Program to check if N is a Myriagon Number

Given a number N, the task is to check if N is a Myriagon Number or not. If the number N is an Myriagon Number then print “Yes” else print “No”.

Myriagon Number is a polygon with 10000 sides. The first few Myriagon numbers are 1, 10000, 29997, 59992, 99985, 149976 …

Examples:

Input: N = 10000 
Output: Yes 
Explanation:
Second Myriagon number is 10000.
Input: N = 300 
Output: No 

Approach:

1. The K^th term of the Myriagon number is given as: 
   
   
2. As we have to check that the given number can be expressed as a Myriagon Number or not. This can be checked as: 
   

=> 
=> 

1. 
   
2. If the value of K calculated using the above formula is an integer, then N is a Myriagon Number.
3. Else N is not a Myriagon Number.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if N is a
// Myriagon Number
bool isMyriagon(int N)
{
    float n
        = (9996 + sqrt(79984 * N + 99920016))
          / 19996;

    // Condition to check if the
    // number is a Myriagon number
    return (n - (int)n) == 0;
}

// Driver Code
int main()
{
    // Given Number
    int N = 10000;

    // Function call
    if (isMyriagon(N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG {

// Function to check if N
// is a myriagon number
static boolean isMyriagon(int N)
{
    double n = (9996 + Math.sqrt(79984 * N +
                                 99920016)) / 19996;

    // Condition to check if the
    // number is a myriagon number
    return (n - (int)n) == 0;
}

// Driver Code
public static void main (String[] args)
{

    // Given Number
    int N = 10000;

    // Function call
    if (isMyriagon(N))
    {
        System.out.println("Yes" );
    }
    else
    {
        System.out.println("No" );
    }
}
}

// This code is contributed by ShubhamCoder

Python3

# Python3 implementation to check that
# a number is a myriagon number or not
import math

# Function to check that the
# number is a myriagon number
def isMyriagon(N):

    n = (9996 + math.sqrt(79984 * N +
                          99920016)) / 19996

    # Condition to check if the
    # number is a myriagon number
    return (n - int(n)) == 0


# Driver Code
n = 10000

# Function call
if (isMyriagon(n)):
    print("Yes")
else:
    print("No")

# This code is contributed by ShubhamCoder

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if N
// is a myriagon number
static bool isMyriagon(int N)
{
    double n = (9996 + Math.Sqrt(79984 * N +
                                 99920016)) / 19996;

    // Condition to check if the
    // number is a myriagon number
    return (n - (int)n) == 0;
}

// Driver Code
static public void Main ()
{

    // Given Number
    int N = 10000;

    // Function call
    if (isMyriagon(N))
    {
        Console.Write( "Yes" );
    }
    else
    {
        Console.Write( "No" );
    }
}
}

// This code is contributed by ShubhamCoder

Javascript

<script>

// Javascript program for the above approach

// Function to check if N is a
// Myriagon Number
function isMyriagon(N)
{
    n = (9996 + Math.sqrt(79984 * N + 99920016))
          / 19996;

    // Condition to check if the
    // number is a Myriagon number
    return (n - parseInt(n)) == 0;
}

// Driver Code
// Given Number
N = 10000;
// Function call
if (isMyriagon(N)) {
    document.write("Yes");
}
else {
    document.write("No");
}

</script>

Output:

Yes

Time Complexity: O(logN) because inbuilt sqrt function is being used

Auxiliary Space: O(1)