首页 > 代码库 > The area 积分积分
The area 积分积分
The area
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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
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
1 #include <iostream> 2 #include <math.h> 3 #include <stdio.h> 4 #include <string.h> 5 using namespace std; 6 struct point 7 { 8 double x,y; 9 } p1,p2,p3;10 double a,b,c,a1,b1;11 double F(double x)12 {13 return fabs(a*(x-b)*(x-b)+c-a1*x-b1);14 }15 void init()16 {17 b = p1.x;18 c = p1.y;19 a = (p2.y - c) / (p2.x - b) / (p2.x - b);20 a1 = (p3.y - p2.y) / (p3.x - p2.x);21 b1 = p2.y - a1 * p2.x;22 //cout<<a<<" "<<b<<" "<<c<<" "<<a1<<" "<<b1<<" "<<endl;23 }24 //三点辛普森公式25 double simpson(double width,double fa,double fb,double fc)26 {27 return (fb+fa+4*fc)*width/6;28 }29 30 //自适应simpson公式递归过程31 double asr(double a,double b,double eps,double A)32 {33 double c=(a+b)/2;34 double fa,fb,fc,L,R;35 fa=F(a);36 fb=F(b);37 fc=F(c);38 L=simpson(c-a,fa,fc,F((c+a)/2));39 R=simpson(b-c,fc,fb,F((b+c)/2));40 if(fabs(L+R-A)<=15*eps) return L+R+(L+R-A)/15;41 return asr(a,c,eps/2,L)+asr(c,b,eps/2,R);42 }43 double asr1(double a,double b,double eps)44 {45 return asr(a,b,eps,simpson(b-a,F(a),F(b),F((b+a)/2)));46 }47 int main()48 {49 int t;50 scanf("%d",&t);51 while(t--)52 {53 scanf("%lf%lf",&p1.x,&p1.y);54 scanf("%lf%lf",&p2.x,&p2.y);55 scanf("%lf%lf",&p3.x,&p3.y);56 init();57 printf("%.2lf\n",asr1(p2.x,p3.x,0.0000001));58 }59 }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。