TOJ 1210 The area(微积分)

Posted 2020-12-29 taozi1115402474

描述

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 a random point on the parabola, and P2, P3 are the intersectant points. Any two points of P1, P2 and P3 are not coincidence each other.

输入

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).

输出

For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.

样例输入

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

样例输出

33.33
40.69

提示

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

题意

题意很简单，给你抛物线上三个点，p2和p3是线上的点，保证三点不共线，求上图红线画出的面积

题解

抛物线设y=ax^2+bx+c，把三个点代入公式，化简可得a，b，c

直线设y=a₁x+b₁，a₁为斜率，易得b₁

然后求积分∫(x₂,x₃)  ax²+bx+c-a₁x-b₁

可得[1/3*ax³+1/2*bx²+cx-1/2*a₁x²-b₁x](x2,x3)  ==  [1/3*ax₃³+1/2*bx₃²+cx₃-1/2*a₁x₃²-b₁x₃]-[1/3*ax₂³+1/2*bx₂²+cx₂-1/2*a₁x₂²-b₁x₂] 即为答案

代码

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int T;
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         double x1,y1,x2,y2,x3,y3,a1,b1,a,b,c,ans;
11         scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
12         a1=(y3-y2)/(x3-x2);
13         b1=y3-a1*x3;
14         a=((y3-y1)-a1*(x3-x1))/((x3*x3-x1*x1)-(x3+x2)*(x3-x1));
15         b=((y3-y2)-a*(x3*x3-x2*x2))/(x3-x2);
16         c=y3-a*x3*x3-b*x3;
17         ans=((1.0/3)*a*x3*x3*x3+0.5*(b-a1)*x3*x3+(c-b1)*x3)-((1.0/3)*a*x2*x2*x2+0.5*(b-a1)*x2*x2+(c-b1)*x2);
18         printf("%.2f
",ans);
19     }
20     return 0;
21 }