Given an integer N, the task is to check if it is a Icosihenagonal number or not.
Icosihenagonal number is class of figurate number. It has 21 – sided polygon called Icosihenagon. The n-th Icosihenagonal number counts the 21 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icosihenagonal numbers are 1, 21, 60, 118, 195, 291, 406…
Examples:
Input: N = 21
Output: Yes
Explanation:
Second icosihenagonal number is 21.Input: N = 30
Output: No
Approach:
1. The Kth term of the icosihenagonal number is given as
2. As we have to check whether the given number can be expressed as a icosihenagonal number or not. This can be checked as follows –
=>
=>
3. Finally, check the value of computed using this formulae is an integer, which means that N is a icosihenagonal number.
Below is the implementation of the above approach:
// C++ implementation to check that // a number is icosihenagonal number or not #include <bits/stdc++.h> using namespace std;
// Function to check that the // number is a icosihenagonal number bool isicosihenagonal( int N)
{ float n
= (17 + sqrt (152 * N + 289))
/ 38;
// Condition to check if the
// number is a icosihenagonal number
return (n - ( int )n) == 0;
} // Driver Code int main()
{ int i = 21;
// Function call
if (isicosihenagonal(i)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
} |
// Java implementation to check that a // number is icosihenagonal number or not class GFG{
// Function to check that the number // is a icosihenagonal number static boolean isicosihenagonal( int N)
{ float n = ( float ) (( 17 + Math.sqrt( 152 * N +
289 )) / 38 );
// Condition to check if the number
// is a icosihenagonal number
return (n - ( int )n) == 0 ;
} // Driver Code public static void main(String[] args)
{ int i = 21 ;
// Function call
if (isicosihenagonal(i))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
} } // This code is contributed by 29AjayKumar |
# Python3 implementation to check that # a number is icosihenagonal number or not import math
# Function to check that the number # is a icosihenagonal number def isicosihenagonal(N):
n = ( 17 + math.sqrt( 152 * N + 289 )) / 38
# Condition to check if the number
# is a icosihenagonal number
return (n - int (n)) = = 0
# Driver Code i = 21
# Function call if isicosihenagonal(i):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by divyamohan123 |
// C# implementation to check that a // number is icosihenagonal number or not using System;
class GFG{
// Function to check that the number // is a icosihenagonal number static bool isicosihenagonal( int N)
{ float n = ( float )((17 + Math.Sqrt(152 * N +
289)) / 38);
// Condition to check if the number
// is a icosihenagonal number
return (n - ( int )n) == 0;
} // Driver Code public static void Main()
{ int i = 21;
// Function call
if (isicosihenagonal(i))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
} } // This code is contributed by Code_Mech |
<script> // JavaScript implementation to check that // a number is icosihenagonal number or not // Function to check that the // number is a icosihenagonal number function isicosihenagonal(N)
{ var n
= (17 + Math.sqrt(152 * N + 289))
/ 38;
// Condition to check if the
// number is a icosihenagonal number
return (n - parseInt(n)) == 0;
} // Driver Code var i = 21;
// Function call if (isicosihenagonal(i)) {
document.write( "Yes" );
} else {
document.write( "No" );
} </script> |
Output:
Yes
Time Complexity: O(logN) because sqrt() function is being used
Auxiliary Space: O(1)