杭电1071 The area

最新推荐文章于 2020-07-28 09:31:33 发布

DeterminedDiligent

最新推荐文章于 2020-07-28 09:31:33 发布

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本文介绍了一个计算由抛物线和直线围成土地面积的方法。通过给定三个交点坐标，首先确定抛物线方程，然后利用定积分计算出精确到两位小数的土地面积。

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

 
 
 
 
 
/*
利用两个点把a求出来，要用一个抛物线的顶点，比较好算
 其他用含a的式子写出来，当 x1,x2,y1,y2,x3,y3，k,h都已知
 的时候，就可以用定积分了 



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