程序检查N是否为正二十进制数字

给定整数N ，任务是检查它是否为正二十进制数。

Icositrigonal number is a class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …

例子：

Input: N = 23 
Output: Yes 
Explanation: 
Second icositrigonal number is 23.
Input: N = 30 
Output: No 
 

方法：

1. 二十面三角数的第K^个项为
   $K^{th} Term = \frac{21*K^{2} - 19*K}{2}$
2. 因为我们必须检查给定的数字是否可以表示为二十面三角数。可以检查如下–

=> $N = \frac{21*K^{2} - 19*K}{2}$
=> $K = \frac{19 + \sqrt{168*N + 361}}{42}$
 

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为二十进制三角数。

下面是上述方法的实现：

C++

// C++ implementation to check that
// a number is a icositrigonal number or not
 
#include 
 
using namespace std;
 
// Function to check that the
// number is a icositrigonal number
bool isicositrigonal(int N)
{
    float n
        = (19 + sqrt(168 * N + 361))
          / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation to check that
// a number is a icositrigonal number or not
import java.util.*;
class GFG{
 
// Function to check that the
// number is a icositrigonal number
static boolean isicositrigonal(int N)
{
    float n = (float)(19 + Math.sqrt(168 * N + 361)) / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String args[])
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Akanksha_Rai

Python3

# Python3 implementation to check that a 
# number is a icositrigonal number or not
import math
 
# Function to check that the number
# is a icositrigonal number
def isicositrigonal(N):
 
    n = (19 + math.sqrt(168 * N + 361)) / 42
 
    # Condition to check if the number 
    # is a icositrigonal number
    return (n - int(n)) == 0
 
# Driver Code
i = 23
 
# Function call
if (isicositrigonal(i)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by divyamohan123

C#

// C# implementation to check that
// a number is a icositrigonal number or not
using System;
class GFG{
 
// Function to check that the
// number is a icositrigonal number
static bool isicositrigonal(int N)
{
    float n = (float)(19 + Math.Sqrt(168 * N + 361)) / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main()
{
    int i = 23;
 
    // Function call
    if (isicositrigonal(i))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed by Nidhi_Biet

Javascript

 
// JavaScript implementation to check that
// a number is a icositrigonal number or not
 
// Function to check that the
// number is a icositrigonal number
function isicositrigonal(N)
{
    var n
        = (19 + Math.sqrt(168 * N + 361))
          / 42;
 
    // Condition to check if the
    // number is a icositrigonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
var i = 23;
// Function call
if (isicositrigonal(i)) {
    document.write("Yes");
}
else {
    document.write("No");
}
    

输出：

Yes