C. OKEA
一道简单的数学问题，一行只能有奇数或者偶数，进行判断就行了
#include<bits/stdc++.h>
using namespace std;
const int N=1e6;
int main()
{int t,n,k;cin>>t;while (t--&&cin>>n>>k){int cnt = (n*k+1)/2;//奇数的数量if (cnt%k==0){cout<<"YES"<<endl;for (int i=1;i<=n*k;i+=2) {cout<<i<<" ";if ((i+1)/2%k==0) cout<<endl;}for (int i=2;i<=n*k;i+=2){cout<<i<<" ";if (i/2%k==0) cout<<endl;}}else cout<<"NO"<<endl;}
}
)
)
(A-D))
