Pseudoprime numbers poj3641（快速幂+素数判定）

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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
题意：给一个 p 和 一个 a，如果这个p 本身就是一个素数，就输出 no，如果不是素数，那么计算  ( a ^ p) % p 如果结果等于 a 那么输出 yes 否则输出 no
#include<stdio.h>
#include<iostream>
#include<memory.h>
#include<algorithm>
#include<math.h>
#define N 1000000000
using namespace std;
typedef long long ll;
int pri(ll n)
{
   int i;
   if(n==2)
    return 0;
   for(i=2;i*i<=n;i++)
    if(n%i==0)
    return 0;
   return 1;
}
ll pow_mod(ll x,ll n,ll mod)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=ans*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    return ans;
}
int main()
{
    ll p,a;
    while(scanf("%I64d%I64d",&p,&a),p&&a)
    {
        if(pri(p)==1)
        {
            printf("no\n");
            continue;
        }
        ll ans=pow_mod(a,p,p);
        if(ans==a)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;

}