GCD and LCM Codechef Solution

GCD and LCM Codechef Solution

Problem

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

Input

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

Output 

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 <= 100000

Example

Input:

3
120 140
10213 312
10 30

Output:

20 840
1 3186456
10 30

GCD and LCM – CodeChef Solution in Python

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))

GCD and LCM – CodeChef Solution in CPP

#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;
}

GCD and LCM – CodeChef Solution in 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 Chase2learn.This tutorial is only for Educational and Learning purpose.

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