Program to check if N is a Icositrigonal number
Last Updated :
02 Dec, 2022
Given an integer N, the task is to check if it is an Icositrigonal number or not.
Icositrigonal number is a class of figurate number. It has 23 – sided polygon called Icositrigon. The N-th Icositrigonal number count’s the 23 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Icositrigonol numbers are 1, 23, 66, 130, 215, 321, 448 …
Examples:
Input: N = 23
Output: Yes
Explanation:
Second icositrigonal number is 23.
Input: N = 30
Output: No
Approach:
1. The Kth term of the icositrigonal number is given as
2. As we have to check whether the given number can be expressed as an icositrigonal number or not. This can be checked as follows –
=>
=>
3. Finally, check the value computed using this formula is an integer, which means that N is an icositrigonal number.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool isicositrigonal( int N)
{
float n
= (19 + sqrt (168 * N + 361))
/ 42;
return (n - ( int )n) == 0;
}
int main()
{
int i = 23;
if (isicositrigonal(i)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
}
|
Java
import java.util.*;
class GFG{
static boolean isicositrigonal( int N)
{
float n = ( float )( 19 + Math.sqrt( 168 * N + 361 )) / 42 ;
return (n - ( int )n) == 0 ;
}
public static void main(String args[])
{
int i = 23 ;
if (isicositrigonal(i))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
}
}
|
Python3
import math
def isicositrigonal(N):
n = ( 19 + math.sqrt( 168 * N + 361 )) / 42
return (n - int (n)) = = 0
i = 23
if (isicositrigonal(i)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG{
static bool isicositrigonal( int N)
{
float n = ( float )(19 + Math.Sqrt(168 * N + 361)) / 42;
return (n - ( int )n) == 0;
}
public static void Main()
{
int i = 23;
if (isicositrigonal(i))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
}
}
|
Javascript
<script>
function isicositrigonal(N)
{
var n
= (19 + Math.sqrt(168 * N + 361))
/ 42;
return (n - parseInt(n)) == 0;
}
var i = 23;
if (isicositrigonal(i)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
</script>
|
Time Complexity: O(log N) because sqrt() function is being used
Auxiliary Space: O(1)
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