POJ 3518

D - Prime Gap

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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 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

//D
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>

#include <iostream>
#define N 1299709
using namespace std;

int p[1299709];
int pn[100000];

//素数打表
void prim_num()
{
    int i,j,n;
    for(i=1; i<=N; i++)
        p[i]=true;
    n=(int)sqrt(N);
    for(i=2; i<=n; i++)
    {
        for(j=i+i; j<=N; j+=i)
        {
            p[j]=false;
        }
    }
    j=1;
    for(i=1; i<=N; i++)
    {
        if(p[i])
        {
            pn[j++]=i;
        }
    }
}
int f(int x)
{
    if(x == 0|| x == 1 )
          return 0;
    int w;
    for( w = 2 ; w<= sqrt(x);w++)
    {
        if(x % w == 0)
            break;
    }
    if(w > sqrt(x))
        return 1 ;
    else return 0;
}
int main()
{
    prim_num();
    int n ;
    while(cin >> n)
    {
        if(n == 0)
            break;
        if(f(n))
            cout << "0\n";
        else

        {
            int m;
            m = lower_bound(pn,pn+100000,n) - pn;

            cout << pn[m]-pn[m-1]<< endl;
        }
    }
    return 0;
}