Problem:
Problem B: Prime Gap

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

Output for the Sample Input

4
0
6
0
114

Solution in C++ (judge verdict: Accepted)

//suhan lee,saitama university

#include <vector>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <ctime>
#include <climits>
#include <cassert>
#include <complex>
using namespace std;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;

#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define repn(i,n) for(int i=0;i<n;i++)
#define EACH(t,i,c) for(t::iterator i=(c).begin(); i!=(c).end(); ++i)
#define pb push_back 
#define pf push_front

int main(){
  bool prime[1300000];
  int tmp,tmp2;
  int k;
  for(int i=0;i<1300000;i++)
    prime[i]=0;
  for(int i=2;i<1300000;i++){
    if(!prime[i]){
      tmp=i;
      //  cout <<i<<endl;
	tmp+=i;
      while(tmp<1300000){

	prime[tmp]=1;
	tmp+=i;	
      }
    }
  }
      while(1){
	cin>>k;
	if(k==0) return 0;
	if(!prime[k])
	  cout<<"0\n";
	else{
	  tmp2=k;
	  tmp=k;
	  while(prime[tmp2])
	    tmp2--; 
	  while(prime[tmp])
	    tmp++; 
	  //	  cout<<tmp<<" "<<tmp2<<" "<<tmp-tmp2<<endl;
	  cout<<tmp-tmp2<<endl;
	}

      }

}