Dreaming of Freedom(数论,贪心)

用nsqrt(n)的时间复杂度就能过

//Dreaming of Freedom:https://codeforces.com/problemset/problem/1826/C

#include <bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e5+10,mod=1e9+7;
string s;
int n,t,a[N],f[N],res,num,ans,m;
bool vis[N];
int main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        cin>>n>>m;
        if(n>=2&&m>=n){
            cout<<"NO"<<endl;
            continue;
        }
        if(!(n%2)&&m>=2){
            cout<<"NO"<<endl;
            continue;
        }
        if(n>m){
            bool f=false;
            for(int i=2;i<=n/i&&i<=m;++i){
                if(n%i==0){
                    f=true;
                    break;
                }
            }
            if(f){
                cout <<"NO"<<endl;
                continue;
            }
        }
        cout<<"YES"<<endl;
        
    }
    return 0;
}