十二进制数

给定一个数 n，任务是找到第 n 个 Icosidigonal 数 (Isdn)。
多边形有许多多边形，取决于它们的共形数列。在数学中，有许多gonal数，icosidigonal数就是其中之一，这些数有22边的多边形（icosidigon）。二十四角数属于比喻数。它们有一个公共点，其他点图案排列在第 n 个嵌套的 icosidigon 图案中。
例子 ：

Input : 2 
Output :22
Input :6 
Output :306 
 

第 n 个 Icosidigonal 数的公式：

$\begin{math} Isd_{n}=((20n^2)-18n)/2 \end{math}$

C++

// C++ program to find nth Icosidigonal
// number
#include 
using namespace std;
 
// Function to calculate Icosidigonal number
int icosidigonal_num(long int n)
{
    // Formula for finding
    // nth Icosidigonal number
    return (20 * n * n - 18 * n) / 2;
}
 
// Driver function
int main()
{
    long int n = 4;
    cout << n << "th Icosidigonal number :" << icosidigonal_num(n);
    cout << endl;
    n = 8;
    cout << n << "th Icosidigonal number:" << icosidigonal_num(n);
    return 0;
}

Java

// Java program to find nth
// Icosidigonal number
import java.io.*;
 
class GFG
{
     
    // Function to calculate
    // Icosidigonal number
    static int icosidigonal_num(int n)
    {
         
    // Formula for finding
    // nth Icosidigonal number
     
    return (20 * n * n - 18 * n) / 2;
    }
 
// Driver Code
public static void main (String[] args)
{
 
    int n = 4;
    System.out.print (n + "th Icosidigonal number :");
    System.out.println (icosidigonal_num(n));
 
    n = 8;
    System.out.print (n + "th Icosidigonal number :");
    System.out.println (icosidigonal_num(n));
}
}
 
// This code is contributed by ajit

Python 3

# python 3 program to find
# nth Icosidigonal number
 
# Function to calculate
# Icosidigonal number
def icosidigonal_num(n) :
     
    # Formula for finding
    # nth Icosidigonal number
    return (20 * n * n -
            18 * n) // 2
 
# Driver Code
if __name__ == '__main__' :
         
    n = 4
    print(n,"th Icosidigonal " +
                    "number : ",
            icosidigonal_num(n))
    n = 8
    print(n,"th Icosidigonal " +
                    "number : ",
            icosidigonal_num(n))
     
# This code is contributed m_kit

C#

// C# program to find nth
// Icosidigonal number
using System;
 
class GFG
{
     
    // Function to calculate
    // Icosidigonal number
    static int icosidigonal_num(int n)
    {
         
    // Formula for finding
    // nth Icosidigonal number
    return (20 * n * n -
            18 * n) / 2;
    }
 
// Driver Code
static public void Main ()
{
int n = 4;
Console.Write(n + "th Icosidigonal " +
                          "number :");
Console.WriteLine(icosidigonal_num(n));
 
n = 8;
Console.Write (n + "th Icosidigonal "+
                          "number :");
Console.WriteLine(icosidigonal_num(n));
}
}
 
// This code is contributed by ajit

Javascript

 
// JavaScript program to find nth Icosidigonal
// number
 
// Function to calculate Icosidigonal number
function icosidigonal_num(n)
{
    // Formula for finding
    // nth Icosidigonal number
    return parseInt((20 * n * n - 18 * n) / 2);
}
 
// Driver function
let n = 4;
document.write(n + "th Icosidigonal number :" + icosidigonal_num(n));
document.write("<br>");
n = 8;
document.write(n + "th Icosidigonal number :" + icosidigonal_num(n));
 

输出 ：

4th Icosidigonal number :124
8th Icosidigonal number:568

时间复杂度： O(1)
辅助空间： O(1)

参考：https://en.wikipedia.org/wiki/Polygonal_number

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