程序检查N是否为二十六面体数字

给定整数N ，任务是检查它是否为二十六边形数字。

Icosihexagonal number is class of figurate number. It has 26 – sided polygon called Icosihexagon. The N-th Icosihexagonal number count’s the 26 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Icosihexagonol numbers are 1, 26, 75, 148 … 
 

例子：

Input: N = 26 
Output: Yes 
Explanation: 
Second icosihexagonal number is 26.
Input: 30 
Output: No 
 

方法：

1. 二十进制六边形数的第K^个项为
   $K^{th} Term = \frac{24*K^{2} - 22*K}{2}$
2. 因为我们必须检查给定的数字是否可以表示为二十面体六边形。可以检查如下–

=> $N = \frac{24*K^{2} - 22*K}{2}$
=> $K = \frac{22 + \sqrt{192*N + 484}}{48}$
 

2. 最后，检查使用此公式计算出的值是否为整数，这意味着N为二十六边形数字。

下面是上述方法的实现：

C++

// C++ program to check whether
// a number is a icosihexagonal number
// or not
 
#include 
 
using namespace std;
 
// Function to check whether the
// number is a icosihexagonal number
bool isicosihexagonal(int N)
{
    float n
        = (22 + sqrt(192 * N + 484))
          / 48;
 
    // Condition to check if the
    // number is a icosihexagonal number
    return (n - (int)n) == 0;
}
 
// Driver code
int main()
{
    int i = 26;
 
    if (isicosihexagonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program to check whether the
// number is a icosihexagonal number
// or not
class GFG{
 
// Function to check whether the
// number is a icosihexagonal number
static boolean isicosihexagonal(int N)
{
    float n = (float) ((22 + Math.sqrt(192 * N +
                                       484)) / 48);
     
    // Condition to check if the number
    // is a icosihexagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number
    int N = 26;
     
    // Function call
    if (isicosihexagonal(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 a icosihexagonal number
# or not
import numpy as np
 
# Function to check whether the
# number is a icosihexagonal number
def isicosihexagonal(N):
 
    n = (22 + np.sqrt(192 * N + 484)) / 48
 
    # Condition to check if the
    # number is a icosihexagonal number
    return (n - (int(n))) == 0
 
# Driver code
i = 26
 
if (isicosihexagonal(i)):
    print ("Yes")
else:
    print ("No")
 
# This code is contributed by PratikBasu

C#

// C# program to check whether the
// number is a icosihexagonal number
// or not
using System;
class GFG{
 
// Function to check whether the
// number is a icosihexagonal number
static bool isicosihexagonal(int N)
{
    float n = (float)((22 + Math.Sqrt(192 * N +
                                      484)) / 48);
     
    // Condition to check if the number
    // is a icosihexagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
     
    // Given number
    int N = 26;
     
    // Function call
    if (isicosihexagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program to check whether
// a number is a icosihexagonal number
// or not
 
// Function to check whether the
// number is a icosihexagonal number
function isicosihexagonal(N)
{
    let n
        = (22 + Math.sqrt(192 * N + 484))
          / 48;
 
    // Condition to check if the
    // number is a icosihexagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver code
let i = 26;
 
if (isicosihexagonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by rishavmahato348.

输出：

Yes