3518 Prime Gap

最新推荐文章于 2022-08-23 15:03:11 发布

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最新推荐文章于 2022-08-23 15:03:11 发布

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原文链接：http://www.cnblogs.com/372465774y/archive/2012/07/05/2578601.html

本文提供了一个算法，用于计算给定整数与相邻素数之间的差值，若该整数为合数。通过筛选法求素数，实现对PrimeGap的高效计算。

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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

Sample Output

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;
}