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  

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.

Output 

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.

Sample Input 

10
11
27
2
492170
0

Sample Output

4
0
6
0
114

题意+思路：

先构造素数表，然后判断输入的n是否为素数，如果是素数的话，直接输出0，如果不是素数的话找到两个连续的素数，分别大于n和小于n输出这两个素数的差即为答案。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#define LL long long
using namespace std;
#define N 5000005
LL prime[N];
int cnt=1;
bool bprime[N];
void make_prime()
{
      memset(bprime,true,sizeof(bprime));
      bprime[0]=false;
      bprime[1]=false;
      for(LL i=2;i<=N;i++)
      {
           if(bprime[i])
           {
                prime[cnt++]=i;
                for(LL j=i*i;j<=N;j+=i)
                bprime[j]=false;
           }
       }
   }
int main()
{
    make_prime();
    LL n;
    while(scanf("%lld",&n)!=EOF&&n)
    {
          if(bprime[n])
           printf("0\n");
          else
          {
               for(LL i=1;i<=cnt;i++)
               if(prime[i]<n&&prime[i+1]>n)
               printf("%lld\n",prime[i+1]-prime[i]);
          }
     }
 return 0;
}