Description：
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
解题思路：
这个题要求求出若p不是素数并且p^a是否等于a，若等于则输出yes，否则输出no。
这个题主要用到快速幂，其他就比较简单了，注意类型一定要是long long！！！（因为结果非常大）
程序代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
long long prime(long long x)
{
for(int i=2;i<=sqrt(x);i++)
if(x%i==0)
return 0;
return 1;
}
long long quickmul(long long a,long long b)
{
long long ans=1;
long long m=b;
while(m)
{
if(m&1)//等价于m%2==1
ans=(ans*a)%b;
a=(a*a)%b;
m>>=1;//等价于m=m/2
}
return ans;
}
int main()
{
long long p,a;
while(scanf("%lld %lld",&p,&a)&&(p+a))
{
if(prime(p))
printf("no\n");
else if(quickmul(a,p)==a)
printf("yes\n");
else
printf("no\n");
}
return 0;
}