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:
- The Kth term of the Myriagon number is given as:
- As we have to check that the given number can be expressed as a Myriagon Number or not. This can be checked as:
=>
=>
- If the value of K calculated using the above formula is an integer, then N is a Myriagon Number.
- 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)