求解最小公倍数思路:

1.最小公倍数=两个数的乘机 / 两个数的最大公约数

2.最大公约数(辗转相除法)gcd(a,b)

gcd(a,b)=gcd(b,r)

将a为较大的数,b为余数,直到b等于0

import java.util.*;

public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt();
        int b = sc.nextInt();
        if(a == b){
            System.out.println(a);
        }else{
            int res = (a*b)/gcd(a,b);
            System.out.println(res);
        }
    }
    public static int gcd(int a,int b){
        if(a<b){
            int t = a;
            a = b;
            b = t;
        }
        while(b != 0){
            int r = a%b;
            a = b;
            b = r;
        }
        return a;
    }
}