Prime Gap （素数）

Problem Description

The sequence of n 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n. For example, 24, 25, 26, 27, 28 between 23 and 29 is a prime gap of length 6.

Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.

Input

<p>The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.</p>

Output

<p>The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.</p>

Sample Input

10
11
27
2
492170
0

Sample Output

4
0
6
0
114

题目大概：

求素数间隙，主要是要把素数打表。

思路：

晒素数。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n;
int su[1300000];

int sushu()
{   su[0]=su[1]=1;
    for(int i=2;i<=1300000;i++)
    {
        if(su[i]==0)
        {
            for(int j=i+i;j<=1300000;j=j+i)
            {
                su[j]=1;
            }
        }
    }

}
int main()
{  memset(su,0,sizeof(su));
   sushu();

   while(scanf("%d",&n)&&n)
   {
       if(!su[n])
       {
           printf("0\n");
       }
       else
       {
           int x=n-1;
           int y=n+1;
           while(su[x])x--;
           while(su[y])y++;
           printf("%d\n",y-x);
       }

   }
    return 0;
}