三角形数

给定一个数N ，任务是找到^第N^个Triacontagon 数。

An Triacontagon number is class of figurate number. It has 30 – sided polygon called triacontagon. The N-th triacontagonal number count’s the 30 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few triacontagonol numbers are 1, 30, 87, 172 … 
 

例子：

Input: N = 2 
Output: 30 
Explanation: 
The second triacontagonol number is 30. 
Input: N = 3 
Output: 87 
 

方法：第N个三角数由下式给出：

• s 边多边形的第 N 项 = $\frac{((s-2)n^2 - (s-4)n)}{2}$
• 因此30边多边形的第N项是

$Tn =\frac{((30-2)n^2 - (30-4)n)}{2} =\frac{(28n^2 - 26n)}{2}$

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
// Finding the nth triacontagonal number
int triacontagonalNum(int n)
{
    return (28 * n * n - 26 * n) / 2;
}
 
// Driver code
int main()
{
    int n = 3;
     
    cout << "3rd triacontagonal Number is = "
         << triacontagonalNum(n);
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

// C program for above approach
#include 
#include 
 
// Finding the nth triacontagonal Number
int triacontagonalNum(int n)
{
    return (28 * n * n - 26 * n) / 2;
}
 
// Driver program to test above function
int main()
{
    int n = 3;
    printf("3rd triacontagonal Number is = %d",
           triacontagonalNum(n));
 
    return 0;
}

Java

// Java program for above approach
import java.io.*;
import java.util.*;
 
class GFG {
     
// Finding the nth triacontagonal number
static int triacontagonalNum(int n)
{
    return (28 * n * n - 26 * n) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
     
    System.out.println("3rd triacontagonal Number is = " +
                                    triacontagonalNum(n));
}
}
 
// This code is contributed by coder001

Python3

# Python3 program for above approach
 
# Finding the nth triacontagonal Number
def triacontagonalNum(n):
 
    return (28 * n * n - 26 * n) // 2
 
# Driver Code
n = 3
print("3rd triacontagonal Number is = ",
                   triacontagonalNum(n))
 
# This code is contributed by divyamohan123

C#

// C# program for above approach
using System;
 
class GFG{
     
// Finding the nth triacontagonal number
static int triacontagonalNum(int n)
{
    return (28 * n * n - 26 * n) / 2;
}
 
// Driver code
public static void Main()
{
    int n = 3;
     
    Console.Write("3rd triacontagonal Number is = " +
                               triacontagonalNum(n));
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

 
 
// JavaScript program for above approach
 
// Finding the nth triacontagonal number
function triacontagonalNum(n)
{
    return (28 * n * n - 26 * n) / 2;
}
 
// Driver code
var n = 3;
document.write("3rd triacontagonal Number is = " + triacontagonalNum(n));
 
 

输出：

3rd triacontagonal Number is = 87

时间复杂度： O(1)

辅助空间： O(1)

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

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