给定周长可能的最大不同矩形计数

给定一个整数N表示矩形的周长。任务是找出给定周长可能存在的不同矩形的数量。

例子

Input: N = 10
Output: 4
Explanation: All the rectangles with perimeter 10 are following in the form of (length, breadth):
(1, 4), (4, 1), (2, 3), (3, 2)

Input: N = 8
Output: 3

方法：这个问题可以通过使用矩形的属性来解决。请按照以下步骤解决给定的问题。

• 矩形的周长是2*(length + width) 。
• 如果N是奇数，则不可能有矩形。因为周长永远不会是奇数。
• 如果N小于4 ，那么也不可能存在任何矩形。由于边的最小可能长度为1 ，因此即使所有边的长度为1 ，那么周长也是4 。
• 现在N = 2*(l + b)和(l + b) = N/2 。
• 因此，需要找到总和为N/2的所有对，即(N/2) – 1 。

下面是上述方法的实现。

C++

#include 
using namespace std;
 
// Function to find the maximum number
// of distinct rectangles with given perimeter
void maxRectanglesPossible(int N)
{
    // Invalid case
    if (N < 4 || N % 2 != 0) {
        cout << -1 << "\n";
    }
    else
        // Number of distinct rectangles.
        cout << (N / 2) - 1 << "\n";
}
 
// Driver Code
int main()
{
 
    // Perimeter of the rectangle.
    int N = 20;
 
    maxRectanglesPossible(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
// Function to find the maximum number
// of distinct rectangles with given perimeter
static void maxRectanglesPossible(int N)
{
   
    // Invalid case
    if (N < 4 || N % 2 != 0) {
        System.out.println(-1);
    }
    else
        // Number of distinct rectangles.
       System.out.println((N / 2) - 1);
}
 
// Driver Code
    public static void main (String[] args) {
          // Perimeter of the rectangle.
        int N = 20;
 
        maxRectanglesPossible(N);
    }
}
 
// This code is contributed by hrithikgarg0388.

Python3

# Function to find the maximum number
# of distinct rectangles with given perimeter
def maxRectanglesPossible (N):
 
    # Invalid case
    if (N < 4 or N % 2 != 0):
        print("-1");
    else:
        # Number of distinct rectangles.
        print(int((N / 2) - 1));
 
 
# Driver Code
 
# Perimeter of the rectangle.
N = 20;
maxRectanglesPossible(N);
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG {
 
// Function to find the maximum number
// of distinct rectangles with given perimeter
static void maxRectanglesPossible(int N)
{
   
    // Invalid case
    if (N < 4 || N % 2 != 0) {
        Console.WriteLine(-1);
    }
    else
        // Number of distinct rectangles.
       Console.WriteLine((N / 2) - 1);
}
 
// Driver Code
    public static void Main () {
          // Perimeter of the rectangle.
        int N = 20;
 
        maxRectanglesPossible(N);
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
    // Function to find the maximum number
    // of distinct rectangles with given perimeter
    const maxRectanglesPossible = (N) => {
     
        // Invalid case
        if (N < 4 || N % 2 != 0) {
            document.write("-1<br/>");
        }
        else
            // Number of distinct rectangles.
            document.write(`${(N / 2) - 1}<br/>`);
    }
 
    // Driver Code
 
    // Perimeter of the rectangle.
    let N = 20;
    maxRectanglesPossible(N);
 
// This code is contributed by rakeshsahni
 

输出

9

时间复杂度： O(1)
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