POJ 3641 Pseudoprime numbers 【快速幂】

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8785		Accepted: 3695

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

Sample Output

no
no
yes
no
yes
yes

Source

Waterloo Local Contest, 2007.9.23

如果a^p%p == a,并且p是素数，输出yes，否则输出no。

#include <stdio.h>
#include <math.h>
__int64 pow(__int64 x, __int64 y, __int64 mod) {
    __int64 res = 1;
    while (y) {
        if (y&1) res = x*res%mod;
        x = x*x%mod;
        y >>= 1;
    }
    return res;
}

bool prime(long long a)  {
    int i;
    if(a == 2)
        return true;
    for(i = 2; i*i<=a; i++)
        if(a%i == 0)
            return false;
    return true;
}

int main() {
    __int64 n, p;
    while (scanf("%I64d%I64d", &p, &n) != EOF) {
        if (p == 0 && n == 0) break;
        if (!prime(p) && n == pow(n, p, p)) {
            printf("yes
");
        }
        else printf("no
");
    }
    return 0;
}