【数论 快速幂 费马小定理】poj 3641 Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16389		Accepted: 7082

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

伪素数条件：1.p不是质数 2.a^p ≡ a (mod p)

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long ll;
int is_prime(ll x)
{
    for(int i=2;i*i<=x;i++)
        if(x%i==0)
            return false;
    return true;
}
ll qpow(ll a,ll b,ll mod)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=(res*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return res;
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    ll p,a;
    while(scanf("%lld%lld",&p,&a) && (a || p))
    {
        if(is_prime(p))
        {
            printf("no\n");
            continue;
        }
        if(qpow(a,p,p)==a%p)
            printf("yes\n");
        else printf("no\n");
    }
    return 0;    
}