一个积分问题
来源:互联网 发布:asp.net 4.5 高级编程 编辑:程序博客网 时间:2024/04/28 07:32
The area
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9036 Accepted Submission(s): 6361
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.
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).
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
33.3340.69HintFor float may be not accurate enough, please use double instead of float.#include <iostream>#include <iomanip>using namespace std;int main(){ int T; double x1,x2,x3,y1,y2,y3,a,b,c,k,t,x,s1,s2; cin>>T; while(T--) { cin>>x1>>y1>>x2>>y2>>x3>>y3; a=(y2-y1)/((x1-x2)*(x1-x2)); b=-2*a*x1; c=y1+a*x1*x1; k=(y3-y2)/(x3-x2); t=y3-(k*x3); s1=a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3; s2=a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2; cout<<setiosflags(ios::fixed)<<setprecision(2)<<s1-s2<<endl; } return 0;}注意:设直线方程:y=kx+t…………………………………………………………(1) 抛物线方程:y=ax^2+bx+c……………………………………………………(2)已知抛物线顶点p1(x1,y1),两线交点p2(x2,y2)和p3(x3,y3)斜率k=(y3-y2)/(x3-x2)……………………………………………………(3)把p3点代入(1)式结合(3)式可得:t=y3-(k*x3)又因为p1是抛物线的顶点,可得关系:x1=-b/2a即b=-2a*x1………………(4)把p1点代入(2)式结合(4)式可得:a*x1*x1-2a*x1*x1+c=y1化简得c=y1+a*x1*x1……(5)把p2点代入(2)式结合(4)式和(5)式可得:a=(y2-y1)/((x1-x2)*(x1-x2))于是通过3点求出了k,t,a,b,c即两个方程式已求出题目时求面积s通过积分可知:s=f(x2->x3)(积分符号)(ax^2+bx+c-(kx+t)) =f(x2->x3)(积分符号)(ax^2+(b-k)x+c-t) =[a/3*x^3+(b-k)/2*x^2+(c-t)x](x2->x3) =a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3-(a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2)化简得:面积公式:s=-(y2-y1)/((x2-x1)*(x2-x1))*((x3-x2)*(x3-x2)*(x3-x2))/6;
1 0
- 一个积分问题
- 积分问题
- 随机过程课中的一个复积分问题
- 一个积分的缘分~~
- 我想要一个积分
- 求助 积分问题
- csdn积分问题
- 积分办法的问题
- 关于CSDN积分问题
- 关于网站积分问题
- 关于CSDN积分问题
- 定积分的问题
- 蒙特卡洛积分的一个例子
- 多参数的积分问题。。。
- 从一个椭圆积分看matlab求定积分
- 积分
- 积分
- 积分
- 面向对象设计原则---里氏代换原则
- iOS:iOS9关于HTTP和HTTPS的问题
- 使用Java调用控制台实现任意视频文件转换Gif图片
- 全面解析Linux 内核 3.10.x - Pid hash 链表
- 词法分析的各类用途2
- 一个积分问题
- 数字图像处理8--Canny算子中涉及到的几个问题
- 生成在嵌入式设备上运行的程序需要进行交叉编译
- linux实现系统调用打印进程信息
- uva11100
- 数据文件误删导致数据库用户不能登陆
- 训练深度模型的优化问题(十二)
- poj 2362 Square
- stm32f10x.h 是库3.0以后的; stm32f10x_lib.h 这个是库2.0的吧 哎,打开看看不就知道了啊!