Program to check if N is a Tetradecagonal Number
Given an integer N, the task is to check if N is a Tetradecagonal Number or not. If the number N is a Tetradecagonal Number then print “Yes” else print “No”.
Tetradecagonal Number is 14-sided polygon called Tetrakaidecagon or Tetradecagon and belongs to the figurative number. The nth tetradecagonal number dotted with some dots and create a series of the pattern. They have a common sharing corner point and dotted with their spaces to each other. The dots continue with nth nested loop.The first few Tetradecagonal Numbers are 1, 14, 39, 76, 125, 186, …
Examples:
Input: N = 14
Output: Yes
Explanation:
Second tetradecagonal number is 14.Input: N = 40
Output: No
Approach:
- The Kth term of the tetradecagonal number is given as
- As we have to check whether the given number can be expressed as a Tetradecagonal 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 Tetradecagonal Number.
- Else N is not a Tetradecagonal 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 // Tetradecagonal Number bool istetradecagonal( int N) { float n = (10 + sqrt (96 * N + 100)) / 24; // Condition to check if the // number is a tetradecagonal number return (n - ( int )n) == 0; } // Driver Code int main() { // Given Number int N = 11; // Function call if (istetradecagonal(N)) { cout << "Yes" ; } else { cout << "No" ; } return 0; } |
Java
// Java program for the above approach import java.lang.Math; class GFG{ // Function to check if N is a // tetradecagonal number public static boolean istetradecagonal( int N) { double n = ( 10 + Math.sqrt( 96 * N + 100 )) / 24 ; // Condition to check if the number // is a tetradecagonal number return (n - ( int )n) == 0 ; } // Driver Code public static void main(String[] args) { // Given number int N = 11 ; // Function call if (istetradecagonal(N)) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } // This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program for the above approach import math # Function to check if N is a # Tetradecagonal Number def istetradecagonal(N): n = ( 10 + math.sqrt( 96 * N + 100 )) / 24 # Condition to check if the # number is a tetradecagonal number if (n - int (n)) = = 0 : return True return False # Driver Code # Given Number N = 11 # Function call if (istetradecagonal(N)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by shubhamsingh10 |
C#
// C# program for the above approach using System; class GFG{ // Function to check if N is a // tetradecagonal number public static bool istetradecagonal( int N) { double n = (10 + Math.Sqrt(96 * N + 100)) / 24; // Condition to check if the number // is a tetradecagonal number return (n - ( int )n) == 0; } // Driver Code static public void Main () { // Given number int N = 11; // Function call if (istetradecagonal(N)) { Console.Write( "Yes" ); } else { Console.Write( "No" ); } } } // This code is contributed by shubhamsingh10 |
Javascript
<script> // Javascript program for the above approach // Function to check if N is a // Tetradecagonal Number function istetradecagonal(N) { n = (10 + Math.sqrt(96 * N + 100)) / 24; // Condition to check if the // number is a tetradecagonal number return (n - parseInt(n)) == 0; } // Driver Code // Given Number N = 11; // Function call if (istetradecagonal(N)) { document.write( "Yes" ); } else { document.write( "No" ); } </script> |
Output:
No
Time Complexity: O(log N) because sqrt() function is being used
Auxiliary Space: O(1)