程序检查N是否为二十进制八角形数

给定整数N ，任务是检查它是否为二十碳八角形数。

An icosikaioctagonal number is class of figurate number. It has 28 – sided polygon called icosikaioctagon. The N-th icosikaioctagonal number count’s the 28 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few icosikaioctagonol numbers are 1, 28, 81, 160 … 

例子：

Input: N = 28 
Output: Yes 
Explanation: 
Second icosikaioctagonal number is 28.

Input: 30 
Output: No 

方法：

1.二十烷八角形数的第K^个项为：

$K^{th} Term = \frac{26*K^{2} - 24*K}{2}$

2.由于我们必须检查给定的数字是否可以表示为二十烷八角形数字。可以检查如下–

=> $N = \frac{26*K^{2} - 24*K}{2}$
=> $K = \frac{24 + \sqrt{208*N + 576}}{52}$

3.最后，检查使用这些公式计算出的值是整数，这意味着N是一个二十烷八角形数。

下面是上述方法的实现：

C++

// C++ program to check whether a
// number is an icosikaioctagonal
// number or not
 
#include 
 
using namespace std;
 
// Function to check whether a number
// is an icosikaioctagonal number or not
bool isicosikaioctagonal(int N)
{
    float n
        = (24 + sqrt(208 * N + 576))
          / 52;
 
    // Condition to check if the
    // number is an
    // icosikaioctagonal number
    return (n - (int)n) == 0;
}
 
// Driver code
int main()
{
    int i = 28;
 
    if (isicosikaioctagonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program to check whether a
// number is an icosikaioctagonal
// number or not
class GFG{
 
// Function to check whether a
// number is an icosikaioctagonal
// number or not
static boolean isicosikaioctagonal(int N)
{
    float n = (float) ((24 + Math.sqrt(208 * N +
                                       576)) / 52);
     
    // Condition to check whether a
    // number is an icosikaioctagonal
    // number or not
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number
    int N = 28;
     
    // Function call
    if (isicosikaioctagonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by shubham

Python3

# Python3 program to check whether a
# number is an icosikaioctagonal
# number or not
import math
 
# Function to check whether a number
# is an icosikaioctagonal number or not
def isicosikaioctagonal(N):
 
    n = (24 + math.sqrt(208 * N +
                        576)) // 52;
 
    # Condition to check if the
    # number is an
    # icosikaioctagonal number
    return (n - int(n)) == 0;
 
# Driver code
i = 28;
 
if (isicosikaioctagonal(i)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by Code_Mech

C#

// C# program to check whether a
// number is an icosikaioctagonal
// number or not
using System;
class GFG{
 
// Function to check whether a
// number is an icosikaioctagonal
// number or not
static bool isicosikaioctagonal(int N)
{
    float n = (float)((24 + Math.Sqrt(208 * N +
                                      576)) / 52);
     
    // Condition to check whether a
    // number is an icosikaioctagonal
    // number or not
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
     
    // Given number
    int N = 28;
     
    // Function call
    if (isicosikaioctagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program to check whether a
// number is an icosikaioctagonal
// number or not
 
// Function to check whether a number
// is an icosikaioctagonal number or not
function isicosikaioctagonal(N)
{
    let n
        = (24 + Math.sqrt(208 * N + 576))
          / 52;
 
    // Condition to check if the
    // number is an
    // icosikaioctagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver code
let i = 28;
 
if (isicosikaioctagonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by rishavmahato348.

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)