HDU 2138 How many prime numbers(Miller_Rabin法判断素数 【*模板】 用到了快速幂算法 )

How many prime numbers

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12955    Accepted Submission(s): 4490

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

 

算法：每次输入一个数直接判断暴力sqrt判断该数字是不是素数，然后累加素数的个数，也可以水过。

但是也可以拿这道水题那练习Miller_Rabin判断素数法！时间复杂度比较低！

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>

using namespace std;

long long pow_mod(long long a, long long i, long long n)
{
	if(i==0) return 1%n;
	long long temp=pow_mod(a, i>>1, n);
	temp = temp*temp%n;
	if(i&1)
		temp = (long long)temp*a%n;
	return temp;
}

bool test(int n, int a, int dd)
{
	if(n==2) return true;
	if(n==a) return true;
	if( (n&1)==0 ) return false;
	while(!(dd&1)) dd=dd>>1;

	int t=pow_mod(a, dd, n); //调用快速幂函数
	while((dd!=n-1) &&(t!=1) && (t!=n-1) )
	{
		t = (long long)t*t%n;
		dd=dd<<1;
	}
    return (t==n-1 || (dd&1)==1 );
}

bool Miller_Rabin_isPrime(int n)  //O(logN)
{
	if(n<2) return false;
	int a[]={2, 3, 61}; //
	for(int i=0; i<3; i++)
		if(!test(n, a[i], n-1) )
			return false;
	return true;
}

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        int dd;
        int cnt=0;
        while(n--)
        {
            scanf("%d", &dd);
            if(Miller_Rabin_isPrime(dd))
                cnt++;
        }
        printf("%d\n", cnt );
    }
    return 0;
}