程序检查N是否为正二十进制数

给定整数N ，任务是检查它是否为正二十进制数。如果数字N是正二十进制数字，则打印“是”，否则打印“否”。

Icosidigonal number: 
The polygon has many gons, depends on their gonal number series. In mathematics, there are a number of gonal numbers and the Icosidigonal Number is one of them and these numbers have 22 -sided polygon(icosidigon). An Icosidigonal Number belong to the class of figurative number. They have one common dots points and other dots pattern is arranged in an n-th nested Icosidigon pattern. 
The first few Icosidigonal numbers are 1, 22, 63, 124, 205, 306…

例子：

Input: N = 22 
Output: Yes 
Explanation: 
Second Icosidigonal number is 22.
Input: 30 
Output: No  

方法：

1.在K的Icosidigonal数^个术语被给定为
$K^{th} Term = \frac{20*K^{2} - 18*K}{2}$

2.由于我们必须检查给定的数字是否可以表示为正二十进制数。可以按以下方式检查–

=> $N = \frac{20*K^{2} - 18*K}{2}$
=> $K = \frac{18 + \sqrt{160*N + 324}}{40}$

3.如果使用上式计算的K值为整数，则N为正二十进制数。

4.其他N不是二十进制的数字。

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 the number N
// is a Icosidigonal number
bool isIcosidigonal(int N)
{
    float n
        = (18 + sqrt(160 * N + 324))
          / 40;
 
    // Condition to check if the
    // number is a Icosidigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 22;
 
    // Function call
    if (isIcosidigonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if the number N
// is a icosidigonal number
static boolean isIcosidigonal(int N)
{
    float n = (float) ((18 + Math.sqrt(160 * N +
                                       324)) / 40);
 
    // Condition to check if the number
    // is a icosidigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number
    int N = 22;
 
    // Function call
    if (isIcosidigonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
import numpy as np
 
# Function to check if the number N
# is a icosidigonal number
def isIcosidigonal(N):
 
    n = (18 + np.sqrt(160 * N + 324)) / 40
 
    # Condition to check if N
    # is a icosidigonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 22
 
# Function call
if (isIcosidigonal(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 the number N
// is a icosidigonal number
static bool isIcosidigonal(int N)
{
    float n = (float) ((18 + Math.Sqrt(160 * N +
                                    324)) / 40);
 
    // Condition to check if the number
    // is a icosidigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given number
    int N = 22;
 
    // Function call
    if (isIcosidigonal(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 the number N
// is a Icosidigonal number
function isIcosidigonal( N)
{
    let n
        = (18 + Math.sqrt(160 * N + 324))
          / 40;
 
    // Condition to check if the
    // number is a Icosidigonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
 
    // Given Number
    let N = 22;
 
    // Function call
    if (isIcosidigonal(N)) {
         document.write( "Yes");
    }
    else {
        document.write( "No");
    }
     
 
    // This code contributed by aashish1995
 

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)