使用正方形上的点形成三角形

给定一个正方形，正方形的每一边都有N个点，并且这些点都不与正方形的角重合。任务是计算使用这4 * N个点（正方形每边的 N 个点）作为三角形的顶点可以形成的三角形总数。
例子：

Input: N = 1 
Output: 4 
There is one point on each side. So we can make three rectangles by leaving one point and picking other three points out of the four. 
Input: N = 2 
Output: 56 
 

方法：在4*N个点中选择3个点的方法数为^(4*N) C ₃ 。然而，它们中的一些不形成三角形。当所有三个选定的点都在正方形的同一侧时，就会发生这种情况。这些三元组的数量对于每一边是^N C ₃ ，即总共4 * ^N C ₃ 。因此，所需的三角形数将为( ^{(4 * N)} C ₃ ) – (4 * ^N C ₃ ) 。
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 possible triangles
int noOfTriangles(int n)
{
    int y = 4 * n;
    return ((y * (y - 2) * (y - 1))
            - (4 * n * (n - 2) * (n - 1)))
           / 6;
}
 
// Driver code
int main()
{
    int n = 1;
 
    cout << noOfTriangles(n);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{
         
    // Function to return the count
    // of possible triangles
    static int noOfTriangles(int n)
    {
        int y = 4 * n;
        return ((y * (y - 2) * (y - 1)) -
                (4 * n * (n - 2) * (n - 1))) / 6;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 1;
     
        System.out.println(noOfTriangles(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
  
# Function to return the count
# of possible triangles
def noOfTriangles(n):
    y = 4 * n
    return ((y * (y - 2) * (y - 1)) -
            (4 * n * (n - 2) * (n - 1)))// 6
 
# Driver code
n = 1
 
print(noOfTriangles(n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the above approach
using System;
     
class GFG
{
         
    // Function to return the count
    // of possible triangles
    static int noOfTriangles(int n)
    {
        int y = 4 * n;
        return ((y * (y - 2) * (y - 1)) -
                (4 * n * (n - 2) * (n - 1))) / 6;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 1;
     
        Console.WriteLine(noOfTriangles(n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// javascript implementation of the approach
 
// Function to return the count
// of possible triangles
function noOfTriangles(n)
{
    var y = 4 * n;
    return ((y * (y - 2) * (y - 1))
            - (4 * n * (n - 2) * (n - 1)))
           / 6;
}
 
// Driver code
var n = 1;
document.write(noOfTriangles(n));
 

输出：

4

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。