POJ - 3641 Pseudoprime numbers

最新推荐文章于 2019-04-14 12:39:21 发布

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最新推荐文章于 2019-04-14 12:39:21 发布

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分类专栏：快速幂运算

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本文介绍如何利用费马小定理来判断一个数是否为针对特定基数的伪素数，并提供了一个使用快速幂算法实现的示例程序。

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Pseudoprime numbers

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Description

Fermat's theorem （费马理论）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（伪素数）, 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

题目要求：p和a要满足(a^p)%p==a且p为非素数！由于数据较大，要用long long型，且要用到快速幂！快速幂简介：点击打开链接，再加上一个判断素数的函数就行了。

#include<cstdio>
int su(long long n){
	for(int i=2;i*i<=n;i++)
		if(n%i==0)
			return 0;//非素数
			return 1;//素数
}
long long pow(long long a,long long p){
	long long ans=1,base=a,mod=p;
	while(p){
		if(p&1)
			ans=(base*ans)%mod;
		base=(base*base)%mod;
		p>>=1;
	}
	return ans;
}
int main(){
	long long p,a;
	while(scanf("%lld %lld",&p,&a)&&(p||a)){
		if(pow(a,p)==a&&su(p)==0)
			printf("yes\n");
		else
			printf("no\n");
	}
	return 0;
}