程序检查N是否为七边形数字

给定整数N ，任务是检查N是否为七边形数。如果数字N是七边形数字，则打印“是”，否则打印“否” 。

Heptagonal Number represents Heptagon and belongs to a figurative number. Heptagonal has seven angles, seven vertices, and seven-sided polygon. The first few Heptagonal Numbers are 1, 7, 18, 34, 55, 81, … 
 

例子：

Input: N = 7 
Output: Yes 
Explanation: 
Second heptagonal number is 7.
Input: N = 30 
Output: No 
 

方法：

1. 七边形数的第K^个项为
   $K^{th} Term = \frac{5*K^{2} - 3*K}{2}$
2. 因为我们必须检查给定的数字是否可以表示为七边形数字。可以检查为：

=> $N = \frac{5*K^{2} - 3*K}{2}$
=> $K = \frac{3 + \sqrt{40*N + 9}}{10}$
 

2. 如果使用上述公式计算出的K的值为整数，则N为七边形数。
3. 其他N不是七边形数字。

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
// Heptagonal number
bool isheptagonal(int N)
{
    float n
        = (3 + sqrt(40 * N + 9))
          / 10;
 
    // Condition to check if the
    // number is a heptagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Nuber
    int N = 7;
 
    // Function call
    if (isheptagonal(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 heptagonal number
public static boolean isheptagonal(int N)
{
    double n = (3 + Math.sqrt(40 * N + 9)) / 10;
     
    // Condition to check if the number
    // is a heptagonal number
    return (n - (int)n) == 0;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Number
    int N = 7;
         
    // Function call
    if (isheptagonal(N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by coder001

Python3

# Python3 program for the above approach
import math
 
# Function to check if N is a
# heptagonal number
def isheptagonal(N):
     
    n = (3 + math.sqrt(40 * N + 9)) / 10
     
    # Condition to check if the
    # number is a heptagonal number
    return (n - int(n)) == 0
     
# Driver Code
N = 7
 
# Function call
if (isheptagonal(N)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by Shubham_Coder

C#

// C# program for the above approach
using System;
 
class GFG {
     
// Function to check if N
// is a heptagonal number
public static bool isheptagonal(int N)
{
    double n = (3 + Math.Sqrt(40 * N + 9)) / 10;
     
    // Condition to check if the number
    // is a heptagonal number
    return (n - (int)n) == 0;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given Number
    int N = 7;
         
    // Function call
    if (isheptagonal(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
// Heptagonal number
function isheptagonal(N)
{
    var n = (3 + Math.sqrt(40 * N + 9))
          / 10;
 
    // Condition to check if the
    // number is a heptagonal number
    return (n - parseInt(n)) == 0;
}
 
// Given Nuber
var N = 7;
// Function call
if (isheptagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)