start&&start+a-1>=j+m) cout<<1;
            else cout<<0;
        }
        start=(start+a-1)%m+1;
        cout<<'\n';
    }

}
int main()
{
    //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    cin>>t;
    while(t--) solve();
    return 0;
}