poj 3518 Prime Gap

Prime Gap
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 9799 Accepted: 5638
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

---

【分析】
对yhx的水题感到无感了…
给一个n，求n前一个素数与后一个素数的差，如果n为素数输出0…
于是变成了素数筛法+二分

---

【代码】

//poj 3518 Prime Gap
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=1299809;
int pri[mxn+1],vis[mxn+1],tot;
inline void shai()
{
    int i,j;
    fo(i,2,mxn)
    {
        if(!vis[i]) pri[++tot]=i;
        fo(j,1,tot)
        {
            if(i*pri[j]>=mxn) break;
            vis[i*pri[j]]=1;
            if(i%pri[j]==0) break;
        }
    }
}
int main()
{
    int i,j,n;
    shai();
    while(scanf("%d",&n) && n)
    {
        if(!vis[n]) {printf("0\n");continue;}
        n=lower_bound(pri+1,pri+tot+1,n)-pri-1;
        printf("%d\n",pri[n+1]-pri[n]);
    }
    return 0;
}