标签：简单的 lse size 函数 names oid for 最小 ber
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; } } } }
【Aizu - ALDS1_1_C】Prime Numbers（素数筛法）
标签：简单的 lse size 函数 names oid for 最小 ber
原文地址：https://www.cnblogs.com/sky-stars/p/11107904.html
