The area

2024-08-24 22:38

文章标签 area

本文主要是介绍The area，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

万恶的高数微分不会啊！！眼馋队友用微分刷刷的做出来的同时，只能用悲催的二分法终于二出来了。。。

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

    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

#include <stdio.h>
#include <math.h>
const double e=1e-3;
double a,h,d;
double fx(double x)
{return a*(x-h)*(x-h)+d;
}
double area(double x1,double y1,double x2,double y2,double x3,double y3)
{double d1,d2,d3,p,s,x4,y4;d1=sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));if (d1<e) return 0;d2=sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));d3=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));p=(d1+d2+d3)/2;s=sqrt(p*(p-d1)*(p-d2)*(p-d3));x4=(x1+x2)/2; y4=fx(x4);s+=area(x4,y4,x1,y1,x2,y2);x4=(x1+x3)/2; y4=fx(x4);s+=area(x4,y4,x1,y1,x3,y3);return s;
}
int main()
{double x1,y1,x2,y2,x3,y3,s;int n,i;scanf("%d",&n);for (i=0;i<n;i++){scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);h=x1; d=y1;a=(y2-d)/(x2-h)/(x2-h);s=area(x1,y1,x2,y2,x3,y3);printf("%.2lf\n",s);}
}