2013
11-26

# The area

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.

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).

For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.

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

33.33
40.69

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


http://acm.hdu.edu.cn/showproblem.php?pid=1071

抛物线方程：y=ax^2+bx+c……………………………………………………（2）

=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)

#include<stdio.h>
int main()
{
double a,b,c,t,k,x1,x2,x3,y1,y2,y3,s1,s2;
int te;
scanf("%d",&te);
while(te--)
{
scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
k=(y3-y2)/(x3-x2);
t=y3-k*x3;
a=(y2-y1)/((x1-x2)*(x1-x2));
b=-x1*2*a;
c=y1-a*x1*x1-b*x1;
s1=1.0/3*a*x2*x2*x2+1.0/2*(b-k)*x2*x2+x2*(c-t);
s2=1.0/3*a*x3*x3*x3+1.0/2*(b-k)*x3*x3+x3*(c-t);
printf("%.2lf\n",s2-s1);
}
return 0;
}

1. 算法是程序的灵魂，算法分简单和复杂，如果不搞大数据类，程序员了解一下简单点的算法也是可以的，但是会算法的一定要会编程才行，程序员不一定要会算法，利于自己项目需要的可以简单了解。