【Project Euler】【Problem 5】Smallest multiple

最新推荐文章于 2022-01-15 10:30:30 发布

原创最新推荐文章于 2022-01-15 10:30:30 发布 · 696 阅读

0 ·

CC 4.0 BY-SA版权

C 同时被 2 个专栏收录

10 篇文章

订阅专栏

Project Euler

10 篇文章

订阅专栏

本文探讨了如何找出能被1至20所有数字整除的最小正数，并提供了一段C语言代码实现，该算法通过分解每个数字的质因数并合并不同质因数来解决问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Smallest multiple

Problem 5

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

答案:	232792560

最小倍数

Problem 5

2520是最小的可以被1到10整除的数

哪个正数可以被1到20整除？

答案:	232792560

#include <stdio.h>
#include <math.h>

//将1-20之间的数全部进行分解因子，剔除相同因子，然后合并。

int main(int argc, char *argv[])
{
    int result = 1;

    int primes[5];
    int primes_length = -1;

    int quot = 0;
    for (int i=2;i<=20;i++)
    {
        primes_length = 0;
        quot = i;
        for (int j = 2; j <= i; j++)
        {
            if (quot % j ==0)
            {
                quot = quot / j;
                primes[primes_length++] = j;
                j = 1;
            }
        }

        for (int j = 0;j<primes_length ;j++ )
        {
            if (result % primes[j] == 0)
            {
                result /= primes[j];
            }
        }
        for (int j = 0;j<primes_length ;j++ )
        {
            result *= primes[j];
        }
    }

    printf("result is : %d",result);
    return 0;
}