📜  异位三角形数

📅  最后修改于: 2021-05-06 18:18:23             🧑  作者: Mango

给定数字N ,任务是找到第N个等位三角形数。

例子:

方法:第N二十四方多边形数由下式给出:

Tn = (22n^2 - 20n)/2

下面是上述方法的实现:

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


Python3
# 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


Javascript


输出:
69
1000

时间复杂度: O(1)

辅助空间: O(1)

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