NBUT-2014暑期集训专题练习1 -> 二分法 A题

DescriptionIgnatius 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.



InputThe 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).
OutputFor 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

就是个数学题，给你三个点，然后求出抛物线和直线的参数，之后积分这里要注意下，抛物线的系数a是不可为0的，还有这题不用考虑直线斜率无穷的情况下面的是代码

#include<stdio.h>
#include<string.h>

double x1,y1,x2,y2,x3,y3;

int main()
{
	int t;
	while(~scanf("%d",&t))
	{
		while(t--)
		{
			scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
			double a,b,c,k,t;
            k=(y3-y2)/(x3-x2);
			t=(x3*y2-x2*y3)/(x3-x2);
			if(y3!=y2)
               a=(y3-y2)/((x3*x3-x2*x2)+2*x1*(x2-x3));
            else
               a=(y2-y1)/((x1-x2)*(x1-x2));
			b=-2*a*x1;
			c=y2-a*x2*x2+2*a*x1*x2;
			double area=(a*x3*x3*x3/3)-(a*x2*x2*x2/3)+((b-k)*x3*x3/2)-((b-k)*x2*x2/2)+(c-t)*(x3-x2);
			printf("%.2f\n",area);
		}
	}
	return 0;
}