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最新推荐文章于 2017-09-30 00:27:15 发布

翻译最新推荐文章于 2017-09-30 00:27:15 发布 · 434 阅读

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#integer #input #output #each #numbers #less

数论同时被 2 个专栏收录

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本文介绍了一个计算特定整数所在素数间隙长度的程序设计方法。该程序通过素数筛选算法预处理所有小于等于第100000个素数（1299709）的整数，并确定这些整数是否位于两个素数之间的连续合数序列中。

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

Sample Output

Source

Japan 2007

素数筛选。

代码如下：

#include<stdio.h>
#include<string.h>
int prime[1299710];
//素数筛选：
void make_prime(void) {
    memset(prime, 1, sizeof (prime));
    prime[0] = prime[1] = 0;
    for (int i = 2; i < 10000; i++)
        if (prime[i])
            for (int j = i * i; j <= 1299710; j += i)
                prime[j] = 0;
}
//找到这个数左右两侧的素数
int work(long x) {
    if (prime[x])
        return 0;
    int i, j;
    for (i = x; !prime[i]; i--);
    for (j = x; !prime[j]; j++);
    return j - i;
}

int main(void) {
    int n;
    make_prime();
    while (scanf("%d", &n), n) {
        int ans = work(n);
        printf("%d\n", ans);//输出结果
    }
    return 0;
}