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

10112724921700

Sample Output

4060114

#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 1300000bool 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[n]){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;}