//题目地址
//http://acm.hdu.edu.cn/showproblem.php?pid=1071
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4
5 int main()
6 {
7 int T;
8 double x1,x2,x3,y1,y2,y3,a,b,c,k,t,x,s1,s2;
9 cin>>T;
10 while(T--)
11 {
12 cin>>x1>>y1>>x2>>y2>>x3>>y3;
13 a=(y2-y1)/((x1-x2)*(x1-x2));
14 b=-2*a*x1;
15 c=y1+a*x1*x1;
16 k=(y3-y2)/(x3-x2);
17 t=y3-(k*x3);
18 s1=a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3;
19 s2=a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2;
20 printf("%.2lf
",s1-s2);
21 }
22 return 0;
23 }
24 /*
25 设直线方程:y=kx+t…………………………………………………………(1)
26 抛物线方程:y=ax^2+bx+c……………………………………………………(2)
27 已知抛物线顶点p1(x1,y1),两线交点p2(x2,y2)和p3(x3,y3)
28 斜率k=(y3-y2)/(x3-x2)……………………………………………………(3)
29 把p3点代入(1)式结合(3)式可得:t=y3-(k*x3)
30 又因为p1是抛物线的顶点,可得关系:x1=-b/2a即b=-2a*x1………………(4)
31 把p1点代入(2)式结合(4)式可得:a*x1*x1-2a*x1*x1+c=y1化简得c=y1+a*x1*x1……(5)
32 把p2点代入(2)式结合(4)式和(5)式可得:a=(y2-y1)/((x1-x2)*(x1-x2))
33 于是通过3点求出了k,t,a,b,c即两个方程式已求出
34 题目时求面积s
35 通过积分可知:s=f(x2->x3)(积分符号)(ax^2+bx+c-(kx+t))
36 =f(x2->x3)(积分符号)(ax^2+(b-k)x+c-t)
37 =[a/3*x^3+(b-k)/2*x^2+(c-t)x](x2->x3)
38 =a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3-(a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2)
39 化简得:
40 面积公式:s=-(y2-y1)/((x2-x1)*(x2-x1))*((x3-x2)*(x3-x2)*(x3-x2))/6;
41 */