hdu 1071 The area(数学－－抛物线)

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8221    Accepted Submission(s): 5772

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

 

Author

Ignatius.L

题目分析：可以可以根据抛物线的性质得，x1 = -b/(2*a) ，然后将其余两个点代入，就能求得抛物线的公式和直线公式，进而积分出面积

#include <algorithm>#include <cstdio>#include <cstring>#include <iostream>using namespace std;int t;double x1 , yy1 , x2 ,y2 , x3 , y3;double a,b,c,k,d;int main ( ){    scanf ( "%d" , &t );    while ( t-- )    {        scanf ( "%lf%lf%lf%lf%lf%lf" , &x1 , &yy1 , &x2 , &y2 , &x3 , &y3 );        a = (yy1-y3)/(x1*x1-x3*x3 + 2*x1*x3 - 2*x1*x1 );        b = -2*x1*a;        k = (y2-y3)/(x2-x3);        c = yy1 - a*x1*x1 - b*x1;        d = y2 - k*x2;        double A = a/3 , B = (b-k)/2 , C = c - d;        printf ( "%.2lf\n" , A*(x3*x3*x3-x2*x2*x2) + B*(x3*x3-x2*x2) + C*(x3-x2) );    }}