给定（M X N）米的矩形地板，应铺设（s X s）的方形瓷砖。任务是找到铺设矩形地板所需的最小瓷砖数量。
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限制条件：
- 允许覆盖比地板大的表面，但必须覆盖地板。
- 不允许打碎瓷砖。
- 瓷砖的侧面应与地板的侧面平行。
例如：
输入： 2 1 2 输出： 1. 楼层长度=2 地板宽度=1 瓷砖侧面的长度=2 铺设所需的瓷砖数量为2。
输入： 222 332 5 输出： 3015
方法： 假设每个瓷砖的边缘必须平行于瓷砖的边缘，这样我们就可以分别分析X轴和Y轴，也就是说，需要多少段长度的“s”来覆盖一段长度的“s”和“N”，并取这两个量的乘积。
ceil(M/s) * ceil(N/s)
，其中ceil（x）是大于或等于x的最小整数。仅使用整数，它通常写为
((M + s - 1) / s)*((N + s - 1) / s)
以下是上述方法的实施情况：
C++
// C++ implementation of above approach #include <bits/stdc++.h> 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
<?php // PHP implementation of above approach // Function to find the number of tiles function 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 ?> |
Javascript
<script> // Javascript implementation // of above approach // Function to find the // number of tiles function 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 square var N = 12, M = 13, s = 4; document.write(solve(M, N, s)); // This code is contributed by todaysgaurav </script> |
输出：
12
使用 细胞功能 :
C++
// C++ implementation of above approach #include <bits/stdc++.h> 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
<?php // PHP implementation of // above approach // Function to find the // number of tiles function 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 ?> |
Javascript
<script> // Javascript implementation of above approach // Function to find the number of tiles function 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 square let N = 12, M = 13, s = 4; document.write(solve(M, N, s)); // This code is contributed by rag2127 </script> |
输出：
12
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