poj3641(快速幂取模)

Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6123 Accepted: 2457

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

3 210 3341 2341 31105 21105 30 0

Sample Output

nonoyesnoyesyes

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;bool isPrime(int a){int num = (int)sqrt(a);for(int i = 2; i<=num; i++){if(a%i==0){return false;}}return true;}int quick_mod(int a, int p){int ret = 1;int mod = p;for( ; p; p>>=1, a = (long long)a*a%mod){if(p&1){ret = (long long)ret*a%mod;}}return ret;}int main(){int p, a;while(cin>>p>>a){if(p==0){break;}bool ok = false;if(!isPrime(p)){int num = quick_mod(a, p);if(num == a){ok = true;}}if(ok){cout<<"yes"<<endl;}elsecout<<"no"<<endl;}return 0;}