The area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7807 Accepted Submission(s): 5483

Problem Description
Ignatius bought a land last week, but he didn’t know the area of the land because the land is enclosed by a parabola and a straight line. The picture below shows the area. Now given all the intersectant points shows in the picture, can you tell Ignatius the area of the land?

Note: The point P1 in the picture is the vertex of the parabola.

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).

Output
For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.

Sample Input
2
5.000000 5.000000
0.000000 0.000000
10.000000 0.000000
10.000000 10.000000
1.000000 1.000000
14.000000 8.222222

Sample Output
33.33
40.69

Hint

For float may be not accurate enough, please use double instead of float.

数学问题，直接上代码：

#include
#include
int main()
{
    int n;
 	double x1,y1,x2,y2,x3,y3;
  	double a,b,c,k,t;
	double area1,area2;
    while(~scanf("%d",&n))
    while(n--)
    {
        scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
        a=(y2-y1)/pow((x2-x1),2);
        b=-2*a*x1;
        c=y1-a*x1*x1-b*x1;
        k=(y3-y2)/(x3-x2);
        t=y3-k*x3;
        area1=1.0/3*a*pow(x3,3)+1.0/2*(b-k)*pow(x3,2)+(c-t)*x3;
        area2=1.0/3*a*pow(x2,3)+1.0/2*(b-k)*pow(x2,2)+(c-t)*x2;
        printf("%.2lfn",area1-area2);
    }
    return 0;
}