程序检查N是否为对角线数

给定整数N ，任务是检查N是否为Chiliagon Number。如果数字N是对角线数字，则打印“是”，否则打印“否”。

Chiliagon Number is class of figurate number. It has 1000 – sided polygon called Chiliagon. The N-th Chiliagon Number counts the 1000 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Chiliagon Numbers are 1, 1000, 2997, 5992, …

例子：

Input: N = 1000
Output: Yes
Explanation:
Second chiliagon number is 1000

Input: 35
Output: No

方法：

1. Chiliagon数的第K^个项为
   $K^{th} Term = \frac{998*K^{2} - 996*K}{2}$
2. 因为我们必须检查给定的数字是否可以表示为Chiliagon数。可以检查如下：

   => $N = \frac{998*K^{2} - 996*K}{2}$
   => $K = \frac{996 + \sqrt{7984*N + 992016}}{1996}$

3. 如果使用上述公式计算出的K的值为整数，则N为Chiliagon数。
4. 其他N不是对角线数字。

下面是上述方法的实现：

C++

// C++ for the above approach
#include 
using namespace std;
  
// Function to check that if N is
// Chiliagon Number or not
bool is_Chiliagon(int N)
{
    float n
        = (996 + sqrt(7984 * N + 992016))
          / 1996;
  
    // Condition to check if N is a
    // Chiliagon Number
    return (n - (int)n) == 0;
}
  
// Driver Code
int main()
{
    // Given Number
    int N = 1000;
  
    // Function call
    if (is_Chiliagon(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
// Function to check that if N is
// Chiliagon Number or not
static boolean is_Chiliagon(int N) 
{
    float n = (float)(996 + Math.sqrt(7984 * N + 
                                      992016)) / 1996;
  
    // Condition to check if N is a
    // Chiliagon Number
    return (n - (int) n) == 0;
}
  
// Driver Code
public static void main(String s[]) 
{
    // Given Number
    int N = 1000;
  
    // Function call
    if (is_Chiliagon(N)) 
    {
        System.out.print("Yes");
    } 
    else 
    {
        System.out.print("No");
    }
}
}
  
// This code is contributed by rutvik_56

Python3

# Python3 for the above approach
import math;
  
# Function to check that if N is
# Chiliagon Number or not
def is_Chiliagon(N):
  
    n = (996 + math.sqrt(7984 * N + 
                         992016)) // 1996;
  
    # Condition to check if N is a
    # Chiliagon Number
    return (n - int(n)) == 0;
  
# Driver Code
  
# Given Number
N = 1000;
  
# Function call
if (is_Chiliagon(N)):
    print("Yes");
else:
    print("No");
  
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{
      
// Function to check that if N is
// Chiliagon Number or not
static bool is_Chiliagon(int N) 
{
    float n = (float)(996 + Math.Sqrt(7984 * N + 
                                      992016)) / 1996;
  
    // Condition to check if N is a
    // Chiliagon Number
    return (n - (int) n) == 0;
}
  
// Driver Code
public static void Main() 
{
    // Given Number
    int N = 1000;
  
    // Function call
    if (is_Chiliagon(N)) 
    {
        Console.Write("Yes");
    } 
    else
    {
        Console.Write("No");
    }
}
}
  
// This code is contributed by Code_Mech

输出：

Yes