poj 3641 Pseudoprime numbers 快速幂取模

题目链接：http://poj.org/problem?id=3641

题目描述：

 

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

 

题目大意：

 

输入两个整数p和a，如果p是素数，输出no，否则判断（a^p）%p是否与a相等，相等就输出yes，否则no

 

AC代码：

 

/*
快速幂取模
by_Superxd
*/
#include
#include
#include
#include
#include
#include
#define MOD 1000000000+7
#define LL long long
using namespace std;
LL m,n;
LL su(LL x)
{
    if(x==2)
        return 1;
    for(LL i=2;i<=sqrt(x);i++)
    {
        if(x%i==0)
            return 1;
    }
    return 0;
}
LL quckmod(LL x,LL y,LL z)//快速幂取模x^z
{
    LL cnt=1;
    while(y)
    {
        if(y%2==1)
            {
                cnt*=x;
                cnt%=z;
            }
        x*=x;
        x%=z;
        y/=2;
    }
    return cnt;
}

int main()
{
    while(1)
    {
        cin>>m>>n;
        if(n==0&&m==0)
            break;
        if(!su(m))
           cout<<"no"<