Aizu - ALDS1_1_CPrime Numbers(素数筛法)
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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;
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