Pseudoprime numbers(poj3641)

题意：就是素数p能够满足a^p%p=a%p,但是有些合数p也满足这个公式，这样的合数被称为伪素数，判断p相对于给定的a是否为伪素数。

解题思路：如果p为素数直接输出no，否则判断a^p%p=a%p是否成立，如果成立输出yes,否则no.

Fermat’s theorem states that for any prime number p and for any integer a>1,ap=a(modp). That is, if we raise ato 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.

输入

Input contains several test cases followed by a line containing “0 0”. Each test case consists of a line containing p and a.

输出

For each test case, output “yes” if p is a base-a pseudoprime; otherwise output “no”.

样例

input

3 2
10 3
341 2
341 3
1105 2
1105 3
0 0

output

no
no
yes
no
yes
yes

#include<stdio.h>
typedef long long ll;
int prim(int p)
{
	for(int i=2;i*i<=p;i++)
	{
		if(p%i==0) return 0;
	}
	return 1;
}
ll mult(int p,int a)
{
	ll ans=1,w=a;
	int mod=p;
	while(p)
	{
		if(p&1) ans=ans*w%mod;
		p>>=1;
		w=w*w%mod;
		
	}
	return ans;
}
int main()
{
	int p,a;
	while(scanf("%d%d",&p,&a),p+a)
	{
		if(!prim(p))
		{
			if(mult(p,a)%p==a)
			 printf("yes\n");
			 else printf("no\n");
		}
		else printf("no\n");
	}
	return 0;
}