计算可以放置在满足给定条件的M * N矩形板上的尺寸为2 * 1的瓷砖

给定两个整数M和N ，任务是找到可放置在M * N网格上的最小2 * 1大小的图块，从而满足以下条件：

1. 每个瓦片必须完全覆盖木板的2个正方形。
2. 一对瓷砖都不能重叠。
3. 每个瓷砖躺椅必须完全放置在木板内部。可以触摸电路板的边缘。

如果无法覆盖整个电路板，请打印-1

Input: N = 2, M = 4
Output: 4
Explanation: 4 tiles of dimension 2 * 1. Place each tile in one column.

Input: N = 3, M = 3
Output: -1

方法：按照以下步骤解决问题

1. 如果N是偶数，则可以放置(N / 2)* M瓦片以覆盖整个电路板。
2. 如果N为奇数，则为2 * 1个图块，因为长度为奇数，不能表示为2的倍数

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to count tiles of dimensions
// 2 x 1 that can be placed in a grid of
// dimensions M * N as per given conditions
int numberOfTiles(int N, int M)
{
    if (N % 2 == 1) {
        return -1;
    }
 
    // Number of tiles required
    return (N * 1LL * M) / 2;
}
 
// Driver Code
int main()
{
    int N = 2, M = 4;
    cout << numberOfTiles(N, M);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to count tiles of dimensions
    // 2 x 1 that can be placed in a grid of
    // dimensions M * N as per given conditions
    static int numberOfTiles(int n, int m)
    {
        if (n % 2 == 1) {
            return -1;
        }
        // Number of tiles required
        return (m * n) / 2;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 2, m = 4;
        System.out.println(
            numberOfTiles(n, m));
    }
}

Python

# Python Program to implement
# the above approach
 
# Function to count tiles of dimensions
# 2 x 1 that can be placed in a grid of
# dimensions M * N as per given conditions
def numberOfTiles(N, M):
    if (N % 2 == 1):
        return -1
     
    # Number of tiles required
    return (N * M) // 2
 
# Driver Code
N = 2
M = 4
print(numberOfTiles(N, M))
 
# This code is contributed by shubhamsingh10

C#

// C# Program to implement
// the above approach
 
using System;
 
public class GFG {
 
    // Function to tiles of size 2 x 1
    // find the number of tiles that can
    // be placed as per the given conditions
    static int numberOfTiles(int n, int m)
    {
        if (n % 2 == 1) {
            return -1;
        }
        // Number of tiles required
        return (m * n) / 2;
    }
 
    // Driver Code
    static public void Main()
    {
        int n = 2, m = 4;
        Console.WriteLine(
            numberOfTiles(n, m));
    }
}

输出：

4

时间复杂度： O(1)
辅助空间： O(1)