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最新推荐文章于 2021-03-26 20:37:11 发布

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本文介绍了一种计算平面上两个圆公共区域面积的方法，并提供了一段C语言程序代码实现。该程序能根据输入的圆心坐标及半径，判断两圆的位置关系并计算其交叠部分的面积。

Tell me the area

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1382 Accepted Submission(s): 430

Problem Description

There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.

Input

There are many cases. In each case, there are two lines. Each line has three numbers: the coordinates (X and Y) of the centre of a circle, and the radius of the circle.

Output

For each case, you just print the common area which is rounded to three digits after the decimal point. For more details, just look at the sample.

Sample Input

  
   0 0 2
2 2 1

Sample Output

  
   0.108

Author

wangye

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

Recommend

wangye

//分三种情况考虑两个圆关系：1.两圆相离、外切或至少有一圆半径为0; 
//                          2.两圆内切、内含；
//                          3.相交。
//这里PI用acos(-1.0)，用3.1415926的话会WA.
#include <stdio.h>
#include <math.h>
#define PI acos(-1.0)
int main()
{
	double x1, y1, r1, x2, y2, r2, dis, s, angle1, angle2, s1, s2, s3, s4;
	while (~scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &r1, &x2, &y2, &r2))
	{
		dis = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
		if(dis >= r1 + r2 || r1 == 0 || r2 == 0)
			s = 0;
		else if(dis <= fabs(r1 - r2))
		{
			r1 = r1 > r2 ? r2 : r1;
			s = PI * r1 * r1;
		}
		else
		{
			angle1 = acos((r1 * r1 + dis * dis - r2 * r2) / (2 * r1 * dis));
			angle2 = acos((r2 * r2 + dis * dis - r1 * r1) / (2 * r2 * dis));
			s1 = angle1 * r1 * r1;
			s2 = r1 * r1 * sin(angle1) * cos(angle1);
			s3 = angle2 * r2 * r2;
			s4 = r2 * r2 * sin(angle2) * cos(angle2);
			s = s1 - s2 + s3 - s4;
		}
		printf("%.3lf\n", s);
	}
	return 0;
}