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

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 ;

 }

巴特西

poj 3641 快速幂

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