hdu1071-The area 积分求解与x轴形成的面积

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10810    Accepted Submission(s): 7620

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

25.000000 5.0000000.000000 0.00000010.000000 0.00000010.000000 10.0000001.000000 1.00000014.000000 8.222222

 

Sample Output

33.3340.69
题意：求阴影部分的面积
利用高中的积分知识可得：
抛物线得到的面积减去直线得到的面积，就是阴影部分的面积
求一般方程如下
求抛物线方程：
已知三点求解抛物线方程：
x1^2a+x1b+c=y1
x2^2a+x2b+c=y2
x3^2a+x3b+c=y3
D=x1^2*x2-x1^2*x3-x2^2*x1+x2^2*x3+x3^2*x1-x3^2*x2
D1=y1*x2-y1*x3-y2*x1+y2*x3+y3*x1-y3*x2
D2=x1^2*y2-x1^2*y3-x2^2*y1+x2^2*y3+x3^2*y1-x3^2*y2
D3=x1^2*y3*x2-x1^2*y2*x3-x2^2*y3*x1+x2^2*y1*x3+x3^2*y2*x1-x3^2*y1*x2
A=D1/D
B=D2/D
C=D3/D
已知直线求方程：
A=(Y1-Y2)/(X1-X2)   B=Y1-A*X1

得到参数后直接求解面积就可以了

代码如下：
#include<cstdio>#include<iostream>#include<cmath>using namespace std;double x1,x2,x3;double y,y2,y3;double a,b,c;double fun(double p,double q){double A=1.0*a*p*p*p/3+1.0*b*p*p/2+c*p;double B=1.0*a*q*q*q/3+1.0*b*q*q/2+c*q;return B-A;}void init(){double d=x1*x1*x2-x1*x1*x3-x2*x2*x1+x2*x2*x3+x3*x3*x1-x3*x3*x2;double d1=y*x2-y*x3-y2*x1+y2*x3+y3*x1-y3*x2;double d2=x1*x1*y2-x1*x1*y3-x2*x2*y+x2*x2*y3+x3*x3*y-x3*x3*y2;double d3=x1*x1*y3*x2-x1*x1*y2*x3-x2*x2*y3*x1+x2*x2*y*x3+x3*x3*y2*x1-x3*x3*y*x2;a=d1/d;b=d2/d;c=d3/d;//printf("a=%.2f\nb=%.2f\nc=%.2f\n",a,b,c);}int main(){int t;scanf("%d",&t);while(t--){scanf("%lf%lf",&x1,&y);scanf("%lf%lf",&x2,&y2);scanf("%lf%lf",&x3,&y3);init();double b1=(y2-y3)/(x2-x3);double c1=y2-b1*x2;//printf("b1=%.2f\nc1=%.2f\n",b1,c1);b=b-b1; c=c-c1;//printf("b=%.2f\nc=%.2f\n",b,c);printf("%.2f\n",fun(x2,x3)); } return 0; } 

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