Given an integer N, the task is to check if N is a Hendecagonal Number or not. If the number N is a Hendecagonal Number then print “Yes” else print “No”
Hendecagonal Number is a figurate number that extends the concept of triangular and square numbers to the decagon(11-sided polygon). The nth hendecagonal number counts the number of dots in a pattern of n nested decagons, all sharing a common corner, where the ith hendecagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few hendecagonal numbers are 1, 11, 30, 58, 95, 141…
Examples:
Input: N = 11
Output: Yes
Explanation:
Second hendecagonal number is 11.Input: N = 40
Output: No
Approach:
1. The Kth term of the Hendecagonal Number is given as
2. As we have to check whether the given number can be expressed as a Hendecagonal Number or not. This can be checked as:
=>
=>
3. If the value of K calculated using the above formula is an integer, then N is a Hendecagonal Number.
4. Else N is not a Hendecagonal Number.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if N is a // Hendecagonal Number bool ishendecagonal( int N)
{ float n
= (7 + sqrt (72 * N + 49))
/ 18;
// Condition to check if the
// number is a hendecagonal number
return (n - ( int )n) == 0;
} // Driver Code int main()
{ // Given Number
int N = 11;
// Function call
if (ishendecagonal(N)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
} |
// Java program for the above approach import java.lang.Math;
class GFG{
// Function to check if N is a // hendecagonal number public static boolean ishendecagonal( int N)
{ double n = ( 7 + Math.sqrt( 72 * N + 49 )) / 18 ;
// Condition to check if the
// number is a hendecagonal number
return (n - ( int )n) == 0 ;
} // Driver code public static void main(String[] args)
{ // Given number
int N = 11 ;
// Function call
if (ishendecagonal(N))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
} } // This code is contributed by divyeshrabadiya07 |
# Python3 program for the above approach import math
# Function to check if N is a # Hendecagonal Number def ishendecagonal(N):
n = ( 7 + math.sqrt( 72 * N + 49 )) / / 18 ;
# Condition to check if the
# number is a hendecagonal number
return (n - int (n)) = = 0 ;
# Driver Code # Given Number N = 11 ;
# Function call if (ishendecagonal(N)):
print ( "Yes" );
else :
print ( "No" );
# This code is contributed by Nidhi_biet |
// C# program for the above approach using System;
class GFG{
// Function to check if N is a // hendecagonal number public static bool ishendecagonal( int N)
{ double n = (7 + Math.Sqrt(72 * N + 49)) / 18;
// Condition to check if the
// number is a hendecagonal number
return (n - ( int )n) == 0;
} // Driver code public static void Main( string [] args)
{ // Given number
int N = 11;
// Function call
if (ishendecagonal(N))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No\n" );
}
} } // This code is contributed by rutvik_56 |
<script> // javascript program for the above approach // Function to check if N is a // Hendecagonal Number function ishendecagonal( N)
{ let n
= (7 + Math.sqrt(72 * N + 49))
/ 18;
// Condition to check if the
// number is a hendecagonal number
return (n - parseInt(n)) == 0;
} // Driver Code // Given Number
let N = 11;
// Function Call
if (ishendecagonal(N)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
// This code contributed by gauravrajput1 </script> |
Output
Yes
Time Complexity: O(log N) because sqrt function is being used
Auxiliary Space: O(1)