Pseudoprime numbers

最新推荐文章于 2020-10-22 21:33:13 发布

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本文介绍了一个基于费马小定理的应用——如何判断一个合数是否为特定基底的伪素数。通过快速幂算法实现，用于解决特定数学问题。

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Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8649		Accepted: 3633

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

快速幂求解，题看懂很容易，但是我的英语水平太差劲了，没有看懂题意。费马定理。

题意：费马定理，费费马定理马小定理：若p是素数且a是正整数，那么a^p=a(mod p)
若a是正整数，p是合数且满足a^p=a(mod p)，那么称p为以a为基的伪素数。
给你p和a，让你判断p是不是伪素数。
 
 
#include<cstdio>
#include<cstring>
#include<cmath>
__int64 sushu(__int64 n)//判断是否是素数
{
	__int64 i;
	for(i=2;i*i<n;i++)
	{
	if(n%i==0)
	return 0;
	}
	return 1;
}
__int64 qpow(__int64 a,__int64 b,__int64 c)//快速幂求解
{
	__int64 ans=1,base=a;
	while(b)
	{
	if(b&1)
	{
		ans=(base*ans)%c;
	}
	base=(base*base)%c;
	b=b/2;
    }
    return ans;
}
int main()
{
	__int64 p,a,l;
	while(scanf("%I64d%I64d",&p,&a)&&(p||a))
	{
		if(sushu(p))
		printf("no\n");
		else
		{
			l=qpow(a,p,p);
			if(a%p==l)
			printf("yes\n");
			else
			printf("no\n");
		}
	}
	return 0;
}