检查 N 是否为二十边形数的程序

给定一个整数N ，任务是检查它是否是二十四边形数。

Icosihenagonal number is class of figurate number. It has 21 – sided polygon called Icosihenagon. The n-th Icosihenagonal number counts the 21 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icosihenagonal numbers are 1, 21, 60, 118, 195, 291, 406…

例子：

Input: N = 21 
Output: Yes 
Explanation: 
Second icosihenagonal number is 21.
Input: N = 30 
Output: No 
 

方法：

1. ^第i个的icosihenagonal数项K被给定为
   $K^{th} Term = \frac{19*K^{2} - 17*K}{2}$
2. 因为我们必须检查给定的数字是否可以表示为 icosihenagonal 数字。这可以检查如下 –

=> $N = \frac{19*K^{2} - 17*K}{2}$
=> $K = \frac{17 + \sqrt{152*N + 289}}{38}$
 

2. 最后，检查使用这个公式计算的值是一个整数，这意味着 N 是一个 icosihenagonal 数。

下面是上述方法的实现：

C++

// C++ implementation to check that
// a number is icosihenagonal number or not
 
#include 
 
using namespace std;
 
// Function to check that the
// number is a icosihenagonal number
bool isicosihenagonal(int N)
{
    float n
        = (17 + sqrt(152 * N + 289))
          / 38;
 
    // Condition to check if the
    // number is a icosihenagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    int i = 21;
 
    // Function call
    if (isicosihenagonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation to check that a
// number is icosihenagonal number or not
class GFG{
 
// Function to check that the number
// is a icosihenagonal number
static boolean isicosihenagonal(int N)
{
    float n = (float) ((17 + Math.sqrt(152 * N +
                                       289)) / 38);
 
    // Condition to check if the number
    // is a icosihenagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int i = 21;
 
    // Function call
    if (isicosihenagonal(i))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to check that
# a number is icosihenagonal number or not
import math
 
# Function to check that the number
# is a icosihenagonal number
def isicosihenagonal(N):
 
    n = (17 + math.sqrt(152 * N + 289)) / 38
 
    # Condition to check if the number
    # is a icosihenagonal number
    return (n - int(n)) == 0
 
# Driver Code
i = 21
 
# Function call
if isicosihenagonal(i):
    print("Yes")
else :
    print("No")
     
# This code is contributed by divyamohan123

C#

// C# implementation to check that a
// number is icosihenagonal number or not
using System;
 
class GFG{
 
// Function to check that the number
// is a icosihenagonal number
static bool isicosihenagonal(int N)
{
    float n = (float)((17 + Math.Sqrt(152 * N +
                                      289)) / 38);
 
    // Condition to check if the number
    // is a icosihenagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
    int i = 21;
 
    // Function call
    if (isicosihenagonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// JavaScript implementation to check that
// a number is icosihenagonal number or not
 
// Function to check that the
// number is a icosihenagonal number
function isicosihenagonal(N)
{
    var n
        = (17 + Math.sqrt(152 * N + 289))
          / 38;
 
    // Condition to check if the
    // number is a icosihenagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
var i = 21;
// Function call
if (isicosihenagonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)