[数学] AOJ 0009 素数筛选 Prime Number

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最新推荐文章于 2024-06-27 09:18:33 发布

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分类专栏： ————ACM训练———— 4、数学题文章标签： ACM 素数数学

本文链接：https://blog.youkuaiyun.com/tlonline/article/details/46974499

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4、数学题

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本文介绍了一个简单的素数筛选算法，该算法能够高效地找出1到n之间的所有素数，并给出了一段C++实现代码示例。

Prime Number

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

Description

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, 7.

Input

Input consists of several datasets. Each dataset has an integer n (n ≤ 999999) in a line.

The number of datasets ≤ 30.

Output

For each dataset, prints the number of prime numbers.

Sample Input

10
3
11

Output for the Sample Input

4
2
5

给定n，求1~n中素数的个数

非常简单的素数筛选，套模板直接1A。

最近做的这几题都是在AC找手感~~明天开始做多校联合

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

const int MAXN = 10000007;

int prime[MAXN+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2; i<=MAXN; i++)
    {
        if (!prime[i]) prime[++prime[0]] = i;
        for(int j = 1;j<=prime[0] &&prime[j] < MAXN/i;j++ )
        {
            prime[prime[j]*i] = 1;
            if(i%prime[j]==0) break;
        }
    }
}

int main()
{
    getPrime();
    int n,i;
    while(cin>>n)
    {
        for(i = 1; ;i++)
            if (prime[i] > n) break;
        cout<<i-1<<endl;
    }
}