题解：Prime Gap（素数间隙长度）

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

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <algorithm>
typedef long long ll;
using namespace std; 
#define N 2000010
ll n,a[N],biao[N];
void prime()
{
 memset(a,0,sizeof(a));
 a[0]=a[1]=1;
 for(ll i=2;i<=N/2;i++)
  if(a[i]==0)
  {
   for(ll j=2*i;j<=N;j+=i)
    a[j]=1;	
  }	
}
int main()
{
    prime();
    ll n,sum=0;
    while(cin>>n&&n)
    {
     sum=0;
     if(!a[n]) 
	 {
	  cout<<0<<endl;continue;	
	 }
	 int sum=2;
	 int t=n-1,t1=n+1;
	 while(t&&a[t])
	 {
	  sum++;t--;	
	 } 
	 while(a[t1])
	 {
	 	sum++;t1++;
	 }
	 cout<<sum<<endl;
	}
    return 0;
}