hdu 2138 How many prime numbers （Miller Rabbin大质数判定）

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本文介绍了一个算法问题，即如何从给定的一系列正整数中找出素数的数量，并采用Miller-Rabin素性测试算法进行高效判断。通过案例输入输出样例展示了算法的应用效果。

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How many prime numbers

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18144 Accepted Submission(s): 6180

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

Author

wangye

Source

HDU 2007-11 Programming Contest_WarmUp

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题解：Miller Rabbin大质数判定

控制一下尝试的次数，此题容易TLE

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
int n;
LL quickpow(LL num,LL x,LL p)
{
	LL base=num%p; LL ans=1;
	while (x) {
		if (x&1) ans=ans*base%p;
		x>>=1;
		base=base*base%p;
	}
	return ans; 
}
bool miller_rabbin(LL n)
{
	if (n==2) return true;
	if (n<2||!(n&1)) return false;
	LL t=0;
	LL a,x,y,u=n-1;
	while (!(u&1)) t++,u>>=1;
	for (int i=0;i<=100;i++) {
		a=rand()*rand()%(n-1)+1;
		x=quickpow(a,u,n);
		for (int j=0;j<t;j++) {
			y=x*x%n;
			if (y==1&&x!=1&&x!=n-1) return false;
			x=y;
		}
		if (x!=1) return false;
	} 
	return true;
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("my.out","w",stdout);
	while (scanf("%d",&n)!=EOF){
	    int cnt=0;
		for (int i=1;i<=n;i++) {
			int x; scanf("%d",&x);
			cnt+=miller_rabbin((LL)x);
		} 
		printf("%d\n",cnt);
    }
}