庚烷对角线数

给定一个数N ，任务是找到^第N^个庚庚二酮数。

A Heptacontadigon Number is a class of figurate numbers. It has a 72-sided polygon called Heptacontadigon. The N-th Heptacontadigon number count’s the 72 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Heptacontadigonol numbers are 1, 72, 213, 424, …

例子：

Input: N = 2
Output: 72
Explanation:
The second Heptacontadigonol number is 72.

Input: N = 3
Output: 213

方法：第N个庚庚二酮数由下式给出：

• S面多边形的第N个项= $\frac{((S - 2)N^2 - (S - 4)N)}{2}$
• 因此，第72个面的第N个项由下式给出：

  $Tn =\frac{((72 - 2)N^2 - (72 - 4)N)}{2} =\frac{(70N^2 - 68N)}{2}$

下面是上述方法的实现：

C/C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to find the N-th
// Heptacontadigon Number
int HeptacontadigonNum(int N)
{
    return (70 * N * N - 68 * N)
           / 2;
}
  
// Driver Code
int main()
{
    // Given number N
    int N = 3;
  
    // Function Call
    cout << HeptacontadigonNum(N);
  
    return 0;
}

Java

// Java program for the above approach
class GFG{ 
  
// Function to find the N-th
// Heptacontadigon Number
static int HeptacontadigonNum(int N)
{
    return (70 * N * N - 68 * N) / 2;
}
  
// Driver code 
public static void main(String[] args) 
{ 
    int N = 3;
      
    System.out.println(HeptacontadigonNum(N));
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
  
# Function to find the N-th
# Heptacontadigon Number
def HeptacontadigonNum(N):
  
    return (70 * N * N - 68 * N) // 2;
  
# Driver Code
  
# Given number N
N = 3;
  
# Function Call
print(HeptacontadigonNum(N));
  
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{ 
  
// Function to find the N-th
// Heptacontadigon Number
static int HeptacontadigonNum(int N)
{
    return (70 * N * N - 68 * N) / 2;
}
  
// Driver code 
public static void Main() 
{ 
    int N = 3;
      
    Console.Write(HeptacontadigonNum(N));
} 
} 
  
// This code is contributed by Code_Mech

输出：

213

时间复杂度： O(1)