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. 

InputThe 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). 
OutputFor 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         

Hint

数学 几何

定积分真™美妙，美得我都看哭了。

二次函数顶点式f(x)=a(x-h)^2+c  一次函数y=kx+b

然后求一波定积分，面积就出来了。

刚开始我用f(x)=ax^2+bx+c求定积分，三项化简快哭了，然后换成顶点式，又算了好久。

看了题解才意识到可以算[0,x3]的定积分，减去[0,x2]的。 而我之前的是[x2,x3]，怪不得超麻烦……

花了一晚上终于搞出来了。

 1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdio> 6 using namespace std; 7 double a,h,c,k,b; 8 double x1,x2,x3,y1,y2,y3; 9 double f(double x){10     return a*x*x*x/3+c*x+x*a*h*h-a*h*x*x - ( k*x*x/2+b*x);11             //二次函数定积分             //一次函数求梯形面积 12 }13 int main(){14     int T,i,j;15     scanf("%d",&T);16     while(T--){17         scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);18         a=(y2-y1)/(x2-x1)/(x2-x1);19         c=y1;20         h=x1;21         k=(y3-y2)/(x3-x2);22         b=y2-k*x2;23         printf("%.2f\n",f(x3)-f(x2));24     }25     return 0;26 }

HDU1071 The area