poj 3641 快速幂

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

 #include <iostream>
 #include <cstdio>
 #include <algorithm>

 using namespace std;
 typedef long long ll;
 ll quick_pow(ll a,ll b,ll mod) {
     ll ans=;
     while(b) {
         if(b&)
             ans=(ans*a)%mod;
         a=(a*a)%mod;
         b>>=;
     }
     return ans;
 }
 bool isprime(int x) {
     for(int i=;i*i<=x;i++) {
         if(x%i==) return false;
     }
     return true;
 }
 int main() {
 //  freopen("input.txt","r",stdin);
     int a,p;
     while(scanf("%d%d",&p,&a)!=EOF) {
         if(a==&&p==) break;
         if(isprime(p)) {
             printf("no\n");
             continue;

         }
         ll z=quick_pow(a,p,p);
         if(z==a) printf("yes\n");
         else printf("no\n");
     }
     return ;
 }