Program to check if N is a Icosihexagonal Number
Given an integer N, the task is to check if it is a Icosihexagonal number or not.
Icosihexagonal number is class of figurate number. It has 26 – sided polygon called Icosihexagon. The N-th Icosihexagonal number count’s the 26 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few Icosihexagonol numbers are 1, 26, 75, 148 …
Examples:
Input: N = 26
Output: Yes
Explanation:
Second icosihexagonal number is 26.
Input: 30
Output: No
Approach:
- The Kth term of the icosihexagonal number is given as
- As we have to check that the given number can be expressed as a icosihexagonal number or not. This can be checked as follows –
=>
=>
- Finally, check the value of computed using this formulae is an integer, which means that N is a icosihexagonal number.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool isicosihexagonal( int N)
{
float n
= (22 + sqrt (192 * N + 484))
/ 48;
return (n - ( int )n) == 0;
}
int main()
{
int i = 26;
if (isicosihexagonal(i)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
}
|
Java
class GFG{
static boolean isicosihexagonal( int N)
{
float n = ( float ) (( 22 + Math.sqrt( 192 * N +
484 )) / 48 );
return (n - ( int )n) == 0 ;
}
public static void main(String[] args)
{
int N = 26 ;
if (isicosihexagonal(N))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
}
}
|
Python3
import numpy as np
def isicosihexagonal(N):
n = ( 22 + np.sqrt( 192 * N + 484 )) / 48
return (n - ( int (n))) = = 0
i = 26
if (isicosihexagonal(i)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
static bool isicosihexagonal( int N)
{
float n = ( float )((22 + Math.Sqrt(192 * N +
484)) / 48);
return (n - ( int )n) == 0;
}
public static void Main()
{
int N = 26;
if (isicosihexagonal(N))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
}
}
|
Javascript
<script>
function isicosihexagonal(N)
{
let n
= (22 + Math.sqrt(192 * N + 484))
/ 48;
return (n - parseInt(n)) == 0;
}
let i = 26;
if (isicosihexagonal(i)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
</script>
|
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
Last Updated :
23 Jun, 2021
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