Pseudoprime numbers

最新推荐文章于 2022-02-09 09:33:23 发布

goodluckwj

最新推荐文章于 2022-02-09 09:33:23 发布

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本文链接：https://blog.youkuaiyun.com/qq_35634181/article/details/52003450

本文介绍了一个程序，该程序基于费马小定理来判断给定的整数对是否构成一个特定类型的伪素数——即对于一些整数a，p是否为a的基伪素数。此外，还提供了判断一个数是否为素数的方法。

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Problem 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 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 ≤ 1,000,000,000 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：给出两个数p，a 如果p是素数输出no 否则判断a的p次方对p取模之后是不是等于a

#include<cstdio>
__int64 quickpow(__int64 a,__int64 p)
{
	__int64 ans=1,base=a,mod=p;	
	while(p)
	{
		if(p&1)
		  ans=((ans%mod)*(base%mod))%mod;
		base=((base%mod)*(base%mod))%mod;
		p>>=1;				
	}
	return ans;
}
__int64 prime(__int64 p)
{
	__int64 i; 
	if(p==2)
	  return 1;
	for(i = 2; i*i<=p; i++)  
        if(p%i == 0)  
            return 0;
	  return 1;	  
}
int main()
{
	__int64 a,p,i;
	__int64 sum;
	while(scanf("%I64d %I64d",&p,&a)&&a||p)
	{
		sum=0;
		sum=quickpow(a,p);	
		if(prime(p))
		  	printf("no\n");	
		else
		{
	       if(sum==a)
	         printf("yes\n");
	       else printf("no\n");
	    }
	}
	return 0;
}