U - Primes (素数筛)

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

1
2
3
4
5
17
0

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 
#include 
#include 
#include 
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;
}