The area
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7611 Accepted Submission(s): 5332
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
题意：
给3个点p1,p2,p3的坐标，求图中斜线覆盖部分。
思路：
设二次函数为y=a*x^2+b*x+c（1）。
由于p1为二次函数顶点，所以b=-2*a*x1 （2）代入二次函数式子 1 里得y=a*x^2-2*a*x1*x+c（3），把p1,p2坐标代入 3 中，算出a，然后由 2 算出b ,然后a、b、p3坐标代入 1 中算出c。这样下来就算出a、b、c了，那么ans=微积分求面积-梯形面积。
代码：
1 #include <iostream>
2 #include <string>
3 #include <map>
4 #include <stdio.h>
5 #include <math.h>
6 using namespace std;
7 #define pi acos(-1)
8
9 double a, b, c;
10 double S(double x){
11 return a*x*x*x/3+b*x*x/2+c*x;
12 }
13
14 main()
15 {
16 int t;
17 double x1, y1, x2, y2, x3, y3;
18 double s;
19 cin>>t;
20 while(t--){
21 scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3);
22 a=(y2-y1)/(x2*x2-2*x1*x2+x1*x1);
23 b=-2*a*x1;
24 c=y3-a*x3*x3-b*x3;
25 s=S(x3)-S(x2)-(y2+y3)*(x3-x2)/2;
26 printf("%.2lf\n",s);
27 }
28 }
HDU 1071 微积分