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)//定义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 ≤ 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
//题意：
给你两个圆的半径，r,R;
再给你这两个圆的圆心点，让你求出这两个圆的不相交面积。
