GCD and LCM | CodeChef Solution

Hello coders, today we are going to solve GCD and LCM CodeChef Solution whose Problem code is FLOW016.

Task

Two integers A and B are the inputs. Write a program to find GCD and LCM of A and B.

Input Format

The first line contains an integer T, total number of testcases. Then follow T lines, each line contains an integer A and B.

Output Format

Display the GCD and LCM of A and B separated by space respectively. The answer for each test case must be displayed in a new line.

Constraints

• 1 ≤ T ≤ 1000
• 1 ≤ A,B ≤ 1000000

Example

Sample Input

3 
120 140
10213 312
10 30

Sample Output

20 840
1 3186456
10 30

Solution – GCD and LCM | CodeChef Solution

C++

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
	// your code goes here
	int t=0;
	cin>>t;
	while(t--)
	{
	    long long int a=0,b=0;
	    cin>>a>>b;
	    cout<<__gcd(a,b)<<" "<<((a*b)/__gcd(a,b))<<endl;
	}
	return 0;
}

Python

#Solution Provided by CodingBroz
T = int(input())
for i in range(T):
    num = [int(x) for x in input().split(' ')]
    a = min(num)
    b = max(num)
    multiply = a*b
    while(b):
        a,b = b,a%b
    hcf = a
    lcm = multiply/hcf
    print(hcf,int(lcm))

Java

import java.util.*;
public class Main {
    
    public static void main(String[] args) {
        Scanner input =new Scanner(System.in);
        int t=input.nextInt();
        long a,b,Gcd,Lcm;
        while(t-->0) {
            a=input.nextInt();
            b=input.nextInt();
            Gcd=getGcd(a,b);
            Lcm=(a*b)/Gcd;
            System.out.println(Gcd+" "+Lcm);
        }

    }
    public static long getGcd(long a, long b) {
        while(a != b) {
            if(a>b) a=a-b;
            else b=b-a;
        }
        return a;
    }

}

Disclaimer: The above Problem (GCD and LCM) is generated by CodeChef but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.