Problem P:Prime Gap（素数打表）

POJ3518http://poj.org/problem?id=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 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

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2
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6
10
11
27
2
492170
0

Sample Output

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2
3
4
5
4
0
6
0
114

Source Code

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#include<iostream>
#include<cstring>
#define N 5000005
#define ll long long
using namespace std;
ll n;
ll primes[N];
ll cnt;
bool prime[N];
void make_prime()
{
    memset(prime,true,sizeof(prime));
    prime[0]=false;
    prime[1]=false;
    for(ll i=2;i<=N;i++)
    {
        if(prime[i])
        {
            primes[cnt++]=i;
            for(ll k=i*i;k<=N;k+=i)
                prime[k]=false;
        }
    }
}
int main()
{
    make_prime();//打表
    while(cin>>n&&n)
    {
        if(prime[n])//本身是素数，长度为0
            cout<<0<<endl;
        else
        {
            for(ll i=0;i<cnt;i++)
            {
                if(primes[i]<n&&primes[i+1]>n)
                    cout<<primes[i+1]-primes[i]<<endl;//找到n两边的素数，相减得长度
            }
        }
    }
    return 0;
}

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