hdu 1071 The area

The area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10676 Accepted Submission(s): 7516

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.

解题思路：很明显的数学题，根据题上给出的3个点我们可以求出直线方程和抛物线方程，然后再用上函数减下函数在x2——x3上求积分；

可以很容易求出

k = (y3-y2)/(x3-x2);

h = y2 - x2*k;

然后根据定点坐标公式顶点（-b/2a，（4ac-b^2）/4a）；将x1，y1，x2，y2带入y=ax^2+bx+c利用x1=-b/2a;可得a=(y2-y1)/(x2-x1)^2;然后可以求出b=-2ax1;c=y1-ax1^2-bx1;

最后求积分；积分公式可化简为s = ((a*x3*x3*x3)/3+((b-k)*x3*x3)/2+(c-h)*x3)-((a*x2*x2*x2)/3+((b-k)*x2*x2)/2+(c-h)*x2);

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

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