poj Pseudoprime numbers 3641

Pseudoprime numbers

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

3 2
10 3
341 2
341 3
1105 2
1105 3
0 0

Sample Output

no
no
yes
no
yes
yes

题意：判断伪质数，即非质数，并且满足:存在a,使得a^p==a mod(p)的p称为伪质数。思路：快速幂运算验证即可。AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<string>
#include<bitset>
using namespace std;
#define INF 0x3f3f3f3f
#define MOD 1000000000
typedef long long ll;
const int N_MAX = 10000;
ll p, a;
ll ll_mult(ll a,ll x,ll p) {
    ll res = 0;

        bitset<32>tmp = static_cast<bitset<32>>(x);//前面低位
        for (int i = 0; i < tmp.size();i++) {
            if (tmp[i])res += a*(1 << i);
            res %= p;
        }

    return res;
}

ll mod_pow(ll x,ll n,ll p) {
    ll res = 1;
    while (n) {
        if (n & 1)res = ll_mult(res,x,p);
        x = ll_mult(x, x,p);
        n >>= 1;
    }
    return res;
}

bool is_prime(ll n) {
    for (int i = 2; i*i <= n;i++) {
        if (n%i == 0)return false;
    }
    return n!=1;
}

int main() {
    while (scanf("%lld%lld",&p,&a)&&(p||a)) {
        if (is_prime(p)) { puts("no"); continue; }
        if (mod_pow(a, p, p) == a)puts("yes");
        else puts("no");
    }

     return 0;
}