| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6018 | Accepted: 2407 |
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
Source
//384K 16MS
#include<stdio.h>
#include<math.h>
bool isprime(long long x)//判断x是不是素数，如果是素数，肯定不是伪素数
{
if(x==1||x==2)return true;
long long tmp=sqrt(x);
for(long long i=2;i<=tmp;i++)
if(x%i==0)return false;
return true;
}
long long quick_mod(long long a,long long b,long long m)//快速幂求a^b%m
{
long long ans=1;
while(b)
{
if(b&1){ans=(ans*a)%m;b--;}
b/=2;
a=a*a%m;
}
return ans;
}
int main()
{
long long p,a;
while(scanf("%lld%lld",&p,&a),p|a)
{
if(isprime(p)){printf("no\n");continue;}
long long mol=quick_mod(a,p,p);
if(mol==a%p)printf("yes\n");
else printf("no\n");
}
return 0;
}
POJ 3641 Pseudoprime numbers 测试费马小定理伪素数
原文：http://blog.csdn.net/crescent__moon/article/details/19397947