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 JAVA (judge verdict: Accepted)

import java.util.Scanner;
public class Main
{
	public static void main(String[] args) 
	{
		Scanner in=new Scanner(System.in);
		while(in.hasNext())
		{
			Point2 p[]=new Point2[4];
			for(int i=0;i<4;i++)
				p[i]=new Point2(in.nextDouble(), in.nextDouble());
			boolean judge=false;
			if(ccw(p[0],p[3],p[1])==1&&ccw(p[1],p[3],p[2])==1&&ccw(p[2],p[3],p[0])==1
					||ccw(p[0],p[3],p[1])==-1&&ccw(p[1],p[3],p[2])==-1&&ccw(p[2],p[3],p[0])==-1)
				judge=true;
			System.out.println(judge ? "YES":"NO");
		}
	}
	//counterClockWise p1->p2->p3が反時計周りなら正 時計周りなら負を返す 
	static double ccw(Point2 p1,Point2 p2,Point2 p3)
	{
		Point2 a=new Point2(p2.x-p1.x, p2.y-p1.y);
		Point2 b=new Point2(p3.x-p1.x, p3.y-p1.y);
		 if(crossProduct(a, b)>0) return 1;//counter clockwise
		 if(crossProduct(a, b)<0) return -1;//clockwise
		 else return 0;
	}
	//外積
	static double crossProduct(Point2 p2,Point2 p3)
	{
		return p2.y*p3.x-p2.x*p3.y;
	}
}
class Point2
{
	double x;
	double y;
	Point2(double x,double y)
	{
		this.x=x;
		this.y=y;
	}
}