二十边形数

原文:https://www.geeksforgeeks.org/icosihexagonal-number/

给定一个编号 N ，任务是找到NthT5Icosihexagon 编号。

一个 Icosihexagon 数是一类图形数。它有一个 26 边的多边形，叫做 Icosihexagon。第 N 个 Icosihexagonal 数字计数是 26 个点的数量，所有其他点被一个公共共享角包围并形成一个图案。前几个 Icosihexagonol 数字是 1、26、75、148……

例:

输入: N = 2 输出: 26 解释: 第二个 Icosihexagonol 数是 26。 输入: N = 3 输出: 75

方法:第 N 个正交数由公式给出:

  • 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 项为

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

以下是上述方法的实现:

C++

// C++ program for above approach
#include <bits/stdc++.h>
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 <stdio.h>
#include <stdlib.h>

// 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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

java 描述语言

<script>

// 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

</script>

Output: 

3rd Icosihexagonal Number is = 75

时间复杂度: O(1)

辅助空间: O(1)

参考资料:https://en . Wikipedia . org/wiki/icsihasexabgon

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