Prime Gap

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 numbersp 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 containsk.

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

10112724921700

Sample Output

4060114

大体的题意：

给定一个数字，如果是素数那么就输出0就ok了，如果不是，就找到两边的第一个素数边界，然后输出左边的与右边素数的差。

打个素数表，然后直接找到，竟然没有超时。

#if 0#include<iostream>#include<cstring>using namespace std;const int MAX=1299999;int a[MAX],k;bool vis[MAX];void fun(){memset(vis,1,sizeof(vis));for(int i=2; i<=1299709; i++){if(vis[i]==1){for(int j=2; j*i<=1299709; j++){vis[i*j]=0;}}}for(int i=1; i<=1299709; i++){if(vis[i]==1){a[k]=i;k++;}}}int main(){int n;fun();while(cin>>n&&n){if(vis[n]==1) {cout<<"0"<<endl;}else{for(int i=1; i<k; i++){if(a[i]>n&&a[i-1]<n){cout<<a[i]-a[i-1]<<endl;break;}}}}}#endif