埃氏筛

最新推荐文章于 2024-09-20 21:43:34 发布

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本文介绍了一种通过埃氏筛法实现的程序，用于计算小于等于给定整数n的所有质数的数量。该程序首先使用埃氏筛法预处理出一定范围内的所有质数，并存储这些质数的状态。然后对于每个输入的n，程序能够快速地输出小于等于n的质数个数。

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Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.

Input

Input consists of several datasets. Each dataset has an integer n (1 ≤ n ≤ 999,999) in a line.

The number of datasets is less than or equal to 30.

Output

For each dataset, prints the number of prime numbers.

Sample Input

10
3
11

Output for the Sample Input

4
2
5

题目大意：求出1~n的质数个数。

方法：打表，埃氏筛。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 1000000
using namespace std;
int n,m;
int prime(int n)
{
    for(int i=2;i*i<=n;i++)
        if(n%i==0) return 0;
    return 1;
}
int vis[N]; //是否已经被访问过，访问过就是素数
int p[N];   //存n之前的素数个数
int main()
{
    int m=sqrt(N);
    for(int i=2;i<=m;i++)
    {
        if(!vis[i])
        {
            for(int j=i*i;j<=N;j+=i)
                vis[j]=1;
        }
    }
    for(int i=2;i<=N;i++)
    {
        p[i]=p[i-1];
        if(!vis[i]) p[i]++;
    }
   while(cin>>n)
   {

       cout<<p[n]<<endl;
   }
    return 0;
}