3518 Prime Gap

Prime Gap

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6597		Accepted: 3775

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

Source

Japan 2007

//给你一个数，是素数的话输出0，否则输出 与之相邻的两个素数差，筛选法求素数

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <queue>

#include <cmath>

#include <string.h>

#define N 1299709

using namespace std;

bool b[1300000];

bool is_prime(int n)

{

    int i,m=sqrt(double(n));

		for(i=2;i<=m;i++)

			if(n%i==0)

				return 0;

	return 1;

}

int main()

{

   int j,i;

   b[1]=1;

   for(i=4;i<=N;i+=2)

      b[i]=1;

   for(i=3;i<=N;i+=2)

	   if(is_prime(i))

	   {

	      for(j=i+i;j<=N;j+=i)

			  b[j]=1;

	   }

	   int k;

   while(scanf("%d",&k),k)

   {

       if(is_prime(k))

		   printf("0\n");

	   else

	   {

	      j=k+1;

		  i=k-1;

		  while(b[j])

			  j++;

         while(b[i])

			 i--;

		 printf("%d\n",j-i);



	   }

   }

   return 0;

}