Given an integer N, the task is to check if it is a Heptadecagonal Number or not. If the number N is an Heptadecagonal Number then print “Yes” else print “No”.
Heptadecagonal Number is class of figurate number. It has 17-sided polygon called heptadecagon. The N-th heptadecagonal number counts the seventeen number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few heptadecagonal numbers are 1, 17, 48, 94, 155, 231…
Examples:
Input: N = 17
Output: Yes
Explanation:
Second heptadecagonal number is 17.Input: N = 30
Output: No
Approach:
1. The Kth term of the heptadecagonal number is given as
2. As we have to check that the given number can be expressed as a heptadecagonal 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 Heptadecagonal Number.
4. Else N is not a Heptadecagonal 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 the number N // is a heptadecagonal number bool isheptadecagonal( int N) { float n = (13 + sqrt (120 * N + 169)) / 30; // Condition to check if number N // is a heptadecagonal number return (n - ( int )n) == 0; } // Driver Code int main() { // Given Number int N = 17; // Function call if (isheptadecagonal(N)) { cout << "Yes" ; } else { cout << "No" ; } return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to check if the number N // is a heptadecagonal number static boolean isheptadecagonal( int N) { float n = ( float ) (( 13 + Math.sqrt( 120 * N + 169 )) / 30 ); // Condition to check if number N // is a heptadecagonal number return (n - ( int )n) == 0 ; } // Driver Code public static void main(String[] args) { // Given Number int N = 17 ; // Function call if (isheptadecagonal(N)) { System.out.print( "Yes" ); } else { System.out.print( "No" ); } } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach import numpy as np # Function to check if the number N # is a heptadecagonal number def isheptadecagonal(N): n = ( 13 + np.sqrt( 120 * N + 169 )) / 30 # Condition to check if number N # is a heptadecagonal number return (n - int (n)) = = 0 # Driver Code N = 17 # Function call if (isheptadecagonal(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 the number N // is a heptadecagonal number static bool isheptadecagonal( int N) { float n = ( float ) ((13 + Math.Sqrt(120 * N + 169)) / 30); // Condition to check if number N // is a heptadecagonal number return (n - ( int )n) == 0; } // Driver Code public static void Main( string [] args) { // Given Number int N = 17; // Function call if (isheptadecagonal(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 the number N // is a heptadecagonal number function isheptadecagonal(N) { let n = (13 + Math.sqrt(120 * N + 169)) / 30; // Condition to check if number N // is a heptadecagonal number return (n - parseInt(n)) == 0; } // Driver Code // Given Number let N = 17; // Function call if (isheptadecagonal(N)) { document.write( "Yes" ); } else { document.write( "No" ); } // This code is contributed by subham348. </script> |
Yes
Time Complexity: O(logN) because inbuilt sqrt function has been used
Auxiliary Space: O(1)
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