平面上 N 点之间唯一直接路径的计数

给定平面上的N个点，其中每个点都有一条直接路径将其连接到不同的点，任务是计算点之间唯一直接路径的总数。

注意： N的值将始终大于2 。

例子：

Input: N = 4
Output: 6
Explanation: Think of 4 points as a 4 sided polygon. There will 4 direct paths (sides of the polygon) as well as 2 diagonals (diagonals of the polygon). Hence the answer will be 6 direct paths.

Input: N = 3
Output:  3
Explanation: Think of 3 points as a 3 sided polygon. There will 3 direct paths (sides of the polygon) as well as 0 diagonals (diagonals of the polygon). Hence the answer will be 3 direct paths.

方法：给定的问题可以通过观察来解决，即对于任何N边都有（边数 + 对角线数）直接路径。对于任何N边多边形，都有N条边和N*(N – 3)/2条对角线。因此，直接路径的总数由N + (N * (N – 3))/2给出。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the total number
// of direct paths
int countDirectPath(int N)
{
    return N + (N * (N - 3)) / 2;
}
 
// Driver Code
int main()
{
 
    int N = 5;
    cout << countDirectPath(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
   
// Function to count the total number
// of direct paths
static int countDirectPath(int N)
{
    return N + (N * (N - 3)) / 2;
}
 
// Driver Code
public static void main(String []args)
{
 
    int N = 5;
    System.out.print(countDirectPath(N));
 
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# python program for the above approach
 
# Function to count the total number
# of direct paths
def countDirectPath(N):
 
    return N + (N * (N - 3)) // 2
 
# Driver Code
if __name__ == "__main__":
 
    N = 5
    print(countDirectPath(N))
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
 
public class GFG
{
   
    // Function to count the total number
    // of direct paths
    static int countDirectPath(int N)
    {
        return N + (N * (N - 3)) / 2;
    }
     
    // Driver Code
    public static void Main(string []args)
    {
     
        int N = 5;
        Console.Write(countDirectPath(N));
     
    }
}
 
// This code is contributed by AnkThon

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to count the total number
        // of direct paths
        function countDirectPath(N) {
            return N + Math.floor((N * (N - 3)) / 2);
        }
 
        // Driver Code
        let N = 5;
        document.write(countDirectPath(N));
 
// This code is contributed by Potta Lokesh
    

输出：

10

时间复杂度： O(1)
辅助空间： O(1)