Fermat's theorem states that for any prime number p and for any integer a > 1, ap = 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 ≤ 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
ac代码：
#include<stdio.h>
typedef long long ll;
int isprem(long long p)
{
for(int i=2;i*i<=p;i++) if(p%i==0) return 0; return 1;
}
int main()
{
ll a,p; while(scanf("%lld %lld",&p,&a)!=EOF && (a||p))
{
ll t,m; t=p; m=a;
if(isprem(p)) printf("no\n");
else
{
ll ans=1; while(p)
{
if(p&1) ans=ans*a%t;
a=a*a%t;
p>>=1;
}
if(ans==m) printf("yes\n");
else printf("no\n");
}
}
return 0;
}