hdu 1071 The area

     

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

25.000000 5.0000000.000000 0.00000010.000000 0.00000010.000000 10.0000001.000000 1.00000014.000000 8.222222

Sample Output

33.3340.69

 数学积分问题，求二次方程和直线相交的面积。求出二次方程a,b,c,以及直线的系数k，h；

 y1=a*x1^2+b*x1+c;

 y2=a*x2^2+b*x2+c;

 y2-y1=a(x2^2-x1^2)+b*(x2-x1) (1)

 x1=-2a*b;

 b=-2*a*x1;

 将b代入1式，可求出a=(y2-y1)/(x2-x1)^2;

 依次求出b,c;

 k=(y3-y2)/(x3-x2)；

 有k可求出h

 之后运用定积分求面积

代码：

#include<stdio.h>#include<string.h>#include<math.h>int main(){double x1,x2,x3,y1,y2,y3;double a,b,c,k,l,s;int t;scanf("%d",&t);while(t--){      scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);      a=(y2-y1)/((x2-x1)*(x2-x1));      b=-2*x1*a;      c=y1-a*x1*x1-b*x1;      k=(y3-y2)/(x3-x2);      l=y3-k*x3;      s=a/3*(x3*x3*x3-x2*x2*x2)+(b-k)/2*(x3*x3-x2*x2)+(c-l)*(x3-x2);      printf("%.2lf\n",s);}return 0; }