The area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1850 Accepted Submission(s): 1449 

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.
一道万恶的定积分题目    给你三个点的坐标  求面积
求出直线方程 然后根据公式套就行了
#include<stdio.h>#include<iostream>#include<cmath>using namespace std;int main(){    int n,i;    double x0,y0,x1,y1,x2,y2,k,b,a,c,h,s;    scanf("%d",&n);    while(n--)    {        cin>>x0>>y0>>x1>>y1>>x2>>y2;            k=(y2-y1)/(x2-x1);    //  (x1,y1) (x2,y2)  求直线的斜率        b=y1-k*x1;   //  一般式 y=kx+b    求b        h=x0;                c=y0;        a=(y1-c)/((x1-h)*(x1-h));  // 求一元二次方程的a        s=(a*x2*x2*x2/3-(2*a*h+k)*x2*x2/2+(a*h*h+c-b)*x2)-(a*x1*x1*x1/3-(2*a*h+k)*x1*x1/2+(a*h*h+c-b)*x1);//对方程从x1 积分到 x2         printf("%.2f\n",s);    }    return 0;}