45 lines
1.1 KiB
Markdown
45 lines
1.1 KiB
Markdown
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---
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title: Smallest multiple
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localeTitle: 最小的倍数
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---
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## 问题5：最小的倍数
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### 方法：
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* 在这个挑战中，我们需要找到1到n个数的LCM。
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* 要查找数字的LCM，我们使用以下公式：
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* 
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* 为了找到两个数的GCD（最大公约数），我们使用欧几里德算法。
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* 一旦我们得到两个数字的LCM，我们就可以得到从1到n的数字的LCM。
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### 解：
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```js
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//LCM of two numbers
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function lcm(a, b){
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return (a*b)/gcd(a, b);
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}
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//Euclidean recursive algorithm
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function gcd(a, b){
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if (b === 0) return a;
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return gcd(b, a%b);
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}
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function smallestMult(n){
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let maxLCM = 1;
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//Getting the LCM in the range
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for (let i = 2; i <= n; i++){
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maxLCM = lcm(maxLCM, i);
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}
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return maxLCM;
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}
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```
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* [运行代码](https://repl.it/@ezioda004/Problem-5-Smallest-multiple)
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### 参考文献：
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* [欧几里德算法](https://en.wikipedia.org/wiki/Euclidean_algorithm)
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* [LCM](https://en.wikipedia.org/wiki/Least_common_multiple)
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