Program to check if N is a Icosihenagonal number
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++
#include <bits/stdc++.h>
using namespace std;
bool isicosihenagonal( int N)
{
float n
= (17 + sqrt (152 * N + 289))
/ 38;
return (n - ( int )n) == 0;
}
int main()
{
int i = 21;
if (isicosihenagonal(i)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
}
|
Java
class GFG{
static boolean isicosihenagonal( int N)
{
float n = ( float ) (( 17 + Math.sqrt( 152 * N +
289 )) / 38 );
return (n - ( int )n) == 0 ;
}
public static void main(String[] args)
{
int i = 21 ;
if (isicosihenagonal(i))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
}
}
|
Python3
import math
def isicosihenagonal(N):
n = ( 17 + math.sqrt( 152 * N + 289 )) / 38
return (n - int (n)) = = 0
i = 21
if isicosihenagonal(i):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
static bool isicosihenagonal( int N)
{
float n = ( float )((17 + Math.Sqrt(152 * N +
289)) / 38);
return (n - ( int )n) == 0;
}
public static void Main()
{
int i = 21;
if (isicosihenagonal(i))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
}
}
|
Javascript
<script>
function isicosihenagonal(N)
{
var n
= (17 + Math.sqrt(152 * N + 289))
/ 38;
return (n - parseInt(n)) == 0;
}
var i = 21;
if (isicosihenagonal(i)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
</script>
|
Time Complexity: O(logN) because sqrt() function is being used
Auxiliary Space: O(1)
Last Updated :
29 Nov, 2022
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