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=;
     while(b)
     {
         if(b %  != )
         {
             t=t*a%c;
         }
         a=a*a%c;
         b/=;
     }
     return t%c;
 }
 __int64 f2(__int64 a)
 {
     __int64 i;
     if(a <=  || a %  == ) return ;
     for(i=;i<=sqrt(a);i++)
     {
         if(a % i == ) return ;
     }
     return ;
 }
 int main()
 {

     __int64 p,a;
     while(scanf("%I64d %I64d",&p,&a) && p && a)
     {
         if(f2(p) == ) printf("no\n");
         else
         {
             if(f(a,p) == a) printf("yes\n");
             else
             printf("no\n");
         }

     }
 }