Hello Programmers In this post, you will know how to solve the GCD and LCM Codechef Solution.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Codechef by yourself. If you find any difficulty after trying several times, then you can look for solutions.
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 Solutions 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 Solutions 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 Solutions 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 BrokenProgrammers. This tutorial is only for Educational and Learning purpose.
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