Given a number N, the task is to check if the number is a Triacontagonal number or not.
A Triacontagonal number is a class of figurate number. It has 30 – sided polygon called triacontagon. The N-th triacontagonal number count’s the 30 number of dots and all other dots are surrounding with a common sharing corner and make a pattern. The first few triacontagonol numbers are 1, 30, 87, 172 …
Examples:
Input: N = 30
Output: Yes
Explanation:
Second triacontagonal number is 30.Input: 32
Output: No
Approach:
1. The Kth term of the triacontagonal number is given as:
2. As we have to check whether the given number can be expressed as a triacontagonal number or not. This can be checked as follows:
=>
=> K = \frac{26 + \sqrt{224*N + 676}}{56}
3. Finally, check the value computed using this formula is an integer, which means that N is a triacontagonal number.
Below is the implementation of the above approach:
// C++ program to check whether a // number is an triacontagonal // number or not #include <bits/stdc++.h> using namespace std;
// Function to check whether a // number is an triacontagonal // number or not bool istriacontagonal( int N)
{ float n
= (26 + sqrt (224 * N + 676))
/ 56;
// Condition to check whether a
// number is an triacontagonal
// number or not
return (n - ( int )n) == 0;
} // Driver Code int main()
{ // Given number
int i = 30;
// Function call
if (istriacontagonal(i)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
} |
// Java program to check whether a // number is an triacontagonal // number or not class GFG{
// Function to check whether a // number is an triacontagonal // number or not static boolean istriacontagonal( int N)
{ float n = ( float ) (( 26 + Math.sqrt( 224 * N +
676 )) / 56 );
// Condition to check whether a
// number is an triacontagonal
// number or not
return (n - ( int )n) == 0 ;
} // Driver code public static void main(String[] args)
{ // Given number
int N = 30 ;
// Function call
if (istriacontagonal(N))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
} } // This code is contributed by shubham |
# Python3 program to check whether a # number is an triacontagonal # number or not import math;
# Function to check whether a # number is an triacontagonal # number or not def istriacontagonal(N):
n = ( 26 + math.sqrt( 224 * N + 676 )) / / 56 ;
# Condition to check whether a
# number is an triacontagonal
# number or not
return (n - int (n)) = = 0 ;
# Driver Code # Given number i = 30 ;
# Function call if (istriacontagonal(i)):
print ( "Yes" );
else :
print ( "No" );
# This code is contributed by Code_Mech |
// C# program to check whether a // number is an triacontagonal // number or not using System;
class GFG{
// Function to check whether a // number is an triacontagonal // number or not static bool istriacontagonal( int N)
{ float n = ( float )((26 + Math.Sqrt(224 * N +
676)) / 56);
// Condition to check whether a
// number is an triacontagonal
// number or not
return (n - ( int )n) == 0;
} // Driver code public static void Main(String[] args)
{ // Given number
int N = 30;
// Function call
if (istriacontagonal(N))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
} } // This code is contributed by sapnasingh4991 |
<script> // javascript program to check whether a // number is an triacontagonal // number or not // Function to check whether a // number is an triacontagonal // number or not function istriacontagonal( N)
{ let n = ((26 + Math.sqrt(224 * N +
676)) / 56);
// Condition to check whether a
// number is an triacontagonal
// number or not
return (n - parseInt(n)) == 0;
} // Driver code // Given number
let N = 30;
// Function call
if (istriacontagonal(N))
{
document.write( "Yes" );
}
else
{
document.write( "No" );
}
// This code is contributed by aashish1995
</script> |
Output
Yes
Time Complexity: O(log(n)), since sqrt() function has been used
Auxiliary Space: O(1)