四边形数

给定一个数字N ，任务是找到^第N^个Tetracontagon 数字。

A Tetracontagon number is class of figurate number. It has 40 – sided polygon called tetracontagon. The N-th tetracontagon number count’s the 40 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few tetracontagonol numbers are 1, 40, 117, 232 … 
 

例子：

Input: N = 2 
Output: 40 
Explanation: 
The second tetracontagonol number is 40. 
Input: N = 3 
Output: 117 
 

方法：第N个四边形数由公式给出：

• s 边多边形的第 N 项 = $\frac{((s-2)n^2 - (s-4)n)}{2}$
• 因此40边多边形的第N项是

$Tn =\frac{((40-2)n^2 - (40-4)n)}{2} =\frac{(38n^2 - 36)}{2}$

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Finding the nth tetracontagon Number
int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}
 
// Driver Code
int main()
{
    int n = 3;
    cout << "3rd tetracontagon Number is = "
         << tetracontagonNum(n);
 
    return 0;
}
 
// This code is contributed by Akanksha_Rai

C

// C program for above approach
 
#include 
#include 
 
// Finding the nth tetracontagon Number
int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}
 
// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd tetracontagon Number is = %d",
           tetracontagonNum(n));
 
    return 0;
}

Java

// Java program for above approach
import java.util.*;
 
class GFG {
 
// Finding the nth tetracontagon number
static int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
     
    System.out.println("3rd tetracontagon Number is = " +
                                    tetracontagonNum(n));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for above approach
 
# Finding the nth tetracontagon Number
def tetracontagonNum(n):
 
    return (38 * n * n - 36 * n) // 2
 
# Driver Code
n = 3
print("3rd tetracontagon Number is = ",
                   tetracontagonNum(n))
 
# This code is contributed by divyamohan123

C#

// C# program for above approach
using System;
 
class GFG {
 
// Finding the nth tetracontagon number
static int tetracontagonNum(int n)
{
    return (38 * n * n - 36 * n) / 2;
}
 
// Driver code
public static void Main(string[] args)
{
    int n = 3;
     
    Console.Write("3rd tetracontagon Number is = " +
                               tetracontagonNum(n));
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// javascript program for above approach
 
// Finding the nth tetracontagon Number
function tetracontagonNum( n)
{
    return (38 * n * n - 36 * n) / 2;
}
 
// Driver code
let n = 3;
document.write("3rd tetracontagon Number is " + tetracontagonNum(n));
 
// This code contributed by gauravrajput1
 

输出：

3rd tetracontagon Number is = 117

时间复杂度： O(1)

辅助空间： O(1)

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

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。