Check-Triangular-perfect-square-number.java

import java.util.*; 
import static java.lang.Math.sqrt;
public class HelloWorld {
public static boolean isSqrNum(int input)
{
	// If the sqrt(input) is an integer, then it is a perfect square or square number
	return (sqrt(input) % 1 == 0);
}
public static boolean isTriangle(int input)
{
  /*This method is probably the most fastest way to check if a number is Triangular or not (At least I can't think of any faster algorithm)
  It basically uses the "S = n(n + 1) / 2" formula which gives the summation of first n natural numbers. By this way we don't have to use a 
  damn slow for loop or while loop which checks each natural number to be the summation of the input.
  From the formula, S = n(n + 1) / 2, we get 2*S = n^2 + n or n^2 + n - 2*S = 0; or n = (-1 + sqrt(1 + 8 * S) )/ 2; [By dhanacharya's formula of quadratic equation]
  */
	return (input > 0 && ((sqrt(1 + 8 * input) - 1) / 2) % 1 == 0);	
}

  public static void main(String[] args) {
	float TriangularNum = 0;
  	for (float i = 0; i < 100; i++)
	{
		TriangularNum += i * 0.1;
		System.out.println(isTriangle(TriangularNum));	
	}
  }
}