Intersection （hdu-5120）

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know. 


A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below. 


Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.

Input

The first line contains only one integer T (T ≤ 10 ⁵), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10). 

Each of the following two lines contains two integers x _i, y _i (0 ≤ x _i, y _i ≤ 20) indicating the coordinates of the center of each ring.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places. 

Sample Input

2
2 3
0 0
0 0
2 3
0 0
5 0

Sample Output

Case #1: 15.707963
Case #2: 2.250778
给出两个内径和外径都相同的圆环，求出他们相交的面积
输入：r代表内径，R代表外径   (x1,y1)（x2,y2）两个圆环的圆心
输出：输出相交的面积



解题思路
这个有了两圆相交的面积的模板，所思路一下就变得非常简单，
就是两个大圆相交的面积（tmp）减去大圆和小圆相交的面积的二倍（tmp1）加上两个小圆相交的面积（tmp2）；
ans=tmp-tmp1*2+tmp2;



#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
double PI=acos(-1.0);
double eps=1e-8;
struct point
{
    double x,y;
}a,b,c,d;
double dist(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double area(point c1,double r1,point c2,double r2)
{
    double d=dist(c1,c2);
    if(r1+r2<d+eps) return 0;
    if(d<fabs(r1-r2)+eps)
    {
        double r=min(r1,r2);
        return PI*r*r;
    }
    double x=(d*d+r1*r1-r2*r2)/(2*d);
    double t1=acos(x/r1);
    double t2=acos((d-x)/r2);
    return r1*r1*t1+r2*r2*t2-d*r1*sin(t1);
}
int main()
{
    int T;
    scanf("%d",&T);
    int cas=0;
    while(T--)
    {
        cas++;
        int r,R;
        scanf("%d%d",&r,&R);
        scanf("%lf%lf",&a.x,&a.y);
        scanf("%lf%lf",&b.x,&b.y);
        r=(double)r*1.0;
        R=(double)R*1.0;
        double ans=area(a,R,b,R);
        //double tmp1=area(a,R,b,r);
        double tmp11=area(a,r,b,R);
        double tmp2=area(a,r,b,r);
        //printf("%.8lf %.8lf %.8lf %.8lf\n",ans,tmp1,tmp11,tmp2);
        printf("Case #%d: %.6lf\n",cas,ans+tmp2-tmp11-tmp11);
    }
    return 0;
}