# 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:

• It’s allowed to cover the surface larger than the rectangle, but the rectangle has to be covered.
• It’s not allowed to break a square.
• The sides of squares should be parallel to the sides of the rectangle.

Examples:

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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

## C++

 `// CPP program to find number of squares ` `// of a*a required to cover n*m rectangle ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python 3

 `# 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 `

## C#

 `// 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 ` `} `

## PHP

 ` `

Output:

```4
``` My Personal Notes arrow_drop_up pawanasipugmailcom

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