数学积分—— The area（求阴影部分的面积）

最新推荐文章于 2020-05-18 12:28:09 发布

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#数学积分

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本文介绍了一种通过解析几何和积分运算求解由抛物线和直线围成的阴影区域面积的方法。通过给出的三个交点坐标，利用二次方程和直线方程，计算出面积。注意使用double类型以确保精度。

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题意描述：
就是求阴影部分的面积，一个公式解决问题（经过推算），实际上也就是积分的加减运算。
注意点：
题目上说用float，结果float过不了，结果还是double解决了问题
原文：
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.
在这里插入图片描述
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
Hint
For float may be not accurate enough, please use double instead of float.
AC代码：

#include<stdio.h>
int main()
{
	int t;
	double a,b,c,x1,x2,x3,y1,y2,y3,k,w,s;
	scanf("%d", &t);
	while(t--)
	{
		s=0;
	scanf("%lf %lf", &x1,&y1);
	scanf("%lf %lf", &x2,&y2);
	scanf("%lf %lf", &x3,&y3);
	a=((y1-y2)/(x1-x2)-(y2-y3)/(x2-x3))/((x1+x2)-(x2+x3));
	b=(y1-y2)/(x1-x2)-a*(x1+x2);
	c=y1-a*x1*x1-b*x1;
	k=(y3-y2)/(x3-x2);
	w=y3-k*x3;
	s=(1.0*1/3)*a*(x3*x3*x3-x2*x2*x2)+(1.0*1/2)*b*(x3*x3-x2*x2)+c*(x3-x2)-(1.0*1/2)*k*(x3*x3-x2*x2)-w*(x3-x2);
	printf("%.2lf\n", s);
	}
	return 0; 
 }