二十六边形数

给定一个数字N ，任务是找到^第N^个二十四边形数字。

An Icosihexagon number is class of figurate number. It has 26 – sided polygon called Icosihexagon. The N-th Icosihexagonal number count’s the 26 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Icosihexagonol numbers are 1, 26, 75, 148 … 
 

例子：

Input: N = 2 
Output: 26 
Explanation: 
The second Icosihexagonol number is 26. 
Input: N = 3 
Output: 75 
 

方法：第N个二十四边形数由以下公式给出：

• s 边多边形的第 N 项 = $\frac{((s-2)n^2 - (s-4)n)}{2}$
• 因此26边多边形的第N项是

$Tn =\frac{((26-2)n^2 - (26-4)n)}{2} =\frac{(24n^2 - 22n)}{2}$

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Finding the nth Icosihexagonal Number
int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << "3rd Icosihexagonal Number is = "
         << IcosihexagonalNum(n);
 
    return 0;
}
 
// This code is contributed by Code_Mech

C

// C program for above approach
#include 
#include 
 
// Finding the nth Icosihexagonal Number
int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}
 
// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd Icosihexagonal Number is = %d",
           IcosihexagonalNum(n));
 
    return 0;
}

Java

// Java program for above approach
class GFG{
     
// Finding the nth icosihexagonal number
public static int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
     
    System.out.println("3rd Icosihexagonal Number is = " +
                                    IcosihexagonalNum(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for above approach
 
# Finding the nth Icosihexagonal Number
def IcosihexagonalNum(n):
 
    return (24 * n * n - 22 * n) // 2
 
# Driver Code
n = 3
print("3rd Icosihexagonal Number is = ",
                   IcosihexagonalNum(n))
 
# This code is contributed by divyamohan123

C#

// C# program for above approach
using System;
 
class GFG{
     
// Finding the nth icosihexagonal number
public static int IcosihexagonalNum(int n)
{
    return (24 * n * n - 22 * n) / 2;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
     
    Console.WriteLine("3rd Icosihexagonal Number is = " +
                                   IcosihexagonalNum(n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// javascript program for above approach
 
 
// Finding the nth Icosihexagonal Number
function IcosihexagonalNum( n)
{
    return (24 * n * n - 22 * n) / 2;
}
 
// Driver code
let n = 3;
document.write("3rd Icosihexagonal Number is " + IcosihexagonalNum(n));
 
// This code contributed by gauravrajput1
 

输出：

3rd Icosihexagonal Number is = 75

时间复杂度： O(1)

辅助空间： O(1)

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

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