The area
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7066 Accepted Submission(s):
4959
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?
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).
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.
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
Hint
For float may be not accurate enough, please use double instead of float.
一道高数题。
回顾一下微积分的知道，曲面的微积分。两个方程相减，对x或者y积分，就是面积。
然后具体的公式详见代码把，
#include<stdio.h>
//#include<math.h>与y1方法重载了、
double x1,y1,x2,y2,x3,y3;
double area;
double l,h,a,k,b; //曲线的方程可用顶点式 y = a(x-h)^2+l, (h,l) 为顶点坐标）
//直线方程 y = kx+b； //图形一定，只是点的位置不定。所以对x积分
double f(double x)
{
return (a*x*x*x/)-(a*h+k/)*x*x+(a*h*h+l-b)*x;//积分,化简
}
int main()
{ int T;
scanf("%d",&T);
while(T--)
{
scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
h=x1;
l=y1;
a=(y2-y1)/((x2-h)*(x2-h));
k=(y3-y2)/(x3-x2);
b=y2-k*x2;
printf("%.2lf\n",f(x3)-f(x2));
}
return ;
}