程序检查N是否为十进制数

给定整数N ，任务是检查N是否为十进制数。如果数字N是十进制数，则打印“是”，否则打印“否” 。

Hendecagonal Number is a figurate number that extends the concept of triangular and square numbers to the decagon(11-sided polygon). The nth hendecagonal number counts the number of dots in a pattern of n nested decagons, all sharing a common corner, where the ith hendecagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few hendecagonal numbers are 1, 11, 30, 58, 95, 141… 
 

例子：

Input: N = 11 
Output: Yes 
Explanation: 
Second hendecagonal number is 11.
Input: N = 40 
Output: No 
 

方法：

1. 十进制数的第K^个项给出为
   $K^{th} Term = \frac{9*K^{2} - 7*K}{2}$
2. 由于我们必须检查给定的数字是否可以表示为十进制数。可以检查为：

=> $N = \frac{9*K^{2} - 7*K}{2}$
=> $K = \frac{7 + \sqrt{72*N + 49}}{18}$
 

2. 如果使用上述公式计算出的K值为整数，则N为一个十进制数。
3. 其他N不是十进制数。

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 N is a
// Hendecagonal Number
bool ishendecagonal(int N)
{
    float n
        = (7 + sqrt(72 * N + 49))
          / 18;
 
    // Condition to check if the
    // number is a hendecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 11;
 
    // Function call
    if (ishendecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.lang.Math;
 
class GFG{
     
// Function to check if N is a
// hendecagonal number
public static boolean ishendecagonal(int N)
{
    double n = (7 + Math.sqrt(72 * N + 49)) / 18;
     
    // Condition to check if the
    // number is a hendecagonal number
    return (n - (int)n) == 0;
}
 
// Driver code   
public static void main(String[] args)
{
         
    // Given number
    int N = 11;
     
    // Function call
    if (ishendecagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
import math
 
# Function to check if N is a
# Hendecagonal Number
def ishendecagonal(N):
 
    n = (7 + math.sqrt(72 * N + 49))// 18;
 
    # Condition to check if the
    # number is a hendecagonal number
    return (n - int(n)) == 0;
 
# Driver Code
 
# Given Number
N = 11;
 
# Function call
if (ishendecagonal(N)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by Nidhi_biet

C#

// C# program for the above approach
using System;
class GFG{
     
// Function to check if N is a
// hendecagonal number
public static bool ishendecagonal(int N)
{
    double n = (7 + Math.Sqrt(72 * N + 49)) / 18;
     
    // Condition to check if the
    // number is a hendecagonal number
    return (n - (int)n) == 0;
}
 
// Driver code
public static void Main(string[] args)
{
         
    // Given number
    int N = 11;
     
    // Function call
    if (ishendecagonal(N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No\n");
    }
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// javascript program for the above approach
 
 
// Function to check if N is a
// Hendecagonal Number
function ishendecagonal( N)
{
    let n
        = (7 + Math.sqrt(72 * N + 49))
          / 18;
 
    // Condition to check if the
    // number is a hendecagonal number
    return (n - parseInt(n)) == 0;
}
 
 
// Driver Code
 
    // Given Number
    let N = 11;
 
    // Function Call
    if (ishendecagonal(N)) {
        document.write( "Yes");
    }
    else {
        document.write( "No");
    }
 
// This code contributed by gauravrajput1
 

输出：

Yes