[BOJ][2609] 최대공약수와 최소공배수

1. 문제 링크

https://www.acmicpc.net/problem/2609

2. 사용 알고리즘

유클리드 호제법

3. 풀이

유클리드 호제법을 이용하여 최대공약수, 최소공배수 계산

LCM = A * B / GCD

4. 소스 코드

4-1. C++

https://github.com/dev-aiden/problem-solving/blob/main/boj/2609.cpp

#include <iostream>

using namespace std;

int gcd(int a, int b) {
    while (b != 0) {
        int r = a % b;
        a = b;
        b = r;
    }
    return a;
}

int lcm(int a, int b) {
    return a * b / gcd(a, b);
}

int main(void) {
    ios_base::sync_with_stdio(false);
    int a, b; cin >> a >> b;
    cout << gcd(a, b) << "\n";
    cout << lcm(a, b) << "\n";
    return 0;
}

4-2. JAVA

https://github.com/dev-aiden/problem-solving/blob/main/boj/2609.java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        System.out.println(gcd(a, b));
        System.out.println(lcm(a, b));
    }

    public static int gcd(int a, int b) {
        while (b != 0) {
            int r = a % b;
            a = b;
            b = r;
        }
        return a;
    }

    public static int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }
}