POJ3641 Pseudoprime numbers(快速幂取模）

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3171		Accepted: 1105

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-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

Waterloo Local Contest, 2007.9.23

#include

bool isprime(int n)

{

    for(int i=2;i*i<=n;i++)

    if(n%i==0)  return true;

    return false;

}

int check(long long a,int b,int p)

{

    long long c=1;

    if(b==0) return 1%p;

    while(b>1)

    {

        if(b%2==0)

        {

            a=(a*a)%p;

            b=b/2;

        }

        else

        {

            c=(c*a)%p;

            b--;

        }

    }

    return a*c%p;

}

int main()

{

    int p,a;

    while(scanf("%d%d",&p,&a))

    {

        if(p==0&&a==0) break;

        if(!isprime(p)||check(a,p,p)!=a)

        {

            printf("no/n");

        }

        else

           printf("yes/n");

    }

    return 0;

}