本文主要是介绍POJ 3518 Prime Gap，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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
给出整数n，求他前一个素数和后一个素数的差。如果n是素数 输出0。
先用素数筛，再二分答案。
#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
int cnt,pri[1300009];
bool vis[1300009];
void euler(int m)
{vis[1]=1;for(int i=2;i<=m;i++){if(!vis[i])pri[++cnt]=i;for(int j=1;i*pri[j]<=m&&j<=cnt;j++){vis[i*pri[j]]=1;if(i%pri[j]==0)break;}}
}
int bin(int x)
{int ans=lower_bound(pri+1,pri+cnt+1,x)-pri;return pri[ans]-pri[ans-1];
}
int main()
{euler(1299709);int x;while(scanf("%d",&x)&&x){if(vis[x])printf("%d\n",bin(x));elseprintf("0\n");}return 0;
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