传送门：poj3641 Pseudoprime numbers
Description
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-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：对于非素数p，如果存在a的p次幂对p取模等于a，那么这个数就是伪素数。
判断伪素数的步骤：1.判断p是否是素数？
2.判断a的p次幂对p取模是否等于a？由于这里p非常大，所以需要用到快速幂。
#include<iostream>
#include<cmath>
using namespace std;bool isPrime(int x){if(x==2) return true;for(int i=2;i<=sqrt(x);i++){if(x%i==0) return false;}return true;
}long long PowerMod(long long a,long long b,long long c){long long ans=1;a=a%c;while(b>0){if(b&1) ans=(ans*a)%c;b>>=1;a=(a*a)%c;}return ans;
}int main(){int p,a;while(cin>>p>>a){if(p==0&&a==0) break;if(isPrime(p)) cout<<"no"<<endl;else if(PowerMod(a,p,p)==a) cout<<"yes"<<endl;else cout<<"no"<<endl;}return 0;
}