Pseudoprime numbers

POJ - 3641 

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 isa. 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-apseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-apseudoprime.

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

题意：判断a的p次幂（mod）p是否==a;&&p不是素数

思路：快速幂

AC 答案：

#include<stdio.h>
#include<math.h>
bool hanshu(__int64 t)
{
    int i;
    for(i=2;i<=sqrt(t);i++)
    {
        if(t%i==0) return 0;//bu shi su shu
    }
    return 1;//shi su shu
}
int main()
{
    __int64 p,a;
    __int64 t,b,sum;
    while(~scanf("%I64d%I64d",&p,&a))
    {
        if(p==0&&a==0) break;
        t=p;
        b=a;


        a=a%p;
        sum=1;
        while(p>0)
        {
            if(p%2==1)
            {
                sum=sum*a%t;
            }
            p/=2;
            a=(a*a)%t;
        }
    sum=sum%t;
    //printf("***%d\n",sum);
        if(sum==b)
        {
            if(hanshu(t)==0)
            {
                printf("yes\n");
            }
            else
                printf("no\n");
        }
        else
        {
            printf("no\n");
        }
    }
    return 0;
}