Kattis - pseudoprime 【快速幂】

题意
给出两个数字 P 和 A 当p 不是素数 并且 满足a^p≡a(mod p) 就输出 yes 否则 输出 no

思路
因为 数据范围较大，用快速幂

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>

using namespace std;
typedef long long LL;

const double PI  = 3.14159265358979323846264338327;
const double E   = 2.718281828459;
const double eps = 1e-6;

const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e5 + 5;
const int MOD  = 1e9 + 7;

LL powerMod(LL x, LL n, LL m)
{
    LL res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = (res * x) % m;
        x = (x * x ) % m;
        n >>= 1;
    }
    return res;
}

bool isPrime(int x)
{
    int flag;
    int n, m;
    if (x <= 1)
        return false;
    if (x == 2 || x == 3)
        return true;
    if (x % 2 == 0)
        return false;
    else
    {
        m = sqrt(x) + 1;
        for (n = 3; n <= m; n += 2)
        {
            if (x % n == 0)
            {
                return false;
            }
        }
        return true;
    }
}

int main()
{
    LL p, a;
    while (cin >> p >> a && (p || a))
    {
        if (powerMod(a, p, p) == a && a % p == a && isPrime(p) == false)
            cout << "yes\n";
        else
            cout << "no\n"; 
    }
}