Icosikaioctagon 或 Icosioctagon 数

给定一个数字N ，任务是找到^第N^个Icosioctagon 数字。

An Icosioctagon number is class of figurate number. It has 28 – sided polygon called icosikaioctagon. The N-th icosikaioctagonal number count’s the 28 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few icosikaioctagonol numbers are 1, 28, 81, 160 … 
 

例子：

Input: N = 2 
Output: 28 
Explanation: 
The second icosikaioctagonol number is 28. 
Input: N = 3 
Output: 81 
 

方法：第 N 个 icosikaiooctagonal 数由以下公式给出：

• s 边多边形的第 N 项 = $\frac{((s-2)n^2 - (s-4)n)}{2}$
• 因此28边多边形的第N项是

$Tn =\frac{((28-2)n^2 - (28-4)n)}{2} =\frac{(26n^2 - 24n)}{2}$

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Finding the nth icosikaioctagonal number
int icosikaioctagonalNum(int n)
{
    return (26 * n * n - 24 * n) / 2;
}
 
// Driver code
int main()
{
    int n = 3;
 
    cout << "3rd icosikaioctagonal Number is = "
         << icosikaioctagonalNum(n);
 
    return 0;
}
 
// This code is contributed by shubhamsingh10

C

// C program for above approach
#include 
#include 
 
// Finding the nth icosikaioctagonal Number
int icosikaioctagonalNum(int n)
{
    return (26 * n * n - 24 * n) / 2;
}
 
// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd icosikaioctagonal Number is = %d",
           icosikaioctagonalNum(n));
 
    return 0;
}

Java

// Java program for above approach
class GFG{
     
// Finding the nth icosikaioctagonal number
public static int icosikaioctagonalNum(int n)
{
    return (26 * n * n - 24 * n) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
    System.out.println("3rd icosikaioctagonal Number is = " +
                                    icosikaioctagonalNum(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for above approach
 
# Finding the nth icosikaioctagonal Number
def icosikaioctagonalNum(n):
 
    return (26 * n * n - 24 * n) // 2
 
# Driver Code
n = 3
print("3rd icosikaioctagonal Number is = ",
                   icosikaioctagonalNum(n))
 
# This code is contributed by divyamohan123

C#

// C# program for above approach
using System;
class GFG{
     
// Finding the nth icosikaioctagonal number
public static int icosikaioctagonalNum(int n)
{
    return (26 * n * n - 24 * n) / 2;
}
 
// Driver code
public static void Main()
{
    int n = 3;
    Console.Write("3rd icosikaioctagonal Number is = " +
                               icosikaioctagonalNum(n));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
 
// JavaScript program for above approach
 
// Finding the nth icosikaioctagonal number
function icosikaioctagonalNum(n)
{
    return (26 * n * n - 24 * n) / 2;
}
 
// Driver code
var n = 3;
document.write("3rd icosikaioctagonal Number is = " + icosikaioctagonalNum(n));
 
 

输出：

3rd icosikaioctagonal Number is = 81

时间复杂度： O(1)

辅助空间： O(1)

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

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