Number of squares of side length required to cover an N*M rectangle

Given three numbers , , . Find Number of squares of dimension required to cover rectangle.

Note:

Examples:

Input: N = 6, M = 6, a = 4
Output: 4

Input: N = 2, M = 3, a = 1
Output: 6

Approach: An efficient approach is to make an observation and find a formula. The constraint that edges of each square must be parallel to the edges of the rectangle that allows to analyze X and Y axes separately, that is, how many squares of length ‘a’ are needed to cover squares of length ‘m’ and ‘n’ and take the product of these two quantities. The number of small squares of side length ‘a’ required to cover ‘m’ sized square are ceil(m/a). Simillary, number of ‘a’ sized squares required to cover ‘n’ sized square are ceil(n/a).

So, the answer will be ceil(m/a)*ceil(n/a).



Below is the implementation of the above approach:

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// CPP program to find number of squares
// of a*a required to cover n*m rectangle
#include <bits/stdc++.h>
using namespace std;
  
// function to find number of squares
// of a*a required to cover n*m rectangle
int Squares(int n, int m, int a)
{
    return ((m + a - 1) / a) * ((n + a - 1) / a);
}
  
// Driver code
int main()
{
    int n = 6, m = 6, a = 4;
  
    // function call
    cout << Squares(n, m, a);
  
    return 0;
}
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// Java program to find number of squares
// of a*a required to cover n*m rectangle
import java.util.*;
  
class solution
{
  
// function to find a number of squares
// of a*a required to cover n*m rectangle
static int Squares(int n, int m, int a)
{
  
    return ((m + a - 1) / a) * ((n + a - 1) / a);
  
}
  
// Driver code
public static void main(String arr[])
{
    int n = 6, m = 6, a = 4;
  
    // function call
    System.out.println(Squares(n, m, a));
  
}
  
}
//This code is contributed by Surendra_Gangwar
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# Python 3 program to find number 
# of squares of a*a required to 
# cover n*m rectangle
  
# function to find number of 
# squares of a*a required to 
# cover n*m rectangle
def Squares(n, m, a):
    return (((m + a - 1) // a) * 
            ((n + a - 1) // a))
  
# Driver code
if __name__ == "__main__":
    n = 6
    m = 6
    a = 4
  
    # function call
    print(Squares(n, m, a))
  
# This code is contributed 
# by ChitraNayal
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// CSHARP program to find number of squares
// of a*a required to cover n*m rectangle
  
using System;
  
class GFG
{
    // function to find a number of squares
    // of a*a required to cover n*m rectangle
    static int Squares(int n, int m, int a)
    {
      
        return ((m + a - 1) / a) * ((n + a - 1) / a);
      
    }
  
    static void Main()
    {
          int n = 6, m = 6, a = 4;
  
         // function call
         Console.WriteLine(Squares(n, m, a));
    }
    // This code is contributed by ANKITRAI1
}
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<?php
// PHP program to find number of squares
// of a*a required to cover n*m rectangle
  
// function to find number of squares
// of a*a required to cover n*m rectangle
function Squares($n, $m, $a)
{
    return ((int)(($m + $a - 1) / $a)) *
           ((int)(($n + $a - 1) / $a));
}
  
// Driver code
$n = 6; $m = 6; $a = 4;
  
// function call
echo Squares($n, $m, $a);
  
// This code is contributed 
// by Akanksha Rai
?>
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Output:
4




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