程序检查N是否为同位四边形数字

给定一个整数N ，任务是检查它是否是二十位三角形。

icositetragonal number: 
s a class of figurate number. It has a 24-sided polygon called Icositetragon. The N-th Icositetragonal number count’s the number of dots and all others dots are surrounding with a common sharing corner and make a pattern 
The first few icositetragonal numbers are 1, 24, 69, 136, 225, 336, …

例子：

Input: N = 24 
Output: Yes 
Explanation: 
Second icositetragonal number is 24.
Input: N = 30 
Output: No 
 

方法：

1. 二十位四边形数的第K^个项为
   $K^{th} Term = \frac{22*K^{2} - 20*K}{2}$
2. 因为我们必须检查给定的数字是否可以表示为二十位三角形。可以检查如下–

=> $N = \frac{22*K^{2} - 20*K}{2}$
=> $K = \frac{10 + \sqrt{44*N + 100}}{22}$
 

2. 最后，检查使用此公式计算的值是一个整数，这意味着N是一个二十四方晶数。

下面是上述方法的实现：

C++

// C++ implementation to check that
// a number is icositetragonal number or not
 
#include 
 
using namespace std;
 
// Function to check that the
// number is a icositetragonal number
bool isicositetragonal(int N)
{
    float n
        = (10 + sqrt(44 * N + 100))
          / 22;
 
    // Condition to check if the
    // number is a icositetragonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    int i = 24;
 
    // Function call
    if (isicositetragonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation to check that
// a number is icositetragonal number or not
class GFG{
 
// Function to check that the
// number is a icositetragonal number
static boolean isicositetragonal(int N)
{
    float n = (float)((10 + Math.sqrt(44 * N +
                                      100)) / 22);
 
    // Condition to check if the
    // number is a icositetragonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int i = 24;
 
    // Function call
    if (isicositetragonal(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 icositetragonal number
# or not
import math
 
# Function to check that the number
# is a icositetragonal number
def isicositetragonal(N):
 
    n = (10 + math.sqrt(44 * N + 100)) / 22
 
    # Condition to check if the number
    # is a icositetragonal number
    return (n - int(n)) == 0
 
# Driver Code
i = 24
 
# Function call
if (isicositetragonal(i)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by divyamohan123

C#

// C# implementation to check that
// a number is icositetragonal number or not
using System;
class GFG{
 
// Function to check that the
// number is a icositetragonal number
static bool isicositetragonal(int N)
{
    float n = (float)((10 + Math.Sqrt(44 * N +
                                      100)) / 22);
 
    // Condition to check if the
    // number is a icositetragonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
    int i = 24;
 
    // Function call
    if (isicositetragonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

 
// JavaScript implementation to check that
// a number is icositetragonal number or not
 
// Function to check that the
// number is a icositetragonal number
function isicositetragonal(N)
{
    var n = (10 + Math.sqrt(44 * N + 100))
          / 22;
 
    // Condition to check if the
    // number is a icositetragonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
var i = 24;
// Function call
if (isicositetragonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
     

输出：

Yes