计算几何专题（计算两圆相交面积）

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

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;#define PI acos(-1.0)int main(){    double a1,b1,r1,a2,b2,r2,d;    double A1,A2,s1,s2,s;                while(~scanf("%lf%lf%lf%lf%lf%lf",&a1,&b1,&r1,&a2,&b2,&r2)){        d=sqrt((a2-a1)*(a2-a1)+(b2-b1)*(b2-b1)); //求圆心距        if(d>=r1+r2)   printf("0.000\n");     //两圆相离或相外切        else if(d<=fabs(r1-r2) && d>=0) {   //内切        if(r1>r2)  printf("%0.3lf\n",PI*r2*r2);            else   printf("%0.3lf\n",PI*r1*r1);        }        else        {             A1=2*acos((d*d+r1*r1-r2*r2)/(2*d*r1)); //求以圆心为顶点与两圆交点连线的角            A2=2*acos((d*d+r2*r2-r1*r1)/(2*d*r2));            s1=0.5*r1*r1*sin(A1)+0.5*r2*r2*sin(A2);            s2=A1/2*r1*r1+A2/2*r2*r2;            s=s2-s1;            printf("%0.3lf\n",s);        }    }    return 0;}