hdu the area

The area

The area

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 35   Accepted Submission(s) : 26

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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

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

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

Author

Ignatius.L

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题意：求阴影部分的面积，由于给出了三个点的坐标，便可确定函数的方程，用积分便可容易得出面积：设抛物线的方程为：a*（x-x1）*(x-x1)+y1=y;直线的方程为：k*x+b=y;

#include <iostream>#include<iomanip>using namespace std;double k,a,x1,x2,x3,y1,y2,y3;double F(double x2,double x3){    double s,d;    s=a/3*(x3-x1)*(x3-x1)*(x3-x1)+y1*x3-k/2*(x3-x2)*(x3-x2)-y2*x3;//抛物线的原函数    d=a/3*(x2-x1)*(x2-x1)*(x2-x1)+y1*x2-y2*x2;直线的原函数    return s-d;}int main(){    int n,i;   cin>>n;   double s;   for(i=1;i<=n;i++)   {       cin>>x1>>y1>>x2>>y2>>x3>>y3;       k=(y3-y2)/(x3-x2);//求出斜率       a=(y2-y1)/((x2-x1)*(x2-x1));//求出系数a       s=F(x2, x3);//积分       cout<<setiosflags(ios::fixed)<<setprecision(2)<<s<<endl;   }    return 0;}