Encuentre el número de rectángulos de tamaño 2*1 que se pueden colocar dentro de un rectángulo de tamaño n*m

Dados dos enteros  n ,  m . Encuentra el número de rectángulos de tamaño 2*1 que se pueden colocar dentro de un rectángulo de tamaño n*m. 
Nota: 

1. No se superponen dos rectángulos pequeños.
2. Cada rectángulo pequeño se encuentra completamente dentro del rectángulo grande. Se permite tocar los bordes del rectángulo grande.

Ejemplos : 

Input : n = 3, m =3
Output : 4

Input : n = 2, m = 4
Output : 4

Acercarse:

1. Si N es par, coloque M filas de N/2 rectángulos pequeños y cubra todo el rectángulo grande.
2. Si M es par, coloque N filas de M/2 rectángulos pequeños y cubra todo el rectángulo grande.
3. Si ambos son impares, cubra N – 1 fila del tablero con pequeños rectángulos y coloque pequeños rectángulos de piso (M/2) en la última fila. En el peor de los casos (N y M son impares), una celda permanece descubierta.

A continuación se muestra la implementación del enfoque anterior:

// CPP program to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
#include <bits/stdc++.h>
using namespace std;
 
// function to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
int NumberOfRectangles(int n, int m)
{
    // if n is even
    if (n % 2 == 0)
        return (n / 2) * m;
 
    // if m is even
    else if (m % 2 == 0)
        return (m / 2) * n;
 
    // if both are odd
    return (n * m - 1) / 2;
}
 
// Driver code
int main()
{
    int n = 3, m = 3;
 
    // function call
    cout << NumberOfRectangles(n, m);
 
    return 0;
}

// C program to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
#include <stdio.h>
 
// function to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
int NumberOfRectangles(int n, int m)
{
    // if n is even
    if (n % 2 == 0)
        return (n / 2) * m;
 
    // if m is even
    else if (m % 2 == 0)
        return (m / 2) * n;
 
    // if both are odd
    return (n * m - 1) / 2;
}
 
// Driver code
int main()
{
    int n = 3, m = 3;
 
    // function call
    printf("%d",NumberOfRectangles(n, m));
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.

// Java program to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
 
public class GFG {
     
    // function to Find the number of
    // rectangles of size 2*1 can be placed
    // inside a rectangle of size n*m
    static int NumberOfRectangles(int n, int m)
    {
        // if n is even
        if (n % 2 == 0)
            return (n / 2) * m;
       
        // if m is even
        else if (m % 2 == 0)
            return (m / 2) * n;
       
        // if both are odd
        return (n * m - 1) / 2;
    }
    public static void main(String args[])
    {
         int n = 3, m = 3;
           
            // function call
            System.out.println(NumberOfRectangles(n, m));
           
    }
    // This Code is contributed by ANKITRAI1
}

# Python 3 program to Find the
# number of rectangles of size
# 2*1 can be placed inside a
# rectangle of size n*m
 
# function to Find the number
# of rectangles of size 2*1
# can be placed inside a
# rectangle of size n*m
def NumberOfRectangles(n, m):
 
    # if n is even
    if (n % 2 == 0):
        return (n / 2) * m
 
    # if m is even
    elif (m % 2 == 0):
        return (m // 2) * n
 
    # if both are odd
    return (n * m - 1) // 2
 
# Driver code
if __name__ == "__main__":
    n = 3
    m = 3
 
    # function call
    print(NumberOfRectangles(n, m))
 
# This code is contributed
# by ChitraNayal

// C# program to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
using System;
 
class GFG
{
     
// function to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
static int NumberOfRectangles(int n, int m)
{
    // if n is even
    if (n % 2 == 0)
        return (n / 2) * m;
 
    // if m is even
    else if (m % 2 == 0)
        return (m / 2) * n;
 
    // if both are odd
    return (n * m - 1) / 2;
}
 
// Driver Code
public static void Main()
{
    int n = 3, m = 3;
 
    // function call
    Console.WriteLine(NumberOfRectangles(n, m));
     
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
}

<?php
 
// PHP program to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
 
// function to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
function NumberOfRectangles($n, $m)
{
     
    // if n is even
    if ($n % 2 == 0)
        return ($n / 2) * $m;
 
    // if m is even
    else if ($m % 2 == 0)
        return ($m / 2) * $n;
 
    // if both are odd
    return ($n * $m - 1) / 2;
}
 
// Driver code
$n = 3;
$m = 3;
 
// function call
echo NumberOfRectangles($n, $m);
 
// This code is contributed
// by Shivi_Aggarwal
?>

<script>
 
// Javascript program to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
 
// Function to Find the number of
// rectangles of size 2*1 can be placed
// inside a rectangle of size n*m
function NumberOfRectangles(n, m)
{
     
    // If n is even
    if (n % 2 == 0)
        return (n / 2) * m;
   
    // If m is even
    else if (m % 2 == 0)
        return (m / 2) * n;
   
    // If both are odd
    return (n * m - 1) / 2;
}
 
// Driver Code
var n = 3, m = 3;
           
// Function call
document.write(NumberOfRectangles(n, m));
 
// This code is contributed by Ankita saini
 
</script>

4

Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)