程序检查N是否为七边形数字

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

Heptadecagonal Number is class of figurate number. It has 17-sided polygon called heptadecagon. The N-th heptadecagonal number counts the seventeen number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few heptadecagonal numbers are 1, 17, 48, 94, 155, 231… 

例子：

Input: N = 17 
Output: Yes 
Explanation: 
Second heptadecagonal number is 17.

Input: N = 30 
Output: No 

方法：

1.七边形数的第K^个项为
$K^{th} Term = \frac{15*K^{2} - 13*K}{2}$

2.由于我们必须检查给定的数字是否可以表示为七边形数。可以按以下方式检查–

=> $N = \frac{15*K^{2} - 13*K}{2}$
=> $K = \frac{13 + \sqrt{120*N + 169}}{30}$

3.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。

4.其他N不是七边形数。

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 the number N
// is a heptadecagonal number
bool isheptadecagonal(int N)
{
    float n
        = (13 + sqrt(120 * N + 169))
          / 30;
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 17;
 
    // Function call
    if (isheptadecagonal(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check if the number N
// is a heptadecagonal number
static boolean isheptadecagonal(int N)
{
    float n = (float) ((13 + Math.sqrt(120 * N +
                                       169)) / 30);
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number
    int N = 17;
 
    // Function call
    if (isheptadecagonal(N))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
import numpy as np
 
# Function to check if the number N
# is a heptadecagonal number
def isheptadecagonal(N):
 
    n = (13 + np.sqrt(120 * N + 169)) / 30
 
    # Condition to check if number N
    # is a heptadecagonal number
    return (n - int(n)) == 0
 
# Driver Code
N = 17
 
# Function call
if (isheptadecagonal(N)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by PratikBasu

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if the number N
// is a heptadecagonal number
static bool isheptadecagonal(int N)
{
    float n = (float) ((13 + Math.Sqrt(120 * N +
                                       169)) / 30);
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - (int)n) == 0;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given Number
    int N = 17;
 
    // Function call
    if (isheptadecagonal(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 the number N
// is a heptadecagonal number
function isheptadecagonal(N)
{
    let n
        = (13 + Math.sqrt(120 * N + 169))
          / 30;
 
    // Condition to check if number N
    // is a heptadecagonal number
    return (n - parseInt(n)) == 0;
}
 
// Driver Code
// Given Number
let N = 17;
 
// Function call
if (isheptadecagonal(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by subham348.

输出：

Yes

时间复杂度： O(1)

辅助空间： O(1)