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，这里用力一个简单的函数，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])
{
j=i+i;
while(j<=x)
{
isprime[j]=false;
j+=i;
}
}
}
}
原文地址：https://www.cnblogs.com/sky-stars/p/11107904.html