poj 3641

根据题目的定义来做，考quick_mod的应用，比较水

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

const int maxn = 1000005;

typedef long long LL;

vector<int>G[maxn];

bool prim(LL n)
{
    int  t = (int)sqrt(n)+1;
    if( n == 2 || n == 1) return 1;
    LL i;
    for(i=2;i<=t;i++){
        if(n%i == 0){
            return 0;
            break;
        }
    }
    return 1;
}

LL quick_mod(LL a,LL b,LL m)
{
    LL ans = 1;
    while(b){
        if(b&1){
            ans = (ans*a)%m;
        }
        b>>=1;
        a = a*a%m;
    }
    return ans;
}

int main()
{
    LL a,n,p,b;
    while(cin>>p>>a&&(a+p)){
        if(prim(p)) puts("no");
        else{
            LL ans = 1;
            ans = quick_mod(a,p,p);
            if(ans == a) puts("yes");
            else puts("no");
        }
    }


    return 0;
}