POJ 3518

D - Prime Gap

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

Submit Status Practice POJ 3518

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

10112724921700

Sample Output

4060114

//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 1299709using 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;}