计算给定大小的矩形内可能的菱形数

给定一个高度H和宽度W的矩形，该矩形的左下角为(0，0) 。任务是计算具有满足以下条件的矩形内部或边界上所有点的不重复Rhombi数量：

• 具有非零面积。
• 对角线平行于x和y轴。
• 具有整数坐标。

例子：

Input: H = 2, W = 2
Output: 2
There is only one rhombus possible with coordinates (0, 1), (1, 0), (2, 1) and (1, 2).

Input: H = 4, W = 4
Output: 16

方法：由于对角线与轴平行，让我们尝试固定对角线并在其上创建菱形。为了使菱形具有整数坐标，对角线的长度必须是偶数。让我们将对角线的长度固定为i和j ，在矩形内以这些对角线长度可以形成的菱形数为(H – i + 1)*(W – j + 1) 。因此，我们遍历i和j的所有可能值并更新计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of rhombi possible
long long countRhombi(int h, int w)
{
    long long ct = 0;
  
    // All possible diagonal lengths
    for (int i = 2; i <= h; i += 2)
        for (int j = 2; j <= w; j += 2)
  
            // Update rhombi possible with
            // the current diagonal lengths
            ct += (h - i + 1) * (w - j + 1);
  
    // Return the total count
    // of rhombi possible
    return ct;
}
  
// Driver code
int main()
{
    int h = 2, w = 2;
  
    cout << countRhombi(h, w);
  
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
// Function to return the count of rhombi possible
static int countRhombi(int h, int w)
{
    int ct = 0;
  
    // All possible diagonal lengths
    for (int i = 2; i <= h; i += 2)
        for (int j = 2; j <= w; j += 2)
  
            // Update rhombi possible with
            // the current diagonal lengths
            ct += (h - i + 1) * (w - j + 1);
  
    // Return the total count
    // of rhombi possible
    return ct;
}
  
    // Driver code
    public static void main (String[] args) 
    {
    int h = 2, w = 2;
    System.out.println (countRhombi(h, w));
    }
}
  
// This code is contributed by jit_t

Python 3

# Python 3 implementation of the approach
  
# Function to return the count of 
# rhombi possible
def countRhombi(h, w):
  
    ct = 0;
  
    # All possible diagonal lengths
    for i in range(2, h + 1, 2):
        for j in range(2, w + 1, 2):
  
            # Update rhombi possible with
            # the current diagonal lengths
            ct += (h - i + 1) * (w - j + 1)
  
    # Return the total count
    # of rhombi possible
    return ct
  
# Driver code
if __name__ == "__main__":
  
    h = 2
    w = 2
  
    print(countRhombi(h, w))
  
# This code is contributed by ita_c

C#

// C# program to find the frequency of 
// minimum element in the array 
using System;
  
class GFG 
{ 
          
    // Function to return the count
    // of rhombi possible 
    static int countRhombi(int h, int w) 
    { 
        int ct = 0; 
      
        // All possible diagonal lengths 
        for (int i = 2; i <= h; i += 2) 
            for (int j = 2; j <= w; j += 2) 
      
                // Update rhombi possible with 
                // the current diagonal lengths 
                ct += (h - i + 1) * (w - j + 1); 
      
        // Return the total count 
        // of rhombi possible 
        return ct; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int h = 2, w = 2; 
          
        Console.WriteLine(countRhombi(h, w)); 
    } 
} 
  
// This code is contributed by Ryuga

输出：

1

时间复杂度： O(H * W)