杭电1071——The area！-优快云博客

本文介绍了一种计算由抛物线与直线围成区域面积的方法。通过给出抛物线顶点及直线与抛物线的两个交点坐标，利用数学公式推导出计算面积的具体步骤。首先确定抛物线与直线的方程参数，进而利用积分简化公式计算出精确到小数点后两位的面积。

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The area

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
000000 5.000000
000000 0.000000
000000 0.000000
000000 10.000000
000000 1.000000
000000 8.222222

Sample Output

33.33
40.69

Hint

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

/*********************************
          已知抛物线与直线相交两点和抛物线顶点，顶点P1(-b/(2a), (4ac-b^2)/4a)。
          抛物线方程：y=ax^2+bx+c;
          直线方程：y=kx+h;
          已知p1,p2,p3可以求出a,b,c,k,h
          y1=ax1^2+bx1+c
          y2=ax2^2+bx2+c
          y1-y2=(x1-x2)(a(x1+x2)+b)
          又x1=-b/2a,
          所以a=(y2-y1)/(x2-x1)^2
          面积用积分公式化简为：
          S=a/3*(x3^3–x2^3)+(b-k)/2*(x3^2–x2^2)+(c-h)*(x3-x2);
**********************************/
#include<stdio.h>

int main()
{
	double x1,x2,x3,y1,y2,y3;
	double a,b,c,k,h;
	double s;
	int num,i;
	scanf("%d",&num);
	for(i=0;i<num;i++)
	{
		    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;
}