求两圆相交面积

最新推荐文章于 2020-11-01 22:05:28 发布

转载最新推荐文章于 2020-11-01 22:05:28 发布 · 125 阅读

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原文链接：http://www.cnblogs.com/zcy19990813/p/9702835.html

本文提供了一种计算平面上两个圆交叠区域面积的方法。通过输入两个圆的坐标及半径，利用数学公式计算并输出交叠部分的面积，结果保留三位小数。

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

π=cos（-1.0）弧度=角度*π/180

#include <stdio.h>
#include <math.h>
#define P acos(-1.0)
int main()
{
    double x1,y1,R,r,x2,y2,z1,z2,s1,s2,ans,m1,m2,d;
    while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&R,&x2,&y2,&r)!=EOF)
    {
        d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
        if(d>=R+r||R==0||r==0)//相离
        {
            printf("0.000\n");
            continue;
        }
        else
        if(d<=fabs(r-R)&&d>=0)//相切
        {
            if(r<R)
                ans=P*r*r;
            else
                ans=P*R*R;
            printf("%.3lf\n",ans);
        }
        else
        {
            z1=acos((R*R+d*d-r*r)/(2*R*d));//余弦定理求第一个扇形一半弧度
            z2=acos((r*r+d*d-R*R)/(2*r*d));
            s1=z1*R*R;//扇形面积
            s2=z2*r*r;
            m1=R*R*sin(z1)*cos(z1);//求三角形面积
            m2=r*r*sin(z2)*cos(z2);
            ans=s1+s2-m1-m2;
            printf("%.3lf\n",ans);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/zcy19990813/p/9702835.html