悲喜剧号

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

给定一个数字 N ，任务是找到 N 第T5【悲喜剧号。

一个二部曲数是一类图形数。它有一个 24 边的多边形，叫做 Icositetragon。第 N 个 Icositetragonal number count 是点的数量，所有其他点都被一个共同的共享角包围并形成一个图案。

示例:

输入:N = 2 T3】输出: 24

输入:N = 6 T3】输出: 336

方法:第 N 个悲喜剧数由公式给出:

下面是上述方法的实现:

C++

// C++ program to find nth
// Icositetragonal number

#include <bits/stdc++.h>
using namespace std;

// Function to find
// Icositetragonal number
int Icositetragonal_num(int n)
{
    // Formula to calculate nth
    // Icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}

// Driver Code
int main()
{
    int n = 3;

    cout << Icositetragonal_num(n) << endl;

    n = 10;

    cout << Icositetragonal_num(n);

    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java program to find nth
// icositetragonal number
import java.util.*;

class GFG {

// Function to find
// icositetragonal number
static int Icositetragonal_num(int n)
{

    // Formula to calculate nth
    // icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}

// Driver code
public static void main(String[] args)
{
    int n = 3;
    System.out.println(Icositetragonal_num(n));

    n = 10;
    System.out.println(Icositetragonal_num(n));
}
}

// This code is contributed by offbeat

Python 3

# Python3 program to find nth
# Icositetragonal number

# Function to find
# Icositetragonal number
def Icositetragonal_num(n):

    # Formula to calculate nth
    # Icositetragonal number
    return (22 * n * n - 20 * n) / 2

# Driver Code
n = 3
print(int(Icositetragonal_num(n)))

n = 10
print(int(Icositetragonal_num(n)))

# This code is contributed by divyeshrabadiya07

C

// C# program to find nth
// icositetragonal number
using System;

class GFG{

// Function to find
// icositetragonal number
static int Icositetragonal_num(int n)
{

    // Formula to calculate nth
    // icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}

// Driver code
public static void Main(string[] args)
{
    int n = 3;
    Console.Write(Icositetragonal_num(n) + "\n");

    n = 10;
    Console.Write(Icositetragonal_num(n) + "\n");
}
}

// This code is contributed by rutvik_56

java 描述语言

<script>

// Javascript program to find nth
// icositetragonal number

// Function to find
// icositetragonal number
function Icositetragonal_num(n)
{

    // Formula to calculate nth
    // icositetragonal number
    return (22 * n * n - 20 * n) / 2;
}

// Driver code
let n = 3;
document.write(Icositetragonal_num(n) + "</br>");

n = 10;
document.write(Icositetragonal_num(n));

// This code is contributed by Ankita saini

</script>

Output: 

69
1000

时间复杂度: O(1)

辅助空间: O(1)

参考:T2】https://en.wikipedia.org/wiki/Polygonal_number

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