hdoj 1071 The area



http://acm.hdu.edu.cn/showproblem.php?pid=1071

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7751    Accepted Submission(s): 5442

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


   

    Hint
   
   

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

    数学题，设抛物线的方程为y=a*x*x+b*x+c,直线的方程为y=k*x+d。此题不考虑抛物线开口向左和向右的情况，下面就是用已知三点求抛物线的a,b,c，已知两点求直线的k,d。代码如下。
   
   

    
#include<stdio.h>
#include<string.h>
#define NUM 3
struct Point
{
    double x;
    double y;
}num[NUM+5];
int main()
{
    int T,i;
    double sum,sum_1,sum_2,a,b,c,d,k;
    scanf("%d",&T);
    while(T--)
    {
        for(i=0;i<NUM;i++)
        scanf("%lf %lf",&num[i].x,&num[i].y);
        a=(num[1].y-num[0].y)/(num[1].x*num[1].x-num[0].x*num[0].x+2*num[0].x*num[0].x-2*num[0].x*num[1].x);
        b=-2*a*num[0].x;
        c=num[0].y-a*num[0].x*num[0].x-b*num[0].x;
        //printf("%lf %lf %lf\n",a,b,c);
        k=(num[2].y-num[1].y)/(num[2].x-num[1].x);
        d=num[1].y-(num[2].y-num[1].y)*num[1].x/(num[2].x-num[1].x);
        //printf("%lf %lf\n",k,d);
        sum_2=(1.0/3.0)*a*(num[2].x)*(num[2].x)*(num[2].x)+(1.0/2.0)*(b-k)*(num[2].x)*(num[2].x)+(c-d)*num[2].x;
        sum_1=(1.0/3.0)*a*(num[1].x)*(num[1].x)*(num[1].x)+(1.0/2.0)*(b-k)*(num[1].x)*(num[1].x)+(c-d)*num[1].x;
        sum=sum_2-sum_1;

        printf("%.2lf\n",sum);
    }
    return 0;
}