POJ3641 (快速幂） 判断a^p = a (mod p)是否成立

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

如果p是素数，输出no；如果p不是素数，判断a^p对p取余是否等于a。

#include<cstdio>
#include<math.h>
__int64 f(__int64 a,__int64 b)
{
    __int64 c=b,t=1;
    while(b)
    {
        if(b % 2 != 0)
        {
            t=t*a%c;
        }
        a=a*a%c;
        b/=2;
    }
    return t%c;
}
__int64 f2(__int64 a)
{
    __int64 i;
    if(a <= 1 || a % 2 == 0) return 0;
    for(i=3;i<=sqrt(a);i++)
    {
        if(a % i == 0) return 0;
    }
    return 1;
}
int main()
{
    
    __int64 p,a;
    while(scanf("%I64d %I64d",&p,&a) && p && a)
    {
        if(f2(p) == 1) printf("no\n");
        else
        {
            if(f(a,p) == a) printf("yes\n");
            else 
            printf("no\n");
        }
        
    }
}