POJ3641_Pseudoprime numbers【快速幂】【伪素数】

Pseudoprime numbers

Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 6544Accepted: 2648

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

Waterloo Local Contest, 2007.9.23

题目大意：费马定理：a^p = a(mod p) (a为大于1的整数，p为素数)，一些非素数p，同样也符合上边的

定理，这样的p被称作基于a的伪素数，给你p和a，判断p是否是基于a的伪素数

思路：很简单的快速幂取余+素性判断

如果p为素数，则直接输出no

如果p不为素数，则进行快速幂取余判断是否为伪素数，若是，输出yes，不是，输出no

#include<stdio.h>
#include<math.h>

__int64 QuickPow(__int64 a,__int64 p)
{
    __int64 r = 1,base = a;
    __int64 m = p;
    while(p!=0)
    {
        if(p&1)
            r = r * base % m;
        base = base * base % m;
        p >>= 1;
    }
    return r;
}

bool IsPrime(__int64 p)
{
    for(__int64 i = 2; i <= sqrt(p) + 1; i++)
    {
        if(p % i == 0)
            return false;
    }
    return true;
}
int main()
{
    __int64 a,p;
    while(~scanf("%I64d %I64d",&p,&a) && (p!=0 || a!=0))
    {
        if(IsPrime(p))
            printf("no\n");
        else
        {
            if(QuickPow(a,p) == a)
                printf("yes\n");
            else
                printf("no\n");
        }
    }
    return 0;
}