最小大小为2的幂的图块可以覆盖整个区域

给定一个NXM区域。您有无数个大小为2 ⁱ X 2 ⁱ的图块，其中i = 0、1、2，…依此类推。任务是找到用瓷砖填充给定区域所需的最小瓷砖数量。

例子：

输入：N = 5，M =6。输出：9 5 X 6的区域至少可以覆盖9个图块。 1 X 1的6个图块2 X 2的2个图块，4 X 4的1个图块。 输入：N = 10，M =5。输出：14

想法是将给定区域划分为最接近的2 ⁱ X 2 ⁱ 。
让我们将问题分为几个案例：
情况1：如果N为奇数而M为偶数，则用M个1 X 1的图块填充行或列。然后计算N / 2 XM / 2面积大小的最小图块数。同样，如果M为奇数而N为偶数，则将N添加到我们的答案中，并找到N / 2 XM / 2区域的最小图块数。

情况2：如果N和M都为奇数，则填充一行和一列，因此将N + M – 1添加到答案中，并找到填充N / 2 XM / 2区域所需的最小图块数。

情况3：如果N和M均为偶数，请计算填充N / 2 XM / 2区域所需的最小块数。因为将两个尺寸减半不会改变所需的图块数量。

以下是此方法的实现：

C++

#include
using namespace std;
  
int minTiles(int n, int m)
{
  // base case, when area is 0.
  if (n == 0 || m == 0)
    return 0;
  
  // If n and m both are even, calculate tiles for n/2 x m/2
  // Halving both dimensions doesn't change the number of tiles
  else if (n%2 == 0 && m%2 == 0)
    return minTiles(n/2, m/2);
    
  // If n is even and m is odd
  // Use a row of 1x1 tiles
  else if (n%2 == 0 && m%2 == 1)
    return (n + minTiles(n/2, m/2));
  
  // If n is odd and m is even
  // Use a column of 1x1 tiles
  else if (n%2 == 1 && m%2 == 0)
    return (m + minTiles(n/2, m/2));
  
  // If n and m are odd
  // add row + column number of tiles
  else
    return (n + m - 1 + minTiles(n/2, m/2)); 
}
  
// Driven Program
int main()
{
  int n = 5, m = 6;
  
  cout << minTiles(n, m) << endl;
  return 0;
}

Java

// Java code for Minimum tiles of 
// sizes in powers of two to cover 
// whole area
  
class GFG {
      
    static int minTiles(int n, int m)
    {
    // base case, when area is 0.
    if (n == 0 || m == 0)
        return 0;
      
    // If n and m both are even, 
    // calculate tiles for n/2 x m/2
    // Halving both dimensions doesn't 
    // change the number of tiles
    else if (n % 2  == 0 && m % 2 == 0)
        return minTiles(n / 2, m / 2);
          
    // If n is even and m is odd
    // Use a row of 1x1 tiles
    else if (n % 2 == 0 && m % 2 == 1)
        return (n + minTiles(n / 2, m / 2));
      
    // If n is odd and m is even
    // Use a column of 1x1 tiles
    else if (n % 2 == 1 && m % 2 == 0)
        return (m + minTiles(n / 2, m / 2));
      
    // If n and m are odd
    // add row + column number of tiles
    else
        return (n + m - 1 + minTiles(n / 2, m / 2)); 
    }
          
    // Driver code
    public static void main (String[] args)
    {
            int n = 5, m = 6;
            System.out.println(minTiles(n, m));
    }
}
  
// This code is contributed by Anant Agarwal.

Python3

def minTiles(n, m):
      
    # base case, when area is 0.
    if n == 0 or m == 0:
        return 0
  
    # If n and m both are even, calculate
    # tiles for n/2 x m/2
    # Halfing both dimensions doesn't
    # change the number of tiles
    elif n%2 == 0 and m%2 == 0:
        return minTiles(int(n/2), int(m/2))
  
    # If n is even and m is odd
    # Use a row of 1x1 tiles
    elif n % 2 == 0 and m % 2 == 1:
        return (n + minTiles(int(n/2), int(m/2)))
  
    # If n is odd and m is even
    # Use a column of 1x1 tiles
    elif n % 2 == 1 and m % 2 == 0:
        return (m + minTiles(int(n/2), int(m/2)))
  
    # If n and m are odd add
    # row + column number of tiles
    else:
        return (n + m - 1 + minTiles(int(n/2), int(m/2))) 
  
# Driven Program
n = 5
m = 6
print (minTiles(n, m))
  
# This code is contributed
# by Shreyanshi Arun.

C#

// C# code for Minimum tiles of
// sizes in powers of two to cover
// whole area
using System;
  
class GFG {
  
    static int minTiles(int n, int m)
    {
          
        // base case, when area is 0.
        if (n == 0 || m == 0)
            return 0;
  
        // If n and m both are even,
        // calculate tiles for n/2 x m/2
        // Halving both dimensions doesn't
        // change the number of tiles
        else if (n % 2 == 0 && m % 2 == 0)
            return minTiles(n / 2, m / 2);
  
        // If n is even and m is odd
        // Use a row of 1x1 tiles
        else if (n % 2 == 0 && m % 2 == 1)
            return (n + minTiles(n / 2, m / 2));
  
        // If n is odd and m is even
        // Use a column of 1x1 tiles
        else if (n % 2 == 1 && m % 2 == 0)
            return (m + minTiles(n / 2, m / 2));
  
        // If n and m are odd
        // add row + column number of tiles
        else
            return (n + m - 1 + minTiles(n / 2, m / 2));
    }
  
    // Driver code
    public static void Main()
    {
        int n = 5, m = 6;
          
        Console.WriteLine(minTiles(n, m));
    }
}
  
// This code is contributed by vt_m.

输出：

9