Program to check if N is a Icosagonal Number
Last Updated :
19 Sep, 2022
Given an integer N, the task is to check if it is a Icosagonal Number or not. If the number N is an Icosagonal Number then print “YES” else print “NO”.
Icosagonal Number is a twenty-sided polygon. The number derived from the figurative class. There are different patterns observed in this series. The dots are countable, arrange in a specific way of position, and create a diagram. All the dots have common dots points, all other dots are connected to these points and except this common point the dots connected to their ith dots with their respective successive layer… The first few Icosagonal numbers are 1, 20, 57, 112, 185, 276…
Examples:
Input: N = 20
Output: Yes
Explanation:
Second Icosagonal Number is 20.
Input: N = 30
Output: No
Approach:
1. The Kth term of the Icosagonal Number is given as
2. As we have to check that the given number can be expressed as a icosagonal 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 an Icosagonal Number.
4. Else the number N is not an Icosagonal Number.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool iicosagonal( int N)
{
float n
= (16 + sqrt (144 * N + 256))
/ 36;
return (n - ( int )n) == 0;
}
int main()
{
int N = 20;
if (iicosagonal(N)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
}
|
Java
import java.util.*;
class GFG{
static boolean iicosagonal( int N)
{
float n = ( float )(( 16 + Math.sqrt( 144 * N +
256 )) / 36 );
return (n - ( int )n) == 0 ;
}
public static void main(String[] args)
{
int N = 20 ;
if (iicosagonal(N))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
}
}
|
Python3
import numpy as np
def iicosagonal(N):
n = ( 16 + np.sqrt( 144 * N + 256 )) / 36
return (n - int (n)) = = 0
N = 20
if (iicosagonal(N)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
static bool iicosagonal( int N)
{
float n = ( float )((16 + Math.Sqrt(144 * N +
256)) / 36);
return (n - ( int )n) == 0;
}
public static void Main( string [] args)
{
int N = 20;
if (iicosagonal(N))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
}
}
|
Javascript
<script>
function iicosagonal(N)
{
var n
= (16 + Math.sqrt(144 * N + 256))
/ 36;
return (n - parseInt(n)) == 0;
}
var N = 20;
if (iicosagonal(N)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
</script>
|
Time Complexity: O(logN) because inbuilt sqrt function is being used
Auxiliary Space: O(1)
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