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# Minimum number of square tiles required to fill the rectangular floor

Given a rectangular floor of (M X N) meters is to be paved with square tiles of (s X s). The task is to find the minimum number of tiles required to pave the rectangular floor.

Constraints:

1. It’s allowed to cover the surface larger than the floor, but the floor has to be covered.
2. It’s not allowed to break the tiles.
3. The sides of tiles should be parallel to the sides of the floor.

Examples:

Input: 2 1 2
Output:
Explanation:
length of floor = 2
length of side of tile = 2
No of tiles required for paving is 2

Input: 222 332 5
Output: 3015

Approach:

It is given that the edges of each tile must be parallel to the edges of the tiles allows us to analyze X and Y axes separately, that is, how many segments of length ‘s’ are needed to cover a segment of length’ and ‘N’ — and take the product of these two quantities.

`ceil(M/s) * ceil(N/s)`

, where ceil(x) is the least integer that is above or equal to x. Using integers only, it is usually written as

`((M + s - 1) / s)*((N + s - 1) / s)`

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to find the number of tiles``int` `solve(``int` `M, ``int` `N, ``int` `s)``{``    ``// if breadth is divisible by side of square``    ``if` `(N % s == 0) {` `        ``// tiles required is N/s``        ``N = N / s;``    ``}``    ``else` `{` `        ``// one more tile required``        ``N = (N / s) + 1;``    ``}` `    ``// if length is divisible by side of square``    ``if` `(M % s == 0) {` `        ``// tiles required is M/s``        ``M = M / s;``    ``}``    ``else` `{``        ``// one more tile required``        ``M = (M / s) + 1;``    ``}` `    ``return` `M * N;``}` `// Driver Code``int` `main()``{``    ``// input length and breadth of``    ``// rectangle and side of square``    ``int` `N = 12, M = 13, s = 4;` `    ``cout << solve(M, N, s);` `    ``return` `0;``}`

## Java

 `// Java implementation``// of above approach``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `GFG``{``    ` `// Function to find the``// number of tiles``static` `int` `solve(``int` `M, ``int` `N, ``int` `s)``{``    ``// if breadth is divisible``    ``// by side of square``    ``if` `(N % s == ``0``)``    ``{` `        ``// tiles required is N/s``        ``N = N / s;``    ``}``    ``else``    ``{` `        ``// one more tile required``        ``N = (N / s) + ``1``;``    ``}` `    ``// if length is divisible``    ``// by side of square``    ``if` `(M % s == ``0``)``    ``{` `        ``// tiles required is M/s``        ``M = M / s;``    ``}``    ``else``    ``{``        ` `        ``// one more tile required``        ``M = (M / s) + ``1``;``    ``}` `    ``return` `M * N;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``// input length and breadth of``    ``// rectangle and side of square``    ``int` `N = ``12``, M = ``13``, s = ``4``;` `    ``System.out.println(solve(M, N, s));``}``}` `// This code is contributed``// by ChitraNayal`

## Python3

 `# Python 3 implementation of``# above approach` `# Function to find the number``# of tiles``def` `solve(M, N, s) :``    ` `    ``# if breadth is divisible``    ``# by side of square``    ``if` `(N ``%` `s ``=``=` `0``) :``        ` `        ``# tiles required is N/s``        ``N ``=` `N ``/``/` `s``        ` `    ``else` `:``        ` `        ``# one more tile required``        ``N ``=` `(N ``/``/` `s) ``+` `1` `    ``# if length is divisible by``    ``# side of square``    ``if` `(M ``%` `s ``=``=` `0``) :``        ` `        ``# tiles required is M/s``        ``M ``=` `M ``/``/` `s``        ` `    ``else` `:``        ` `        ``# one more tile required``        ``M ``=` `(M ``/``/` `s) ``+` `1``    ` `    ``return` `M ``*` `N` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``# input length and breadth of``    ``# rectangle and side of square``    ``N, M, s ``=` `12``, ``13``, ``4` `    ``print``(solve(M, N, s))``            ` `# This code is contributed by ANKITRAI1`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{``    ` `// Function to find the``// number of tiles``static` `int` `solve(``int` `M, ``int` `N, ``int` `s)``{``    ``// if breadth is divisible``    ``// by side of square``    ``if` `(N % s == 0)``    ``{` `        ``// tiles required is N/s``        ``N = N / s;``    ``}``    ``else``    ``{` `        ``// one more tile required``        ``N = (N / s) + 1;``    ``}` `    ``// if length is divisible``    ``// by side of square``    ``if` `(M % s == 0)``    ``{` `        ``// tiles required is M/s``        ``M = M / s;``    ``}``    ``else``    ``{``        ` `        ``// one more tile required``        ``M = (M / s) + 1;``    ``}` `    ``return` `M * N;``}` `// Driver Code``static` `void` `Main()``{``    ``// input length and breadth of``    ``// rectangle and side of square``    ``int` `N = 12, M = 13, s = 4;` `    ``Console.WriteLine(solve(M, N, s));``}``}` `// This code is contributed``// by mits`

## PHP

 ``

## Javascript

 ``

Output:

`12`

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

Using ceil function:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to find the number of tiles``int` `solve(``double` `M, ``double` `N, ``double` `s)``{``    ``// no of tiles``    ``int` `ans = ((``int``)(``ceil``(M / s)) * (``int``)(``ceil``(N / s)));` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``// input length and breadth of``    ``// rectangle and side of square``    ``double` `N = 12, M = 13, s = 4;` `    ``cout << solve(M, N, s);` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach``class` `GFG``{``// Function to find the number of tiles``static` `int` `solve(``double` `M,``                 ``double` `N, ``double` `s)``{``    ``// no of tiles``    ``int` `ans = ((``int``)(Math.ceil(M / s)) *``               ``(``int``)(Math.ceil(N / s)));` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// input length and breadth of``    ``// rectangle and side of square``    ``double` `N = ``12``, M = ``13``, s = ``4``;` `    ``System.out.println(solve(M, N, s));``}``}` `// This Code is contributed by mits`

## Python3

 `# Python 3 implementation of``# above approach``import` `math` `# Function to find the``# number of tiles``def` `solve(M, N, s):` `    ``# no of tiles``    ``ans ``=` `((math.ceil(M ``/` `s)) ``*``           ``(math.ceil(N ``/` `s)));` `    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``# input length and breadth of``    ``# rectangle and side of square``    ``N ``=` `12``    ``M ``=` `13``    ``s ``=` `4` `    ``print``(solve(M, N, s))` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# implementation of above approach``using` `System;``class` `GFG``{``// Function to find the number of tiles``static` `int` `solve(``double` `M,``                ``double` `N, ``double` `s)``{``    ``// no of tiles``    ``int` `ans = ((``int``)(Math.Ceiling(M / s)) *``            ``(``int``)(Math.Ceiling(N / s)));` `    ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``// input length and breadth of``    ``// rectangle and side of square``    ``double` `N = 12, M = 13, s = 4;` `    ``Console.WriteLine(solve(M, N, s));``}``}` `// This Code is contributed by mits`

## PHP

 ``

## Javascript

 ``

Output:

`12`

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

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