POJ 3641 FLY

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
Source

Waterloo Local Contest, 2007.9.23

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#define ll long long
using namespace std;
ll mod_pow(ll a, ll b, ll mod)
{
	if(b==0) return 1;
	ll res=mod_pow(a*a%mod, b/2, mod);
	if(b%2==1) res=res*a%mod;
	return res;
 }                       //模板a的b次幂对mod取余
 ll prime(ll n)
 {
 	if(n==2) return 1;
 	if(n<=1||n%2==0) return 0;
 	long long j=3;  
    while(j<=(long long)sqrt(double(n)))  
    {  
        if(n%j==0)  
            return 0;  
        j+=2;  
    }  
    return 1;  
}                   //判断素数
 int main()
 {
 	ll p,a;
	while(scanf("%lld%lld",&p,&a)!=EOF)
	{
		if(p==0&&a==0) break;
		if(prime(p)==1) printf("no\n");
		else
		{
			if(mod_pow(a,p,p)==a)
				printf("yes\n");
			else
			printf("no\n");
		}
	}
	return 0;
 }                      //主文件