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Find the number of rectangles of size 2*1 which can be placed inside a rectangle of size n*m

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Given two integers nm. Find the number of rectangles of size 2*1 that can be placed inside a rectangle of size n*m. 
Note: 

  1. No two small rectangles overlap.
  2. Each small rectangle lies entirely inside the large rectangle. It is allowed to touch the edges of the large rectangle.

Examples

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

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

Approach:

  1. If N is even, then place M rows of N/2 small rectangles and cover the whole large rectangle.
  2. If M is even, then place N rows of M/2 small rectangles and cover the whole large rectangle.
  3. If both are odd then cover N – 1 row of the board with small rectangles and put floor(M/2) small rectangles to the last row. In the worst case (N and M are odd) one cell remains uncovered.

Below is the implementation of the above approach:

C++




// 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




// 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




// 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
}


Python3




# 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#




// 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
 
// 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
?>


Javascript




<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>


Output

4

Time Complexity: O(1)
Auxiliary Space: O(1)



Last Updated : 09 Jun, 2022
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