Nyist 68 three-point order

Three-point sequential time limit: 1000 MS | memory limit: 65535 kb difficulty: 3

Description

The coordinates of the three non-collocated vertices A, B, and C must be a triangle. Now let you determine the coordinates of A, B, is C provided clockwise or counterclockwise?

For example:

Figure 1: clockwise

Figure 2: counter-clockwise

 

<Figure 1> <Figure 2>

 

Input

Each row is a set of test data, with six integers X1, Y1, X2, Y2, X3, and Y3 representing the horizontal and vertical coordinates of A, B, and C. (The coordinates are between 0 and 10000)
Input 0 0 0 0 0 0 indicates input is complete
Test data cannot exceed 10000 groups

Output

If the three points are clockwise, output 1, and output 0

Sample Input

0 0 1 1 1 30 1 1 0 0 00 0 0 0 0 0

Sample output

01

# Include <stdio. h>
Int main ()
{
Int X1, Y1, X2, Y2, X3, Y3;
Double S;
While (scanf ("% d", & X1, & Y1, & X2, & Y2, & X3, & Y3) & X1 + X2 + X3 + Y1 + y2 + Y3)
{
S = x1 * y2 + X3 * Y1 + x2 * y3-x3 * y2-x2 * y1-x1 * Y3; // The Cross Product of AB and BC vectors ......
If (S <0) printf ("1 \ n ");
Else printf ("0 \ n ");

}
Return 0;
}

 

 

 

Length of the cross product | A × B | it can be interpreted as the area of the parallelogram with a and B as the adjacent edges.

 

 

 

 

 

 

 

 

 

# Include <stdio. h>
# Include <math. h>
Main ()
{
Int x1, x2, X3, Y1, Y2, Y3;
Double S;
While (scanf ("% d", & X1, & Y1, & X2, & Y2, & X3, & Y3 ))
{
If (x1 = 0 & Y1 = 0 & X2 = 0 & y2 = 0 & X3 = 0 & Y3 = 0) break;

S = (x2-x1) * (y3-y1)-(x3-x1) * (y2-y1)/2.0;
If (S> = 0)
Printf ("0 \ n ");
Else
Printf ("1 \ n ");
}
}

 

 

 

 

 

 

# Include <stdio. h>
# Include <math. h>
Main ()
{
Int x1, x2, X3, Y1, Y2, Y3;
Double S;
While (scanf ("% d", & X1, & Y1, & X2, & Y2, & X3, & Y3 ), (X1 | Y1 | X2 | Y2 | X3 | Y3 ))
{

S = (x2-x1) * (y3-y1)-(x3-x1) * (y2-y1)/2.0;
If (S> = 0)
Printf ("0 \ n ");
Else
Printf ("1 \ n ");
}
}

 

 

# Include <iostream>
Using namespace STD;
Int main ()
{
While (1)
{
Int X1, Y1, X2, Y2, X3, Y3;
Cin> x1> Y1> X2> Y2> X3> Y3;

If (x1 = 0 & Y1 = 0 & X2 = 0 & y2 = 0 & X3 = 0 & Y3 = 0) break;

Int AX = x2-x1, ay = y2-y1, BX = x2-x3, by = y2-y3;
If (ax * by-ay * BX <0)
Cout <0 <Endl;

Else
Cout <1 <Endl;

}
}