Intersection

Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)

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 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

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

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注意： Pi 不能自己写，要用acos（-1.0） ，在这里WA了好多次

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
double dis,R,r;
struct point{
	double x,y;
}a,b;
const double pi=acos(-1.0);
double S(double r1,double r2){
	if(dis>=(r1+r2)) return 0; //外切或不相交
	double s1,s2,sit1,sit2;
	if(dis<=fabs(r1-r2)){  //内切或内含
		r1=min(r1,r2);
	    return pi*r1*r1;
	}
	s1=(r1*r1+dis*dis-r2*r2)/(2*r1*dis);
	s2=(r2*r2+dis*dis-r1*r1)/(2*r2*dis);
    sit1=acos(s1);
    sit2=acos(s2);
	return sit1*r1*r1+sit2*r2*r2-r1*dis*sin(sit1);

}
int main(){
	int t,cas=0;
	scanf("%d",&t);
	while(t--){
		++cas;
		scanf("%lf%lf",&r,&R);
		scanf("%lf%lf",&a.x,&a.y);
		scanf("%lf%lf",&b.x,&b.y);
	    dis=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
	    if(fabs(dis)<1e-7){
	    	printf("Case #%d: %.6lf\n",cas,pi*(R*R-r*r));
	    	continue;
	    }
		double ss=0;
		ss=S(R,R);
		ss-=2*S(R,r)-S(r,r);
		printf("Case #%d: %.6lf\n",cas,ss);
	}
	return 0;
}

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