使用2×1尺寸的瓷砖覆盖给定尺寸的地板所需的最大瓷砖数量

给定尺寸为MxN的地板和尺寸为2×1的瓷砖，任务是找到覆盖MxN尺寸的地板所需的最大瓷砖数量。
注意：瓷砖可以水平或垂直放置，并且任何两个瓷砖都不应重叠。

例子：

Input: M = 2, N = 4
Output: 4
Explanation:
4 tiles are needed to cover the floor.

Input: M = 3, N = 3
Output: 4
Explanation:
4 tiles are needed to cover the floor.

方法：

1. 如果N为偶数，则任务是放置m行(N / 2)数量的图块以覆盖整个地板。
2. 否则，如果N为奇数，则以与上述要点相同的方式覆盖M行，直到N – 1 (偶数)列，并在最后一列中放入(M / 2)个图块。如果M和N均为奇数，则地板的一个单元仍然未被覆盖。
3. 因此，瓷砖的最大数量为floor((M * N)/ 2) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
void maximumTiles(int n, int m)
{
    // Print the answer
    cout << (m * n) / 2 << endl;
}
 
// Driver Code
int main()
{
    // Given M and N
    int M = 3;
    int N = 4;
 
    // Function Call
    maximumTiles(N, M);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
static void maximumTiles(int n, int m)
{
     
    // Print the answer
    System.out.println((m * n) / 2);
}
 
// Driver code
public static void main (String[] args)
{
 
    // Given M and N
    int M = 3;
    int N = 4;
     
    // Function call
    maximumTiles(N, M);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to find the maximum number
# of tiles required to cover the floor
# of size m x n using 2 x 1 size tiles
def maximumTiles(n, m):
 
    # Prthe answer
    print(int((m * n) / 2));
 
# Driver code
if __name__ == '__main__':
 
    # Given M and N
    M = 3;
    N = 4;
 
    # Function call
    maximumTiles(N, M);
 
# This code is contributed by sapnasingh4991

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the maximum number
// of tiles required to cover the floor
// of size m x n using 2 x 1 size tiles
static void maximumTiles(int n, int m)
{
     
    // Print the answer
    Console.WriteLine((m * n) / 2);
}
 
// Driver code
public static void Main(String[] args)
{
 
    // Given M and N
    int M = 3;
    int N = 4;
     
    // Function call
    maximumTiles(N, M);
}
}
 
// This code is contributed by amal kumar choubey

输出：

6

时间复杂度： O(1)
辅助空间： O(1)