Given a number N, the task is to check if N is a Nonagonal Number or not. If the number N is an Nonagonal Number then print “Yes” else print “No”.
Nonagonal Number is a figurate number that extends the concept of triangular and square numbers to the Nonagon. Specifically, the nth Nonagonal Numbers count the number of dots in a pattern of n nested nonagons(9 sided polygon), all sharing a common corner, where the ith nonagon in the pattern has sides made of i dots spaced one unit apart from each other. The first few Nonagonal Numbers are 1, 9, 24, 46, 75, 111, 154, …
Examples:
Input: N = 9
Output: Yes
Explanation:
Second Nonagonal Number is 9.
Input: N = 20
Output: No
Approach:
1. The Kth term of the nonagonal number is given as
2. As we have to check that the given number can be expressed as a Nonagonal 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 Nonagonal Number.
4. Else N is not a Nonagonal 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 // is a Nonagonal Number bool isnonagonal( int N)
{ float n
= (5 + sqrt (56 * N + 25))
/ 14;
// Condition to check if the
// number is a nonagonal number
return (n - ( int )n) == 0;
} // Driver Code int main()
{ // Given Number
int N = 9;
// Function call
if (isnonagonal(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 // nonagonal number public static boolean isnonagonal( int N)
{ double n = ( 5 + Math.sqrt( 56 * N + 25 )) / 14 ;
// Condition to check if the
// number is a nonagonal number
return (n - ( int )n) == 0 ;
} // Driver code public static void main(String[] args)
{ // Given number
int N = 9 ;
// Function call
if (isnonagonal(N))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
} } // This code is contributed by divyeshrabadiya07 |
# Python3 program for the above approach # Function to check if N is a # nonagonal number def isnonagonal(N):
n = ( 5 + pow (( 56 * N + 25 ), 1 / 2 )) / 14 ;
# Condition to check if the
# number is a nonagonal number
return (n - int (n)) = = 0 ;
# Driver code if __name__ = = '__main__' :
# Given number
N = 9 ;
# Function call
if (isnonagonal(N)):
print ( "Yes" );
else :
print ( "No" );
# This code is contributed by Rajput-Ji |
// C# program for the above approach using System;
class GFG{
// Function to check if N is a // nonagonal number public static bool isnonagonal( int N)
{ double n = (5 + Math.Sqrt(56 * N + 25)) / 14;
// Condition to check if the
// number is a nonagonal number
return (n - ( int )n) == 0;
} // Driver code public static void Main( string [] args)
{ // Given number
int N = 9;
// Function call
if (isnonagonal(N))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
} } // This code is contributed by rutvik_56 |
<script> // Javascript program for the above approach // Function to check if N is a // is a Nonagonal Number function isnonagonal(N)
{ let n
= (5 + Math.sqrt(56 * N + 25))
/ 14;
// Condition to check if the
// number is a nonagonal number
return (n - parseInt(n)) == 0;
} // Driver Code // Given Number let N = 9; // Function call if (isnonagonal(N))
{ document.write( "Yes" );
} else { document.write( "No" );
} // This code is contributed by subhammahato348. </script> |
Output:
Yes
Time Complexity: O(log N) because sqrt() function is being used
Auxiliary Space: O(1)