题解 | #求最小公倍数#

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;

void async function () {
    // Write your code here
    while(line = await readline()){
        let [a,b] = line.split(' ');
        if(a>b){
            [a,b] = [b,a]
        }
        // 思路：从2往上循环，如果两个a b数除完i之后余数为0，则i为公约数，然后重置 a b 的值，并且将 i--，因为除完了2，可能还有个2的公约数
        let arr = []
        for(let i = 2;i<=a;i++){
            if(a%i == 0 && b%i == 0){
                arr.push(i)
                a = a/i
                b = b/i
                i--;
            }
        }
        arr.push(a,b)
        console.log(arr.reduce((pre,v)=>pre*=v,1)) 
    }
}()