Intersection

Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1643 Accepted Submission(s): 620

#include<stdio.h>
#include<string.h>
#include<math.h>
#define P acos(-1.0)//定義π 
#define E 1e-6
#define min(a,b) (a>b?b:a)
double dis,r,R;
struct zz
{
	double x,y;
}a,b;
double ll(double x1,double y1,double x2,double y2)
{
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double fun(double r1,double r2)
{
	if(dis>=(r1+r2))
		return 0;
	double s1,s2,ss1,ss2;
	if(dis<=abs(r1-r2))
	{
		r1=min(r1,r2);
		return P*r1*r1;
	}
	s1=(r1*r1+dis*dis-r2*r2)/(2*r1*dis);
	s2=(r2*r2+dis*dis-r1*r1)/(2*r2*dis);
	ss1=acos(s1);
	ss2=acos(s2);
	return ss1*r1*r1+ss2*r2*r2-r1*dis*sin(ss1);
}
int main()
{
	int t;
	int T=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf",&r,&R);
		scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
		dis=ll(a.x,a.y,b.x,b.y);
		if(abs(dis)<E)
		{
			printf("Case #%d: %.6lf\n",T++,P*(R*R-r*r));
			continue;
		}
		double ss=0;
		ss=fun(R,R);
		ss-=2*fun(R,r)-fun(r,r);
		printf("Case #%d: %.6lf\n",T++,ss);
	}
	return 0;
}
 

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//題意：給你兩個圓的半徑，r,R;再給你這兩個圓的圓心點，讓你求出這兩個圓的不相交面積。