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++ 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 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 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# 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 |
<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)