Problema de conteo de vueltas del temporizador de arena

Dados dos números enteros A y B que representan el tiempo que tardan dos temporizadores de arena diferentes en vaciarse. La tarea es encontrar el conteo de vueltas de cada temporizador hasta la instancia en la que ambos temporizadores de arena se vacían al mismo tiempo.
Ejemplos: 
 

Entrada: A = 30, B = 15 
Salida: 0 1 
Después de 15 minutos: 15 0 
Flip timer 2: 15 15 
Después de otros 15 minutos: 0 0
Entrada: A = 10, B = 8 
Salida: 3 4 
 

Enfoque: el factor común más bajo (LCM) de dos números determinará el momento en que ambos temporizadores de arena se vaciarán juntos. 
LCM(A, B) = (A * B) / GCD(A, B) 
Dividir el LCM por la entrada dará el número de vueltas para cada reloj de arena respectivamente. 
 

//C++14 implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
//Recursive function to return
//the gcd of a and b
int gcd(int a, int b){
    //Everything divides 0
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
//Function to print the number of
//flips for both the sand timers
void flip(int a,int b){
    int lcm =(a * b)/gcd(a, b);
    a = lcm/a;
    b = lcm/b;
    cout<<a-1<<" "<<b-1;
}
 
//Driver code
int main(){
 
int a = 10;
int b = 5;
flip(a, b);
 
}

// Java implementation of the approach
class GFG
{
 
// Recursive function to return
// the gcd of a and b
static int gcd(int a, int b)
{
    // Everything divides 0
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to print the number of
// flips for both the sand timers
static void flip(int a, int b)
{
    int lcm = (a * b) / gcd(a, b);
    a = lcm / a;
    b = lcm / b;
    System.out.print((a - 1) + " " + (b - 1));
}
 
// Driver code
public static void main(String[] args)
{
    int a = 10;
    int b = 5;
    flip(a, b);
}
}
 
// This code is contributed by 29AjayKumar

# Python3 implementation of the approach
 
# Recursive function to return
# the gcd of a and b
def gcd(a, b):
     
    # Everything divides 0
    if (b == 0):
        return a
    return gcd(b, a % b)
 
# Function to print the number of
# flips for both the sand timers
def flip(a, b):
    lcm = (a * b) // gcd(a, b)
    a = lcm // a
    b = lcm // b
    print(a - 1, b - 1)
 
# Driver code
a = 10
b = 5
flip(a, b)
 
# This code is contributed by Mohit Kumar

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Recursive function to return
    // the gcd of a and b
    static int gcd(int a, int b)
    {
        // Everything divides 0
        if (b == 0)
        return a;
         
        return gcd(b, a % b);
    }
     
    // Function to print the number of
    // flips for both the sand timers
    static void flip(int a, int b)
    {
        int lcm = (a * b) / gcd(a, b);
        a = lcm / a;
        b = lcm / b ;
        Console.WriteLine((a - 1) + " " +
                          (b - 1));
    }
     
    // Driver code
    public static void Main()
    {
        int a = 10;
        int b = 5;
        flip(a, b);
    }
}
 
// This code is contributed by AnkitRai01

<script>
// Javascript implementation of the approach
 
// Recursive function to return
// the gcd of a and b
function gcd(a, b)
{
    // Everything divides 0
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to print the number of
// flips for both the sand timers
function flip(a, b){
    let lcm =parseInt((a * b)/gcd(a, b));
    a = parseInt(lcm/a);
    b = parseInt(lcm/b);
    document.write((a-1)+" "+(b-1));
}
 
// Driver code
let a = 10;
let b = 5;
flip(a, b);
 
// This code is contributed by subham348.
</script>

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