Tell me the area(计算几何--求两圆相交面积)

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

acos（）函数调用时括号内是弧度

弧度=角度*π/180

 

#include <stdio.h>
#include<math.h>
#define PI acos(-1.0)   //PI就是π
int main()
{
    double x1,y1,x2,y2,r1,r2,s,ans,z1,z2,s1,s2,m1,m2;
    while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&r1,&x2,&y2,&r2)!=EOF)
    {
        s=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));   //求两点之间的距离

        if(s>=r1+r2||r1==0||r2==0)      //相离的情况
            printf("0.000\n");
        else if(s<=fabs(r1-r2)&&s>=0)    //相切的情况 面积是小圆
        {
            if(r1<r2)
                ans=PI*r1*r1;
            else
                ans=PI*r2*r2;
            printf("%.3lf\n",ans);
        }
        else                             //相交的情况
        {
            z1=acos((r1*r1+s*s-r2*r2)/(2*r1*s));      //这里是余弦定理（对r1 r2 s这个三角形）z1是扇形圆心角一半的弧度
            z2=acos((r2*r2+s*s-r1*r1)/(2*r2*s));    //同上
            s1=z1*r1*r1;      //这里是求扇形的面积 面积公式s=n（圆心角）*π*r*r/360 
            s2=z2*r2*r2;     //同上
            m1=r1*r1*sin(z1)*cos(z1);   //这里是求三角形面积 s=1/2*a*b*sin c

            m2=r2*r2*sin(z2)*cos(z2);   //同上
            ans=s1+s2-m1-m2;       
            printf("%.3lf\n",ans);
        }
    }

    return 0;
}