Given an integer N, the task is to check if N is a Chiliagon Number or not. If the number N is a Chiliagon Number then print “Yes” else print “No”.
Chiliagon Number is class of figurate number. It has 1000 – sided polygon called Chiliagon. The N-th Chiliagon Number counts the 1000 number of dots and all others dots are surrounding with a common sharing corner and make a pattern. The first few Chiliagon Numbers are 1, 1000, 2997, 5992, …
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Examples:
Input: N = 1000
Output: Yes
Explanation:
Second chiliagon number is 1000
Input: 35
Output: No
Approach:
- The Kth term of the Chiliagon Number is given as
- As we have to check that the given number can be expressed as a Chiliagon 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 a Chiliagon Number.
4. Else N is not a Chiliagon Number.
Below is the implementation of the above approach:
// C++ for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check that if N is // Chiliagon Number or not bool is_Chiliagon( int N)
{ float n
= (996 + sqrt (7984 * N + 992016))
/ 1996;
// Condition to check if N is a
// Chiliagon Number
return (n - ( int )n) == 0;
} // Driver Code int main()
{ // Given Number
int N = 1000;
// Function call
if (is_Chiliagon(N)) {
cout << "Yes" ;
}
else {
cout << "No" ;
}
return 0;
} |
// Java program for the above approach class GFG{
// Function to check that if N is // Chiliagon Number or not static boolean is_Chiliagon( int N)
{ float n = ( float )( 996 + Math.sqrt( 7984 * N +
992016 )) / 1996 ;
// Condition to check if N is a
// Chiliagon Number
return (n - ( int ) n) == 0 ;
} // Driver Code public static void main(String s[])
{ // Given Number
int N = 1000 ;
// Function call
if (is_Chiliagon(N))
{
System.out.print( "Yes" );
}
else
{
System.out.print( "No" );
}
} } // This code is contributed by rutvik_56 |
# Python3 for the above approach import math;
# Function to check that if N is # Chiliagon Number or not def is_Chiliagon(N):
n = ( 996 + math.sqrt( 7984 * N +
992016 )) / / 1996 ;
# Condition to check if N is a
# Chiliagon Number
return (n - int (n)) = = 0 ;
# Driver Code # Given Number N = 1000 ;
# Function call if (is_Chiliagon(N)):
print ( "Yes" );
else :
print ( "No" );
# This code is contributed by Code_Mech |
// C# program for the above approach using System;
class GFG{
// Function to check that if N is // Chiliagon Number or not static bool is_Chiliagon( int N)
{ float n = ( float )(996 + Math.Sqrt(7984 * N +
992016)) / 1996;
// Condition to check if N is a
// Chiliagon Number
return (n - ( int ) n) == 0;
} // Driver Code public static void Main()
{ // Given Number
int N = 1000;
// Function call
if (is_Chiliagon(N))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
} } // This code is contributed by Code_Mech |
<script> // Javascript for the above approach // Function to check that if N is // Chiliagon Number or not function is_Chiliagon(N)
{ let n
= (996 + Math.sqrt(7984 * N + 992016))
/ 1996;
// Condition to check if N is a
// Chiliagon Number
return (n - Math.floor(n)) == 0;
} // Driver Code // Given Number
let N = 1000;
// Function call
if (is_Chiliagon(N)) {
document.write( "Yes" );
}
else {
document.write( "No" );
}
// This code is contributed by Mayank Tyagi </script> |
Output:
Yes
Time Complexity: O(logN)
Auxiliary Space: O(1)