Program to check if N is a Icosidigonal Number
Given an integer N, the task is to check if it is a Icosidigonal Number or not. If the number N is an Icosidigonal Number then print “Yes” else print “No”.
Icosidigonal number:
The polygon has many gons, depends on their gonal number series. In mathematics, there are a number of gonal numbers and the Icosidigonal Number is one of them and these numbers have 22 -sided polygon(icosidigon). An Icosidigonal Number belong to the class of figurative number. They have one common dots points and other dots pattern is arranged in an n-th nested Icosidigon pattern.
The first few Icosidigonal numbers are 1, 22, 63, 124, 205, 306…
Examples:
Input: N = 22
Output: Yes
Explanation:
Second Icosidigonal number is 22.
Input: 30
Output: No
Approach:
1. The Kth term of the Icosidigonal number is given as
2. As we have to check that the given number can be expressed as an Icosidigonal 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 Icosidigonal Number.
4. Else N is not an Icosidigonal Number.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if the number N // is a Icosidigonal number bool isIcosidigonal( int N) { float n = (18 + sqrt (160 * N + 324)) / 40; // Condition to check if the // number is a Icosidigonal number return (n - ( int )n) == 0; } // Driver Code int main() { // Given Number int N = 22; // Function call if (isIcosidigonal(N)) { cout << "Yes" ; } else { cout << "No" ; } return 0; } |
Java
// Java program for the above approach class GFG{ // Function to check if the number N // is a icosidigonal number static boolean isIcosidigonal( int N) { float n = ( float ) (( 18 + Math.sqrt( 160 * N + 324 )) / 40 ); // Condition to check if the number // is a icosidigonal number return (n - ( int )n) == 0 ; } // Driver Code public static void main(String[] args) { // Given number int N = 22 ; // Function call if (isIcosidigonal(N)) { System.out.print( "Yes" ); } else { System.out.print( "No" ); } } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach import numpy as np # Function to check if the number N # is a icosidigonal number def isIcosidigonal(N): n = ( 18 + np.sqrt( 160 * N + 324 )) / 40 # Condition to check if N # is a icosidigonal number return (n - int (n)) = = 0 # Driver Code N = 22 # Function call if (isIcosidigonal(N)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by PratikBasu |
C#
// C# program for the above approach using System; class GFG{ // Function to check if the number N // is a icosidigonal number static bool isIcosidigonal( int N) { float n = ( float ) ((18 + Math.Sqrt(160 * N + 324)) / 40); // Condition to check if the number // is a icosidigonal number return (n - ( int )n) == 0; } // Driver Code public static void Main( string [] args) { // Given number int N = 22; // Function call if (isIcosidigonal(N)) { Console.Write( "Yes" ); } else { Console.Write( "No" ); } } } // This code is contributed by rutvik_56 |
Javascript
<script> // javascript program for the above approach // Function to check if the number N // is a Icosidigonal number function isIcosidigonal( N) { let n = (18 + Math.sqrt(160 * N + 324)) / 40; // Condition to check if the // number is a Icosidigonal number return (n - parseInt(n)) == 0; } // Driver Code // Given Number let N = 22; // Function call if (isIcosidigonal(N)) { document.write( "Yes" ); } else { document.write( "No" ); } // This code contributed by aashish1995 </script> |
Output:
Yes
Time Complexity: O(logN) because the inbuilt sqrt function has been used
Auxiliary Space: O(1)