Prime Gap--素数打表

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 Description
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 Description
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

#include <iostream>
//复数，不是1也不是素数
//如果没有素数间隙包含K的话，那么直接输出0
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
//素数打表得打到130W个
#define maxn 1300000
bool vis[maxn];
int B[120000];
int main()
{
memset(vis,0,sizeof(vis));
vis[0]=vis[1]=1;
for(int i=2;i*i<=maxn;i++)
{
if(!vis[i])
{
for(int j=i*i;j<=maxn;j+=i)
{
vis[j]=1;
}
}
}
int k=1;
for(int i=2;i<=maxn;i++)
{
if(!vis[i])B[k++]=i;
}
int n;
while(scanf("%d",&n)!=EOF&&n)
{
if(n==2||n==1299709||!vis
)
{
printf("%d\n",0);
}
else
{
for(int i=1;i<k;i++)
{
if(B[i]<n&&B[i+1]>n)
{
printf("%d\n",B[i+1]-B[i]);
break;
}
}
}
}
return 0;
}