Pseudoprime numbers 快速幂+伪素数

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a(mod p). That is, if we raise a to the pth 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-apseudoprimes for all a.)

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

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

题目大意：费马定理：a^p = a(mod p) (a为大于1的整数，p为素数)，一些非素数p，同样也符合上边的

定理，这样的p被称作基于a的伪素数，给你p和a，判断p是否是基于a的伪素数

思路：很简单的快速幂取余+素性判断 

则进行快速幂取余判断是否满足条件，然后在判断它不是素数输出yes，否则输出no

#include<iostream>
using namespace std;
typedef long long LL;
LL fast_mod(LL a,LL b,LL mod)
{
    LL ans=1;
    while(b){
        if(b&1)
            ans=ans*a%mod;//写成ans*=a%mod就不对
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
int prime(LL x)
{
    if(x==1) return 0;
    for(int i=2;i*i<=x;i++){
        if(x%i==0) return 0;
    }
    return 1;
}
int main()
{
    LL p,a;
    while(cin>>p>>a,p!=0&&a!=0){
        if(fast_mod(a,p,p)==a&&!prime(p))
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}