Minimum number of square tiles required to fill the rectangular floor

14 May, 2021
Given a rectanglular 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: 
It’s allowed to cover the surface larger than the floor, but the floor has to be covered.It’s not allowed to break the tiles.The sides of tiles should be parallel to the sides of the floor.
It’s allowed to cover the surface larger than the floor, but the floor has to be covered.
It’s not allowed to break the tiles.
The sides of tiles should be parallel to the sides of the floor.
Examples:  
Input: 2 1 2 Output: 1 length of floor = 2 breadth of floor = 1 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 edges of each tile must be parallel to 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 product of these two quantities.  
ceil(M/s) * ceil(N/s)
, where ceil(x) is the least integer which 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 above approach: 
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; // Function to find the number of tilesint 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 Codeint 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 implementation// of above approachimport java.util.*;import java.lang.*;import java.io.*; class GFG{     // Function to find the// number of tilesstatic 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 Codepublic 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
# Python 3 implementation of# above approach # Function to find the number# of tilesdef 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 Codeif __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# implementation of above approachusing System; class GFG{     // Function to find the// number of tilesstatic 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 Codestatic 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// PHP implementation of above approach // Function to find the number of tilesfunction 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 / $s) + 1;    }     return (int)$M * $N;} // Driver Code // input length and breadth of// rectangle and side of square$N = 12;$M = 13;$s = 4; echo solve($M, $N, $s); // This code is contributed by mits?>
<script> // Javascript implementation// of above approach // Function to find the// number of tilesfunction 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 parseInt(M * N);} // Driver Code // Input length and breadth of// rectangle and side of squarevar N = 12, M = 13, s = 4; document.write(solve(M, N, s)); // This code is contributed by todaysgaurav </script>
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Using ceil function: 
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; // Function to find the number of tilesint solve(double M, double N, double s){    // no of tiles    int ans = ((int)(ceil(M / s)) * (int)(ceil(N / s)));     return ans;} // Driver Codeint 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 implementation of above approachclass GFG{// Function to find the number of tilesstatic 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 Codepublic 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
# Python 3 implementation of# above approachimport math # Function to find the# number of tilesdef solve(M, N, s):     # no of tiles    ans = ((math.ceil(M / s)) *           (math.ceil(N / s)));     return ans # Driver Codeif __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# implementation of above approachusing System;class GFG{// Function to find the number of tilesstatic 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 Codepublic 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// PHP implementation of// above approach // Function to find the// number of tilesfunction solve($M, $N, $s){    // no of tiles    $ans = ((int)(ceil($M / $s)) *            (int)(ceil($N / $s)));     return $ans;} // Driver Code // input length and breadth of// rectangle and side of square$N = 12;$M = 13;$s = 4; echo solve($M, $N, $s); // This code is contributed by mits?>
<script> // Javascript implementation of above approach     // Function to find the number of tilesfunction solve(M, N, s){         // No of tiles    let ans = Math.floor(((Math.ceil(M / s)) *                          (Math.ceil(N / s))));         return ans;} // Driver Code // Input length and breadth of// rectangle and side of squarelet N = 12, M = 13, s = 4; document.write(solve(M, N, s)); // This code is contributed by rag2127 </script>
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tufan_gupta2000
Mithun Kumar
ankthon
ukasp
todaysgaurav
rag2127
Modular Arithmetic
Geometric
Mathematical
Mathematical
Modular Arithmetic
Geometric
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