Problem:
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A Point in a Triangle

There is a triangle formed by three points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ on a plain.

Write a program which prints "YES" if a point $P$ $(x_p, y_p)$ is in the triangle and "NO" if not.

Input

Input consists of several datasets. Each dataset consists of:

$x_1$ $y_1$ $x_2$ $y_2$ $x_3$ $y_3$ $x_p$ $y_p$

All the input are real numbers. Input ends with EOF. The number of datasets is less than or equal to 100.

Constraints

You can assume that:

$ -100 \leq x_1, y_1, x_2, y_2, x_3, y_3, x_p, y_p \leq 100$

1.0 $\leq$ Length of each side of a tringle

0.001 $\leq$ Distance between $P$ and each side of a triangle

Output

For each dataset, print "YES" or "NO" in a line.

Sample Input

0.0 0.0 2.0 0.0 2.0 2.0 1.5 0.5
0.0 0.0 1.0 4.0 5.0 3.0 -1.0 3.0

Output for the Sample Input

YES
NO

Solution in C# (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace AOJ
{
	class Program
	{
		static void Main(string[] args)
		{
			var str = "";
			while ((str = Console.ReadLine()) != null)
			{
				var xy = str.Split(' ').Select(d => double.Parse(d)).ToArray();
				double x1 = xy[0], y1 = xy[1], x2 = xy[2], y2 = xy[3], x3 = xy[4], y3 = xy[5], xp = xy[6], yp = xy[7];

				double x = Math.Sqrt(Math.Pow(x1 - xp, 2) + Math.Pow(y1 - yp, 2));
				double y = Math.Sqrt(Math.Pow(x2 - xp, 2) + Math.Pow(y2 - yp, 2));
				double z = Math.Sqrt(Math.Pow(x3 - xp, 2) + Math.Pow(y3 - yp, 2));

				double a = Math.Sqrt(Math.Pow(x1 - x2, 2) + Math.Pow(y1 - y2, 2));
				double b = Math.Sqrt(Math.Pow(x3 - x1, 2) + Math.Pow(y3 - y1, 2));
				double c = Math.Sqrt(Math.Pow(x2 - x3, 2) + Math.Pow(y2 - y3, 2));

				double S1 = area(a, x, y);
				double S2 = area(b, x, z);
				double S3 = area(c, y, z);
				double S = area(a, b, c);

				Console.WriteLine(((S > S1 + S2) && (S > S2 + S3) && (S > S1 + S3)) ? "YES" : "NO");
			}
		}

		static double area(double a, double b, double c)
		{
			double t = (a + b + c) / 2;
			return Math.Sqrt(t * (t - a) * (t - b) * (t - c));
		}
	}
}