The area
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
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
25.000000 5.0000000.000000 0.00000010.000000 0.00000010.000000 10.0000001.000000 1.00000014.000000 8.222222
Sample Output
33.3340.69
#include <iostream>
#include <math.h>
#include <stdio.h>
#include <string.h>
using namespace std;
struct point
{
double x,y;
} p1,p2,p3;
double a,b,c,a1,b1;
double F(double x)
{
return fabs(a*(x-b)*(x-b)+c-a1*x-b1);
}
void init()
{
b = p1.x;
c = p1.y;
a = (p2.y - c) / (p2.x - b) / (p2.x - b);
a1 = (p3.y - p2.y) / (p3.x - p2.x);
b1 = p2.y - a1 * p2.x;
//cout<<a<<" "<<b<<" "<<c<<" "<<a1<<" "<<b1<<" "<<endl;
}
//三点辛普森公式
double simpson(double width,double fa,double fb,double fc)
{
return (fb+fa+*fc)*width/;
} //自适应simpson公式递归过程
double asr(double a,double b,double eps,double A)
{
double c=(a+b)/;
double fa,fb,fc,L,R;
fa=F(a);
fb=F(b);
fc=F(c);
L=simpson(c-a,fa,fc,F((c+a)/));
R=simpson(b-c,fc,fb,F((b+c)/));
if(fabs(L+R-A)<=*eps) return L+R+(L+R-A)/;
return asr(a,c,eps/,L)+asr(c,b,eps/,R);
}
double asr1(double a,double b,double eps)
{
return asr(a,b,eps,simpson(b-a,F(a),F(b),F((b+a)/)));
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&p1.x,&p1.y);
scanf("%lf%lf",&p2.x,&p2.y);
scanf("%lf%lf",&p3.x,&p3.y);
init();
printf("%.2lf\n",asr1(p2.x,p3.x,0.0000001));
}
}
