伪素数判断----Pseudoprime numbers

伪素数定义：伪素数，又叫做伪质数：它满足费马小定理，但其本身却不是素数。最小的伪素数是341。有人已经证明了伪素数的个数是无穷的。事实上，费马小定理给出的是关于素数判定的必要非充分条件。若n能整除2^(n-1)-1，并n是非偶数的合数，那么n就是伪素数。第一个伪素数341 是萨鲁（Sarrus)在1819年发现的。
简单解释一下伪素数：一个数字p本身不是素数但满足费马小定理，即a的p次方模p与a模p同余。

下面看具体题目：

Pseudoprime numbers

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 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：给出p，a,判断p是否为伪素数 ，如果是输出yes否则输出no

AC代码:

#include 
#include 
#include 
#include 


using namespace std;

typedef long long LL;
const int maxm = 1e6;
LL mod;


LL qmod(LL p , LL q)		//快速幂算法求a的p次方
{
	LL ans = 1;
	while(q)
	{
		if(q&1)
			ans = (p*ans) % mod;
		p = (p * p) % mod;
		q >>= 1;
	}
	return ans;
}


bool isprime(int n)
{
	for(int i = 2 ; i <= (int)sqrt(n) ; i++)
	{
		if(n % i == 0)
			return false;
	}
	return true;
}

int main()
{
	int a,p;
	while(cin >> p >> a)
	{
		if(a == 0 && p == 0)
			break;
		if(isprime(p))	//首先判断是否是素数
		{
			cout << "no" << endl;
			continue;
		}
		else
		{
			mod = p;
			if(qmod(a,p) == a % p) 	//判断费马小定律
			{
				cout << "yes" << endl;				
			}
			else
				cout << "no" << endl;
		}
			
	}
	return 0;
}