Maximice el número de hojas posible reduciendo repetidamente su área a la mitad

Dados dos números enteros A y B , que representan el largo y el ancho de una hoja, la tarea es encontrar el número máximo de hojas que se pueden generar reduciendo repetidamente el área a la mitad hasta que no sea divisible por 2 .

Ejemplos:

Entrada : A = 5, B = 10
Salida : 2
Explicación: 

• Área Inicial = 5 * 10. Cuenta = 0.
• Área / 2 = 5 * 5. Cuenta = 2.
   

Entrada : A = 1, B = 8
Salida : 8

Enfoque: siga los pasos a continuación para resolver el problema:

• Calcular el área total de la hoja inicial proporcionada.
• Ahora, sigue dividiendo el área de la hoja por 2 hasta que se vuelva impar.
• Después de cada división, aumente la cuenta al doble de su valor.
• Finalmente, imprima el conteo obtenido.

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the
// maximum number of sheets
// possible by given operations
int maxSheets(int A, int B)
{
    int area = A * B;
 
    // Initial count of sheets
    int count = 1;
 
    // Keep dividing the
    // sheets into half
    while (area % 2 == 0) {
 
        // Reduce area by half
        area /= 2;
 
        // Increase count by twice
        count *= 2;
    }
 
    return count;
}
 
// Driver Code
int main()
{
 
    int A = 5, B = 10;
    cout << maxSheets(A, B);
 
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to calculate the
  // maximum number of sheets
  // possible by given operations
  static int maxSheets(int A, int B)
  {
    int area = A * B;
 
    // Initial count of sheets
    int count = 1;
 
    // Keep dividing the
    // sheets into half
    while (area % 2 == 0)
    {
 
      // Reduce area by half
      area /= 2;
 
      // Increase count by twice
      count *= 2;
    }
    return count;
  }
 
 
  // Driver Code
  public static void main(String args[])
  {
    int A = 5, B = 10;
    System.out.println(maxSheets(A, B));
  }
}
 
// This code is contributed by jana_sayantan.

# Python program for the above approach
 
# Function to calculate the
# maximum number of sheets
# possible by given operations
def maxSheets( A, B):
    area = A * B
 
    # Initial count of sheets
    count = 1
 
    # Keep dividing the
    # sheets into half
    while (area % 2 == 0):
 
        # Reduce area by half
        area //= 2
 
        # Increase count by twice
        count *= 2
    return count
 
# Driver Code
A = 5
B = 10
print(maxSheets(A, B))
 
# This code is contributed by rohitsingh07052.

// C# program for the above approach
using System;
class GFG
{
 
    // Function to calculate the
    // maximum number of sheets
    // possible by given operations
    static int maxSheets(int A, int B)
    {
        int area = A * B;
 
        // Initial count of sheets
        int count = 1;
 
        // Keep dividing the
        // sheets into half
        while (area % 2 == 0)
        {
 
            // Reduce area by half
            area /= 2;
 
            // Increase count by twice
            count *= 2;
        }
 
        return count;
    }
 
    // Driver Code
    public static void Main()
    {
        int A = 5, B = 10;
        Console.WriteLine(maxSheets(A, B));
    }
}
 
// This code is contributed by chitranayal.

<script>
 
    // Javascript program for the above approach
     
    // Function to calculate the
    // maximum number of sheets
    // possible by given operations
    function maxSheets(A, B)
    {
        let area = A * B;
 
        // Initial count of sheets
        let count = 1;
 
        // Keep dividing the
        // sheets into half
        while (area % 2 == 0) {
 
            // Reduce area by half
            area /= 2;
 
            // Increase count by twice
            count *= 2;
        }
 
        return count;
    }
    // Driver Code
     
     let A = 5, B = 10;
    document.write(maxSheets(A, B));
     
</script>
   

2

Complejidad de Tiempo: O(log ₂ (A * B))
Espacio Auxiliar: O(1)