icosikaioctogon 或 Icosioctagon 编号

原文:https://www . geeksforgeeks . org/icosikaioctagon-or-icosioctagon-number/

给定一个编号 N ，任务是找到NthT5icosiocatgon 编号。

一个icosiocatgon 数是一类图形数。它有一个 28 边的多边形，叫做 icosikaioctagon。第 N 个 icosikaioctagonal 数字计数是 28 个点的数量，所有其他点都被一个公共共享角包围并形成一个图案。icosikaioctagonol 的前几个数字是 1、28、81、160……

例:

输入: N = 2 输出: 28 说明: 第二个 icosikaioctagonol 号为 28。 输入: N = 3 输出: 81

方法:第 N 个 icosikaioctagonal 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 28 边多边形的第 n 项为

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

以下是上述方法的实现:

C++

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

// 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 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 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

Python 3

# 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

java 描述语言

<script>

// 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));

</script>

Output: 

3rd icosikaioctagonal Number is = 81

时间复杂度: O(1)

辅助空间: O(1)

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

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