sicily 1500. 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 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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10
11
27
2
492170
0

Sample Output

4
0
6
0
114

分析：首先不可能输入每个数都去找一遍其周围的素数，我比较喜欢的是直接求出所需要的最大范围的素数，然后从这里面再进行下一步的计算。这里使用的是筛选法来得到素数集。

#include <iostream>

using namespace std;

int main(int argc, char const *argv[])
{
    int prime[];
    for (int i = ; i != ; ++i) {
        prime[i] = ;
    }
    for (int i = ; i != ; ++i) {
        if (prime[i] == ) {
            for (int j =  * i; j < ; j += i)
                prime[j] = ;
        }
    } // 筛选法找素数

    int num;
    while (cin >> num && num != ) {
        int length = ;
        if (prime[num] != ) {
            int upperBound = , lowerBound = ;
            for (lowerBound = num - ; lowerBound >= ; --lowerBound) {
                if (prime[lowerBound] == )
                    break;
            }
            for (upperBound = num + ; upperBound < ; ++upperBound) {
                if (prime[upperBound] == )
                    break;
            }
            length = upperBound - lowerBound;
        }
        cout << length << endl;
    }
    return ;
}