Pseudoprime numbers 【poj-3641】【快速幂】

最新推荐文章于 2018-08-10 16:34:36 发布

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最新推荐文章于 2018-08-10 16:34:36 发布

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分类专栏： # 快速幂文章标签： POJ

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本文介绍了一种利用快速幂算法进行伪素数判断的方法，并通过具体题目实例展示了如何编写代码实现这一过程。文章重点讲解了快速幂算法原理及其在素数判定中的应用。

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Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10620		Accepted: 4578

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

Sample Output

no
no
yes
no
yes
yes

题意：输入两个数，判断公式 a^p=a(mod p)是否成立，即a的p次方对p求模是否等于a。只有当p是非素数时且等式成立时，才输出yes。其他情况输出no。

题解：利用快速幂算法，简单的素数判定。

代码如下：

#include<cstdio>
typedef long long ll;
bool Prime_judge(ll q)
{
	for(ll i=2;i*i<=q;i++)
	{
		if(q%i==0) return false;
	}
	return true;
}


ll p,a;
ll mod_pow(ll x,ll n){
	ll res=1;
	while(n>0){
		if(n&1) res=res*x%p;
		x=x*x%p;
		n>>=1;
	}
	return res;
}

int main()
{
	
	while(scanf("%lld%lld",&p,&a)){
		if(p==0&&a==0) break;
		if(Prime_judge(p)){
			printf("no\n");
		}else{
			ll ans=mod_pow(a,p);
			ans=ans%p;
			if(ans==a) printf("yes\n");
			else printf("no\n");
		}
		
		
	}
	
	
	return 0;
}