HDU 5901 Count primes 大素数计数

**题意：**计算1~N间素数的个数（N<=1e11) **题解：**题目要求很简单，作为论文题，模板有两种 \(O（n^frac{3}{4} )\)，另一种lehmer\(O（n^frac{2}{3})\) [link:https://zh.wikipedia.org/wiki/%E7%B4%A0%E6%95%B0%E8%AE%A1%E6%95%B0%E5%87%BD%E6%95%B0](https://zh.wikipedia.org/wiki/%E7%B4%A0%E6%95%B0%E8%AE%A1%E6%95%B0%E5%87%BD%E6%95%B0)

/** @Date    : 2016-11-18-13.59

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @Link    : https://github.com/

  * @Version :

  */

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <math.h>

#include <queue>

//#include<bits/stdc++.h>

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e6+2000;

using namespace std;

//high[i] means pi(n/i),low[i] means pi(i)

LL high[340000];

LL low[340000];

LL n;

LL fun()

{

	LL i,m,p,s,x;

	for(m = 1; m * m <= n; m++)

		high[m] = n/m-1;

	for(i = 1;i <= m; i++)

		low[i] = i-1;

	for(p = 2; p <= m; p++)

	{

		if(low[p] == low[p-1])

			continue;

		s = min(n/(p*p),m-1);

		for(x = 1; x <= s; x++)

		{

			if(x*p <= m-1)

				high[x] -= high[x*p] - low[p-1];

			else

				high[x] -= low[n/(x*p)] - low[p-1];

		}

		for(x = m; x >= p*p; x--)

			low[x] -= low[x/p] - low[p-1];

	}

}



int main()

{

	while(cin>>n)

    {

        fun();

        cout << high[1] << endl;

    }

}