【Aizu

Prime Numbers

 Descriptions:

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.

Write a program which reads a list of N integers and prints the number of prime numbers in the list.

Input

The first line contains an integer N, the number of elements in the list.

N numbers are given in the following lines.

Output

Print the number of prime numbers in the given list.

Constraints

1 ≤ N ≤ 10000

2 ≤ an element of the list ≤ 10⁸

Sample Input 1

5
2
3
4
5
6

Sample Output 1

3

Sample Input 2

11
7
8
9
10
11
12
13
14
15
16
17

Sample Output 2

4

题目链接：

https://vjudge.net/problem/Aizu-ALDS1_1_C

题目大意：
输入n个数，判断里面有几个是素数，逐个枚举是最简单的但是会超时，这里用了一个简单的函数，以后可以套用

bool isprime(int x)
{
    if(x==2)
        return true;
    if(x<2||x%2==0)
        return false;
    int i=3;
    while(i<=sqrt(x))
    {
        if(x%i==0)
            return false;
        i+=2;
    }
    return true;
}

 AC代码：

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
#define ME0(x) memset(x,0,sizeof(x))
using namespace std;
bool isprime(int x)
{
    if(x==2)
        return true;
    if(x<2||x%2==0)
        return false;
    int i=3;
    while(i<=sqrt(x))
    {
        if(x%i==0)
            return false;
        i+=2;
    }
    return true;
}
int n;
int main()
{
    int total=0;
    int m;
    cin>>n;
    while(n--)
    {
        cin>>m;
        if(isprime(m))
          total++;
    }
    cout<<total<<endl;
}

这题比较简单，我之所以要为他写一篇博客，是因为一下这个埃拉托色筛选法：

1.列举大于等于2的整数

2.留下最小的整数2，删除所有2的倍数

3.在剩下的整数中留下最小的3，删除所有3的倍数

4.在剩下的整数中留下最小的5，删除所有5的倍数

5.以下同理，留下仍未被删除的最小的整数，删除该整数的倍数，一直循环到结束

int isprime[100005];
void eratos(int x)
{
    for(int i=0; i<=x; ++i)
        isprime[i]=true;
    isprime[0]=isprime[1]=false;
    for(int i=2; i<=x; ++i)
    {
        if(isprime[i])
        {
            int j=i+i;
            while(j<=x)
            {
                isprime[j]=false;
                j+=i;
            }
        }
    }
}

 这个是kuangbin的素数筛法，直接保存到数组，从1开始。

const int MAXN=100010;
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;
        }
    }
}