U - Primes (素数筛)

最新推荐文章于 2023-03-03 13:32:39 发布

原创最新推荐文章于 2023-03-03 13:32:39 发布 · 539 阅读

0 ·

CC 4.0 BY-SA版权

ACM训练+实训+大学编程练习专栏收录该内容

362 篇文章

订阅专栏

本文介绍了一个简单的程序，用于读取一系列整数并判断每个数是否为素数。素数定义为仅能被1和自身整除的大于1的自然数。程序通过素数筛选算法预先标记所有不大于16000的整数，并针对每个输入整数输出其是否为素数的结果。

Write a program to read in a list of integers and determine whether or not each number is prime. A number, n, is prime if its only divisors are 1 and n. For this problem, the numbers 1 and 2 are not considered primes.
Input
Each input line contains a single integer. The list of integers is terminated with a number<= 0. You may assume that the input contains at most 250 numbers and each number is less than or equal to 16000.
Output
The output should consists of one line for every number, where each line first lists the problem number, followed by a colon and space, followed by “yes” or “no”.
Sample Input

Sample Output

1: no
2: no
3: yes
4: no
5: yes
6: yes

注意题目的条件，1和2不是素数，还要注意输入结束的标志是<=0。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
bool p[17345];
void isprime()//素数筛
{
   memset(p, true, sizeof(p));
   p[1] = false;
   for(int i=2;i<=16000;i++)
   {
      if(p[i]==true)
      {
         for(int j=i+i;j<=16000;j+=i)
         {
           p[j] = false;
         }
      }
   }
}
int main()
{
  int n;
  int t = 1;
  isprime();
  while(1)
  {
     scanf("%d", &n);
     if(n<=0)//结束条件
     break;
     if(p[n]==true&&n!=2)//2不是素数
     printf("%d: yes\n", t);
     else
     printf("%d: no\n", t);
     t++;
  }
   return 0;
}