程序检查N是否是居中的十二边形数

给定整数N ，任务是检查N是否为中心十二边形数。如果数字N是居中的十二边形数字，则打印“是”，否则打印“否” 。

Centered Dodecagonal Number represents a dot in the center and other dots surrounding it in successive Dodecagonal Number(12 sided polygon) layers. The first few Centered Dodecagonal Numbers are 1, 13, 37, 73 … 

例子：

Input: N = 13 
Output: Yes 
Explanation: 
Second Centered dodecagonal number is 13.

Input: N = 30 
Output: No 

方法：

1中心十二边形数的第K^个项为：
$K^{th} Term = 6*K^{2} - 6*K + 1$

2.由于我们必须检查给定的数字是否可以表示为中心十二边形数字。可以检查为：

=> $N = {6*K^{2} - 6*K + 1}$
=> $K = \frac{6 + \sqrt{24*N + 12}}{12}$ 

3.如果使用上述公式计算出的K的值为整数，则N为中心十二边形数。

4.否则，数字N不是居中的十二边形数字。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if number N
// is a Centered dodecagonal number
bool isCentereddodecagonal(int N)
{
    float n
        = (6 + sqrt(24 * N + 12))
          / 12;
 
    // Condition to check if N
    // is a Centered Dodecagonal Number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 13;
 
    // Function call
    if (isCentereddodecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if number N
// is a centered dodecagonal number
static boolean isCentereddodecagonal(int N)
{
    float n = (float) ((6 + Math.sqrt(24 * N +
                                      12)) / 12);
 
    // Condition to check if N is a
    // centered dodecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number
    int N = 13;
 
    // Function call
    if (isCentereddodecagonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
import numpy as np
 
# Function to check if the number N
# is a centered dodecagonal number
def isCentereddodecagonal(N):
 
    n = (6 + np.sqrt(24 * N + 12)) / 12
 
    # Condition to check if N
    # is a centered dodecagonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 13
 
# Function call
if (isCentereddodecagonal(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by PratikBasu

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if number N
// is a centered dodecagonal number
static bool isCentereddodecagonal(int N)
{
    float n = (float) ((6 + Math.Sqrt(24 * N +
                                      12)) / 12);
 
    // Condition to check if N is a
    // centered dodecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Number
    int N = 13;
 
    // Function call
    if (isCentereddodecagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by rutvik_56

Javascript

// Javascript program for the above approach
 
// Function to check if number N
// is a Centered dodecagonal number
function isCentereddodecagonal(N)
{
    let n
        = (6 + Math.sqrt(24 * N + 12))
          / 12;
 
    // Condition to check if N
    // is a Centered Dodecagonal Number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
// Given Number
let N = 13;
 
// Function call
if (isCentereddodecagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by subham348.

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)