cf1091div2 C.Grid Covering（数论）

Problem - C - Codeforces

保证遍历完每行每列所以gcd(n,a)==1,gcd(m,b)==1很好理解

为了遍历所有网格，因为在2*lcm(n,m)次数后会再次踏上轮回重复循环，此时访问了2*lcm(n,m)个格子，于是 2*lcm(n,m)>=n*m，也就是2*lcm>=gcd(n,m)*lcm(n,m)。所以 gcd(n,m)<=2

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

const int N=2e5+10;
int n,m,a,b;

int gcd(int a,int b) {
    return b==0?a:gcd(b,a%b);
}
void solve() {
    cin>>n>>m>>a>>b;

    if (gcd(n, a) == 1 && gcd(m, b) == 1&&gcd(n,m)<=2 ){
        cout<<"YES"<<endl;
    }
    else {
        cout<<"NO"<<endl;
    }
}
int main() {

    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t=1;
    cin>>t;
    while(t--){solve();}

    return 0;
}