二十进制对角线数

二十碳六烯对角数是一类数字。它有29面的多边形，称为icosikaienneagon 。第N个icosikaienneagonal数是29个点，所有其他点都围绕着一个公共的共享角并形成图案。
前几个icosikaienneagonol编号为

1, 29, 84, 166 … 
 

找出第N^个二十烷碳对角线数

给定数N ，任务是找到^第N^个icosikaienneagonal数。
例子：

Input: N = 2 
Output: 29 
Explanation: 
The second icosikaienneagonol number is 29. 
Input: N = 3 
Output: 84 
 

方法：

• 在数学中，第N个s面多边形数由以下公式给出：
  $\text{Nth term of s sided polygon} = \frac{((s-2)n^2 - (s-4)n)}{2}$
• 因此，第29面多边形的第N个项是

$Tn =\frac{((29-2)n^2 - (29-4)n)}{2} =\frac{(27n^2 - 25n)}{2}$

下面是上述方法的实现：

C++

// C++ implementation for
// above approach
 
#include 
using namespace std;
 
// Function to Find the Nth
// icosikaienneagonal Number
int icosikaienneagonalNum(int n)
{
    return (27 * n * n - 25 * n) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << icosikaienneagonalNum(n);
 
    return 0;
}

Java

// Java implementation for
// above approach
class GFG{
 
// Function to Find the Nth
// icosikaienneagonal Number
static int icosikaienneagonalNum(int n)
{
    return (27 * n * n - 25 * n) / 2;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 3;
    System.out.print(icosikaienneagonalNum(n));
}
}
 
// This code is contributed by Code_Mech

Python 3

# Python3 implementation for
# above approach
 
# Function to Find the Nth
# icosikaienneagonal Number
def icosikaienneagonalNum(n):
    return (27 * n * n - 25 * n) // 2
 
# Driver Code
 
# Given N
N = 3
print(icosikaienneagonalNum(N))
 
# This code is contributed by Vishal Maurya

C#

// C# implementation for
// above approach
using System;
class GFG{
 
// Function to Find the Nth
// icosikaienneagonal Number
static int icosikaienneagonalNum(int n)
{
    return (27 * n * n - 25 * n) / 2;
}
 
// Driver Code
public static void Main()
{
    int n = 3;
    Console.Write(icosikaienneagonalNum(n));
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript implementation for
// above approach
 
 
    // Function to Find the Nth
    // icosikaienneagonal Number
    function icosikaienneagonalNum( n) {
        return (27 * n * n - 25 * n) / 2;
    }
 
    // Driver Code
      
    let n = 3;
    document.write(icosikaienneagonalNum(n));
 
// This code is contributed by Rajput-Ji

输出：

84