程序检查N是否为八角形数字

给定数字N ，任务是检查N是否为八角形数。如果数字N是八角形数字，则打印“是”，否则打印“否” 。

Octagonal Number is the figure number that represent octagonal. Octagonal Numbers can be formed by placing triangular numbers on the four sides of a square. The first few Octagonal Numbers are 1, 8, 21, 40, 65, 96 … 
 

例子：

Input: N = 8 
Output: Yes 
Explanation: 
Second octagonal number is 8.
Input: N = 30 
Output: No 
 

方法：

1. 八角数的第K^个项为
   $K^{th} Term = 3*K^{2} - 2*K$
2. 因为我们必须检查给定的数字是否可以表示为八角形数字。可以检查为：

=> $N = 3*K^{2} - 2*K$
=> $K = \frac{2 + \sqrt{12*N + 4}}{6}$
 

2. 如果使用上述公式计算出的K的值为整数，则N为八角形数。
3. 其他N不是八角形数字。

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 N is a
// Octagonal Number
bool isoctagonal(int N)
{
    float n
        = (2 + sqrt(12 * N + 4))
          / 6;
 
    // Condition to check if the
    // number is a octagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 8;
 
    // Function call
    if (isoctagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG {
     
// Function to check if N is a
// octagonal number
public static boolean isoctagonal(int N)
{
    double n = (2 + Math.sqrt(12 * N + 4)) / 6;
     
    // Condition to check if the
    // number is a octagonal number
    return (n - (int)n) == 0;
}
     
// Driver code
public static void main(String[] args)
{
     
    // Given Number
    int N = 8;
     
    // Function call
    if (isoctagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by coder001

Python3

# Python3 program for the above approach
from math import sqrt
 
# Function to check if N is a
# octagonal number
def isoctagonal(N):
 
    n = (2 + sqrt(12 * N + 4)) / 6;
 
    # Condition to check if the
    # number is a octagonal number
    return (n - int(n)) == 0;
     
# Driver Code
if __name__ == "__main__":
 
    # Given number
    N = 8;
 
    # Function call
    if (isoctagonal(N)):
        print("Yes");
     
    else:
        print("No");
 
# This code is contributed by AnkitRai01

C#

// C# program for the above approach
using System;
 
class GFG {
     
// Function to check if N is a
// octagonal number
public static bool isoctagonal(int N)
{
    double n = (2 + Math.Sqrt(12 * N +
                              4)) / 6;
     
    // Condition to check if the
    // number is a octagonal number
    return (n - (int)n) == 0;
}
     
// Driver code
public static void Main(String[] args)
{
     
    // Given number
    int N = 8;
     
    // Function call
    if (isoctagonal(N))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// JavaScript program for the above approach
 
// Function to check if N is a
// Octagonal Number
function isoctagonal(N)
{
    var n = (2 + Math.sqrt(12 * N + 4)) / 6;
 
    // Condition to check if the
    // number is a octagonal number
    return (n - parseInt(n) == 0);
}
 
// Given Number
var N = 8;
 
// Function call
if (isoctagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 

输出：

Yes