程序检查N是否为非对角数

给定数字N ，任务是检查N是否为非对角数。如果数字N是非对角线数字，则打印“是”，否则打印“否” 。

Nonagonal Number is a figurate number that extends the concept of triangular and square numbers to the Nonagon. Specifically, the nth Nonagonal Numbers count the number of dots in a pattern of n nested nonagons(9 sided polygon), all sharing a common corner, where the ith nonagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few Nonagonal Numbers are 1, 9, 24, 46, 75, 111, 154, … 
 

例子：

Input: N = 9 
Output: Yes 
Explanation: 
Second Nonagonal Number is 9.

Input: N = 20 
Output: No 

方法：

1.在K的九边形数的^第术语被给定为
$K^{th} Term = \frac{7*K^{2} - 5*K}{2}$

2.由于我们必须检查给定的数字是否可以表示为非对角线数字。可以检查为：

=> $N = \frac{7*K^{2} - 5*K}{2}$
=> $K = \frac{5 + \sqrt{56*N + 25}}{14}$

3.如果使用上述公式计算的K值为整数，则N为非对角数。

4.其他N不是非对角数。

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
// is a Nonagonal Number
bool isnonagonal(int N)
{
    float n
        = (5 + sqrt(56 * N + 25))
          / 14;
 
    // Condition to check if the
    // number is a nonagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 9;
 
    // Function call
    if (isnonagonal(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
// nonagonal number
public static boolean isnonagonal(int N)
{
    double n = (5 + Math.sqrt(56 * N + 25)) / 14;
     
    // Condition to check if the
    // number is a nonagonal number
    return (n - (int)n) == 0;
}
 
// Driver code    
public static void main(String[] args)
{
         
    // Given number
    int N = 9;
     
    // Function call
    if (isnonagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function to check if N is a
# nonagonal number
def isnonagonal(N):
    n = (5 + pow((56 * N + 25), 1 / 2)) / 14;
 
    # Condition to check if the
    # number is a nonagonal number
    return (n - int(n)) == 0;
 
# Driver code
if __name__ == '__main__':
 
    # Given number
    N = 9;
 
    # Function call
    if (isnonagonal(N)):
        print("Yes");
    else:
        print("No");
 
# This code is contributed by Rajput-Ji

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if N is a
// nonagonal number
public static bool isnonagonal(int N)
{
    double n = (5 + Math.Sqrt(56 * N + 25)) / 14;
     
    // Condition to check if the
    // number is a nonagonal number
    return (n - (int)n) == 0;
}
 
// Driver code    
public static void Main(string[] args)
{
         
    // Given number
    int N = 9;
     
    // Function call
    if (isnonagonal(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 N is a
// is a Nonagonal Number
function isnonagonal(N)
{
    let n
        = (5 + Math.sqrt(56 * N + 25))
          / 14;
 
    // Condition to check if the
    // number is a nonagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
 
// Given Number
let N = 9;
 
// Function call
if (isnonagonal(N))
{
    document.write("Yes");
}
else
{
    document.write("No");
}
 
// This code is contributed by subhammahato348.

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)