Where do I Turn?（叉积）

题目：

Trouble came from the overseas lands: a three-headed dragon Gorynych arrived. The dragon settled at point C and began to terrorize the residents of the surrounding villages.

A brave hero decided to put an end to the dragon. He moved from point A to fight with Gorynych. The hero rode from point A along a straight road and met point B on his way. The hero knows that in this land for every pair of roads it is true that they are either parallel to each other, or lie on a straight line, or are perpendicular to each other. He also knows well that points B and C are connected by a road. So the hero must either turn 90 degrees to the left or continue riding straight ahead or turn 90 degrees to the right. But he forgot where the point C is located.

Fortunately, a Brave Falcon flew right by. It can see all three points from the sky. The hero asked him what way to go to get to the dragon's lair.

If you have not got it, you are the falcon. Help the hero and tell him how to get him to point C: turn left, go straight or turn right.

At this moment the hero is believed to stand at point B, turning his back to point A.

Input

The first input line contains two space-separated integers xa, ya (|xa|, |ya| ≤ 109) — the coordinates of point A. The second line contains the coordinates of point B in the same form, the third line contains the coordinates of point C.

It is guaranteed that all points are pairwise different. It is also guaranteed that either point B lies on segment AC, or angle ABC is right.

Output

Print a single line. If a hero must turn left, print "LEFT" (without the quotes); If he must go straight ahead, print "TOWARDS" (without the quotes); if he should turn right, print "RIGHT" (without the quotes).

input

0 0
0 1
1 1

output

RIGHT

input

-1 -1
-3 -3
-4 -4

output

TOWARDS

input

-4 -6
-3 -7
-2 -6

 output

   LEFT

题意：有三个点A、B、C，给出他们坐标，从A走到B再到C，让你判断位于B点时，想要到C点是左转、右转还是直走。

用向量的叉积来做，还是挺简单的。

#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <limits>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Scc(c) scanf("%c",&c)
#define Scs(s) scanf("%s",s)
#define Sci(x) scanf("%d",&x)
#define Sci2(x, y) scanf("%d%d",&x,&y)
#define Sci3(x, y, z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%I64d",&x)
#define Scl2(x, y) scanf("%I64d%I64d",&x,&y)
#define Scl3(x, y, z) scanf("%I64d%I64d%I64d",&x,&y,&z)
#define Pri(x) printf("%d
",x)
#define Prl(x) printf("%I64d
",x)
#define Prc(c) printf("%c
",c)
#define Prs(s) printf("%s
",s)
#define For(i,x,y) for(int i=x;i<y;i++)
#define For_(i,x,y) for(int i=x;i<=y;i++)
#define FFor(i,x,y) for(int i=x;i>y;i--)
#define FFor_(i,x,y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define LL long long
#define ULL unsigned long long
#define MAXSIZE 1005
#define INF 0x3f3f3f3f
const int mod=1e9+7;
const double PI = acos(-1.0);

using namespace std;

int main()
{
   LL x1,y1,x2,x3,y2,y3;
   Scl2(x1,y1);
   Scl2(x2,y2);
   Scl2(x3,y3);
   x1=x2-x1;
   y1=y2-y1;
   x2=x3-x2;
   y2=y3-y2;
   if(x1*y2-x2*y1>0)
    Prs("LEFT");
   else
    if(x1*y2-x2*y1==0)
    Prs("TOWARDS");
   else
    Prs("RIGHT");
    return 0;
}