Hdoj 1905.Pseudoprime numbers 题解

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

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

Sample Output

no
no
yes
no
yes
yes

Author

Gordon V. Cormack

Source

2008-1杭电公开赛(非原创)

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思路

就是判断\(a^p\%p==a\)，计算\(a^p\)可以用快速幂的方法，快速幂本质也是二分不断加速

代码

#include<bits/stdc++.h>
using namespace std;
typedef __int64 ll;

bool isprime(ll x)
{
	for(int i=2;i<sqrt(x);i++)
		if(x%i==0)
			return false;
	return true;
}//判断是否为质数

ll quickpower(ll a,ll b,ll c)
{
	ll ans =1;
	while(b)
	{
		if(b&1)
			ans = (ans*a) % c;
		a = (a*a) % c;
		b >>= 1;
	}
	return ans;
}//返回a^b%c的结果
int main()
{
	int a,p;
	while(cin>>p>>a)
	{
		if(p==0 && a==0) break;
		if(isprime(p))
			cout << "no" << endl;
		else
		{
			int ans_power = quickpower(a,p,p);
			if(ans_power==a)
				cout << "yes" << endl;
			else
				cout << "no" << endl;
		}
	}
	return 0;
}