快速幂+素数 HDU 1905 Pseudoprime numbers

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 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 ≤ 1,000,000,000 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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include<math.h>
#include <algorithm>
using namespace std;
long long g(long long a, long long n, long long p)//快速幂
{
    if(n == 1) return a % p;
    long long tmp = g(a, n/2, p) % p;
    if(n % 2 == 0)
    {
        return tmp * tmp % p;
    }
    else
    {
        return tmp * tmp % p * a % p;
    }
}
bool prime(long long x)//素数
{
    if(x==1)
        return false;
    if(x==2)
        return true;
    for(long long i=2;i<=sqrt(x);i++)
        if(x%i==0)
        return false;
    return true;
}
int main()
{
    long long a, n;
    while(scanf("%I64d%I64d",&a,&n)&&a+n!=0)
        if(prime(a))
            puts("no");
        else
            printf(g(n, a, a)==n?"yes\n":"no\n");

    return 0;
}