通过N个平面的相交形成的最大线相交数

给定N架飞机。任务是找到可以通过N个平面的相交形成的最大线相交数。
例子：

Input: N = 3 
Output: 3
Input: N = 5 
Output: 10
 

方法：
假设有N个平面，使得没有3个平面在一条相交线上相交，并且没有2个平面彼此平行。在保留上述两个条件的同时，应该可以在此空间中添加第N + 1个平面。在这种情况下，该平面将以“ N”条不同的线与N个平面的每一个相交。
因此，第“ N + 1”个平面最多可以将“ N”个新线添加到相交线的总数中。同样，第N个平面最多可以添加“ N-1”个相交线。因此，很容易看到，对于“ N”个平面，相交的最大线数可能是：

(N-1) + (N-2) +...+ 1 = N*(N-1)/2

下面是上述方法的实现：

C++

// C++ implementation of the above pproach
#include 
using namespace std;
 
// Function to count maximum number of
// intersections possible
int countIntersections(int n)
{
    return n * (n - 1) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
 
    cout << countIntersections(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function to count maximum number of
    // intersections possible
    static int countIntersections(int n)
    {
        return n * (n - 1) / 2;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int n = 3;
     
        System.out.println(countIntersections(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the above pproach
 
# Function to count maximum number of
# intersections possible
def countIntersections(n):
    return n * (n - 1) // 2
 
# Driver Code
n = 3
 
print(countIntersections(n))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to count maximum number of
    // intersections possible
    static int countIntersections(int n)
    {
        return n * (n - 1) / 2;
    }
     
    // Driver Code
    public static void Main (String[] args)
    {
        int n = 3;
     
        Console.WriteLine(countIntersections(n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the above approach
 
// Function to count maximum number of
// intersections possible
function countIntersections(n)
{
    return n * (n - 1) / 2;
}
 
// Driver Code
var n = 3;
document.write(countIntersections(n));
 

输出：

3

时间复杂度： O(1)