Program to check if N is a Dodecagonal Number
Given a number N, the task is to check if N is a Dodecagonal Number or not. If the number N is a Dodecagonal Number then print “Yes” else print “No”.
dodecagonal number represent Dodecagonal(12 sides polygon).The first few dodecagonal numbers are 1, 12, 33, 64, 105, 156, 217…
Examples:
Input: N = 12
Output: Yes
Explanation:
Second dodecagonal number is 12.Input: N = 30
Output: No
Approach:
1. The Kth term of the Dodecagonal Number is given as
2. As we have to check whether the given number can be expressed as a Dodecagonal Number or not. This can be checked as follows:
=>
=>
3. If the value of K calculated using the above formula is an integer, then N is a Dodecagonal Number.
4. Else the number N is not a Dodecagonal 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 number N // is a dodecagonal number or not bool isdodecagonal( int N) { float n = (4 + sqrt (20 * N + 16)) / 10; // Condition to check if the // N is a dodecagonal number return (n - ( int )n) == 0; } // Driver Code int main() { // Given Number int N = 12; // Function call if (isdodecagonal(N)) { cout << "Yes" ; } else { cout << "No" ; } return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to check if number N // is a dodecagonal number or not static boolean isdodecagonal( int N) { float n = ( float ) (( 4 + Math.sqrt( 20 * N + 16 )) / 10 ); // Condition to check if the // N is a dodecagonal number return (n - ( int )n) == 0 ; } // Driver Code public static void main(String[] args) { // Given Number int N = 12 ; // Function call if (isdodecagonal(N)) { System.out.print( "Yes" ); } else { System.out.print( "No" ); } } } // This code is contributed by sapnasingh4991 |
Python3
# Python3 program for the above approach import numpy as np # Function to check if number N # is a dodecagonal number or not def isdodecagonal(N): n = ( 4 + np.sqrt( 20 * N + 16 )) / 10 # Condition to check if the # N is a dodecagonal number return (n - int (n)) = = 0 # Driver Code N = 12 # Function call if (isdodecagonal(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 number N // is a dodecagonal number or not static bool isdodecagonal( int N) { float n = ( float ) ((4 + Math.Sqrt(20 * N + 16)) / 10); // Condition to check if the // N is a dodecagonal number return (n - ( int )n) == 0; } // Driver Code public static void Main( string [] args) { // Given number int N = 12; // Function call if (isdodecagonal(N)) { Console.Write( "Yes" ); } else { Console.Write( "No" ); } } } // This code is contributed by rutvik_56 |
Javascript
<script> // Javascript program for the above approach // Function to check if number N // is a dodecagonal number or not function isdodecagonal(N) { let n = (4 + Math.sqrt(20 * N + 16)) / 10; // Condition to check if the // N is a dodecagonal number return (n - parseInt(n)) == 0; } // Driver Code // Given Number let N = 12; // Function call if (isdodecagonal(N)) { document.write( "Yes" ); } else { document.write( "No" ); } // This code is contributed by subhammahato348. </script> |
Output
Yes
Time Complexity: O(log N), since sqrt() function has been used
Auxiliary Space: O(1)