HDOJ 1071 The area

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

题目：

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7341    Accepted Submission(s): 5138

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
  
  

解题思路：

求抛物线y = ax^2 + bx + c与一条直线y=f(x)围成的区域的面积S。S = SA - SB，SA表示抛物线与直线x = x2, x = x3 , y = 0围成的区域的面积，SB表示直线x = x2, x = x3 , y = 0 与 y = f(x)围成的梯形的面积。SB = 1/2 * (y2+y3) * (x3 - x2), SA = ∫ (ax^2 + bx + c) dx (x2->x3)从x2到x3积分。我们计算一下可以得到，S = 1/6 * a * (x2 - x3)^3, 剩下来的就是要求出a了。由抛物线的性质x1 = -b / (2a), 所以b = -2ax1。建立方程组 y1 = ax1^2 - 2ax1^2 + c, y3 = ax3^2 - a2x1*x3 + c；两式相减并化简得a = (y3 - y1) / (x1 - x3)^2.最后，我们就可以得到公式S = 1/6 *(y3 - y1) * (x2 - x3) / (x1 - x3) ^ 3。ps：纯数学题，高数忘得差不多了，算公式算了好久啊。

代码：

#include <cstdio>

int main()
{
    int t;
    while(~scanf("%d", &t))
    {
        while(t--)
        {
            double x1, y1, x2, y2, x3, y3;
            scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3);
            printf("%.2f\n", (y3-y1)*(x2-x3)*(x2-x3)*(x2-x3)/6.0/(x1-x3)/(x1-x3));
        }
    }
    return 0;
}