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

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\n");

        else

            printf("no\n");

    }

    return 0;

}

</span>

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