POJ 3518 （筛素数）

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

题意：给你任意一个数，求它的下一个素数和上一个素数的差，前提是这个数不是素数。若该数本事就是素数，就输出0.
思路：本题是一个比较简单的晒素数的题，最主要的是要细心。我的想法是记录该数的前一个素数和后一个素数，如果它本身就是一个素数，那么前一个素数和后一个素数都是它自己。
这样做最大的好处在于访问的时候时间复杂度是O(1)


代码如下：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;

const int N=1300000;

struct node
{
    bool v;      //当前素数是不是素数
    int left;    //前一个素数
    int right;    //后一个素数
    int id;
};

node a[N];

void isprime()
{
    for(int i=0;i<N;i++)
    {
        a[i].v=true;
        a[i].id=0;
        a[i].left=a[i].right=i;
    }
    a[0].v=a[1].v=false;
    int k=1;
    for(int i=2;i<N;i++)
    {
        if(a[i].v)     //i表示当前素数
        {
            a[i].id=k;    //上一个素数的位置
            for(int j=2*i;j<N;j+=i)
            {
                a[j].v=false;    //标记不是素数的数
            }
            for(int j=k+1;j<i;j++)
            {
                a[j].left=k;
                a[j].right=i;
            }
            k=i;
        }
    }
}

int main()
{
    isprime();
    int m;
    while(~scanf("%d",&m))
    {
        if(m==0) return 0;
        if(a[m].v==true)
        {
            printf("0\n");
            continue;
        }
        int k=a[m].right-a[m].left;
        printf("%d\n",k);
    }
    return 0;
}