Programa para encontrar la cantidad después del reemplazo de la mezcla

Dado un recipiente que tiene X litros de leche. Se extraen Y litros de leche y se reemplazan con Y litros de agua. Esta operación se realiza Z veces. La tarea es averiguar la cantidad de leche que queda en el recipiente.
Ejemplos: 
 

Input: X = 10 liters, Y = 2 liters, Z = 2 times
Output: 6.4 liters

Input: X = 25 liters, Y = 6 liters, Z = 3 times
Output: 10.97 liters

Fórmula:- 
 

A continuación se muestra la implementación requerida: 
 

// C++ implementation using above formula
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the Remaining amount.
float Mixture(int X, int Y, int Z)
{
    float result = 0.0, result1 = 0.0;
 
    // calculate Right hand Side(RHS).
    result1 = ((X - Y) / (float)X);
    result = pow(result1, Z);
 
    // calculate Amount left by
    // multiply it with original value.
    result = result * X;
 
    return result;
}
 
// Driver Code
int main()
{
    int X = 10, Y = 2, Z = 2;
 
    cout << Mixture(X, Y, Z) << " litres";
    return 0;
}

// Java code using above Formula.
import java.io.*;
 
class GFG
{
// Function to calculate the
// Remaining amount.
static double Mixture(int X, int Y, int Z)
{
    double result1 = 0.0, result = 0.0;
 
    // calculate Right hand Side(RHS).
    result1 = ((X - Y) / (float)X);
    result = Math.pow(result1, Z);
 
    // calculate Amount left by
    // multiply it with original value.
    result = result * X;
 
    return result;
}
 
// Driver Code
public static void main(String[] args)
{
    int X = 10, Y = 2, Z = 2;
 
    System.out.println((float)Mixture(X, Y, Z) +
                                     " litres");
}
}
 
// This code is contributed
// by Naman_Garg

# Python 3 implementation using
# above formula
 
# Function to calculate the
# Remaining amount.
def Mixture(X, Y, Z):
 
    result = 0.0
    result1 = 0.0
 
    # calculate Right hand Side(RHS).
    result1 = ((X - Y) / X)
    result = pow(result1, Z)
 
    # calculate Amount left by
    # multiply it with original value.
    result = result * X
 
    return result
 
# Driver Code
if __name__ == "__main__":
    X = 10
    Y = 2
    Z = 2
 
    print("{:.1f}".format(Mixture(X, Y, Z)) +
                                   " litres")
 
# This code is contributed by ChitraNayal

// C# code using above Formula.
using System;
 
class GFG
{
// Function to calculate the
// Remaining amount.
static double Mixture(int X,
                      int Y, int Z)
{
    double result1 = 0.0, result = 0.0;
 
    // calculate Right hand Side(RHS).
    result1 = ((X - Y) / (float)X);
    result = Math.Pow(result1, Z);
 
    // calculate Amount left by
    // multiply it with original value.
    result = result * X;
 
    return result;
}
 
// Driver Code
public static void Main()
{
    int X = 10, Y = 2, Z = 2;
 
    Console.WriteLine((float)Mixture(X, Y, Z) +
                                    " litres");
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

<?php
// PHP implementation of above formula
 
// Function to calculate the
// Remaining amount.
function Mixture($X, $Y, $Z)
{
    $result = 0.0;
    $result1 = 0.0;
     
    // calculate Right hand Side(RHS).
    $result1 = (($X - $Y) / $X);
    $result = pow($result1, $Z);
 
    // calculate Amount left by
    // multiply it with original value.
    $result = $result * $X;
 
    return $result;
}
 
// Driver Code
$X = 10;
$Y = 2;
$Z = 2;
echo Mixture($X, $Y, $Z), " litres";
 
// This code is contributed
// by Sanjit_Prasad
?>

<script>
 
// Javascript implementation using above formula
 
// Function to calculate the Remaining amount.
function Mixture(X, Y, Z)
{
    var result = 0.0, result1 = 0.0;
 
    // calculate Right hand Side(RHS).
    result1 = ((X - Y) / X);
    result = Math.pow(result1, Z);
 
    // calculate Amount left by
    // multiply it with original value.
    result = result * X;
 
    return result;
}
 
// Driver Code
var X = 10, Y = 2, Z = 2;
document.write( Mixture(X, Y, Z).toFixed(1) + " litres");
 
</script>

6.4 litres

Complejidad de tiempo: O(log(n))
Espacio auxiliar: O(1)