Pseudoprime numbers

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

3 210 3
341 2341 31105 21105 30 0

 

Sample Output

nonoyesnoyesyes

 

Author

Gordon V. Cormack

 

Source

2008-1杭电公开赛(非原创)

 题意：

通过 p的值判断，若p为素数，就断定不是伪素数，若p不是素数，则判断式子(a^n)%p==a;若相等，则输出yes,否则输出no;

由于数据较大，故容易超时，采用分治法求解；

代码：

#include<stdio.h>
int s(__int64 a)
{
int i;
for(i=2;i*i<=a;i++)
if(a%i==0)
return 1;
return 0;
}
__int64 f(__int64 a,__int64 n,__int64 m)
{
__int64 ans;
if(n==0)
return 1;
__int64 x=f(a,n/2,m);
ans=x*x%m;
if(n%2==1)
ans=ans*a%m;
return (__int64)ans;
}
int main(){
__int64 i,j,n,m;
while(scanf("%I64d%I64d",&n,&m))
{
if(n==m&&n==0)
break;
if(!s(n))
printf("no\n");
else if(f(m,n,n)==m)
printf("yes\n");
else printf("no\n");
}
return 0;
}