Problem Description
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 ≤ 105), 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 xi, yi (0 ≤ xi, yi ≤ 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
 
两个圆环相交,然而我只有圆相交的板子
首先拿其中一个大圆A与另一个大圆B和小圆b算交面积,两者相减,求的是A与圆环Bb相交的面积 area1
然后拿小圆a另一个大圆B和小圆b算交面积,两者相减,求的是a与圆环Bb相交的面积,area2
我们输出area1-area2就行了
 #include <bits/stdc++.h>

 using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-;
int dblcmp (double k)
{
if (fabs(k)<eps) return ;
return k>?:-;
}
struct Point
{
double x,y;
};
double dis (Point a,Point b)
{
return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double area_of_overlap (Point c1,double r1,Point c2,double r2)//圆相交模板
{
double d = dis(c1,c2);
if (r1+r2<d+eps) return ;
if (d<fabs(r1-r2)+eps){
double r = min(r1,r2);
return PI*r*r;
}
double x = (d*d+r1*r1-r2*r2)/(*d);
double t1 = acos(x/r1);
double t2 = acos((d-x)/r2);
return r1*r1*t1+r2*r2*t2-d*r1*sin(t1);
}
int t;
int casee = ;
int main()
{
//freopen("de.txt","r",stdin);
scanf("%d",&t);
while (t--){
Point p1,p2;
double r1,r2;
scanf("%lf%lf",&r1,&r2);
scanf("%lf%lf",&p1.x,&p1.y);
scanf("%lf%lf",&p2.x,&p2.y);
if (dblcmp(r1-r2)>) swap(r1,r2);
double ans = area_of_overlap (p1,r2,p2,r2) -area_of_overlap (p1,r2,p2,r1)
-(area_of_overlap (p1,r1,p2,r2) - area_of_overlap(p1,r1,p2,r1) );
/*double ans = area (p1,r2,p2,r2) -area (p1,r2,p2,r1)
-(area(p1,r1,p2,r2) - area(p1,r1,p2,r1) );*/
printf("Case #%d: %.6f\n",++casee,ans);
}
return ;
}
 

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