poj 3641 Pseudoprime numbers 快速幂+素数判定 模板题

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7954		Accepted: 3305

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-apseudoprimes, 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 anda.

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

<span style="font-size:32px;">#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long a,p;
long long power(long long a,long long p)
{
     long long ret=1,temp=p;
     while(temp)
     {
          if(temp&1)
             ret=(ret*a)%p;
          a=(a*a)%p;
          temp>>=1;
     }
     return ret%p;
}
bool prime(long long m)
{
    for(long long i=2;i*i<=m;i++)
        if(m%i==0)
          return false;
    return true;
}
int main()
{
    long long a,p;
    while(~scanf("%lld %lld",&p,&a))
    {
        if(a==0&&p==0)  return 0;
        if(power(a,p)==a%p&&!prime(p))
            printf("yes
");
        else
            printf("no
");
    }
    return 0;
}
</span>