hdu 2138(米勒—拉宾素数测试)

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

  Give you a lot of positive integers, just to find out how many prime numbers there are.

Input

  There are a lot of cases. In each case, there is an integer N representing the number of integers to find. Each integer won’t exceed 32-bit signed integer, and each of them won’t be less than 2.

Output

  For each case, print the number of prime numbers you have found out.

Sample Input

3 2 3 4

Sample Output

2

解题思路：这道题目如果靠筛选法那肯定会超时。米勒—拉宾素数测试可以进行大素数判断。。。

米勒—拉宾素数测试主要依靠两个基本定理：

1）费马小定理：当gcd（a，p）=1且p为素数时，有a^p-1 mod p = 1

2）快速幂取模算法（RSA公钥加密算法）

其实这个算法模板我也没有很明白，先凑合着看吧。。。

AC:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

__int64 qpow(int a,int b,int r)
{
	__int64 ans = 1,buff = a;
	while(b)
	{
		if(b & 1)
			ans = (ans*buff) % r;
		buff = (buff*buff) % r;
		b >>= 1;
	}
	return ans;
}

bool Miller_Rabbin(int n,int a)
{
	int r = 0,s = n-1,j;
	if(!(n % a)) return false;
	while(!(s & 1))
	{
		s >>= 1;
		r++;
	}
	__int64 k = qpow(a,s,n);
	if(k == 1)
		return true;
	for(j = 0; j < r; j++, k = k * k % n)
		if(k == n-1)
			return true;
	return false;
}

bool isPrime(int n)
{
	int tab[4] = {2,3,5,7};
	for(int i = 0; i < 4; i++)
	{
		if(n == tab[i])
			return true;
		if(!Miller_Rabbin(n,tab[i]))
			return false;
	}
	return true;
}

int main()
{
	int n;
	while(cin>>n)
	{
		int a,ans = 0;
		while(n--)
		{
			cin>>a;
			if(isPrime(a))
				ans++;
		}
		cout<<ans<<endl;
	}
	return 0;
}