HDOJ 5120 Intersection【求圆与圆的交叉部分面积】

Intersection

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

Total Submission(s): 1874 Accepted Submission(s): 715



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

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

不理解为什么我保留8位小数就WA，非得6位才行

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
double r,R,xr,yr,xR,yR;
const double pi=acos(-1.0);
double ins(double r,double R)
{
double dis=sqrt((xr-xR)*(xr-xR)+(yr-yR)*(yr-yR));
if(dis+r<=R)
return pi*r*r;
else if(r+R<=dis)
return 0;
else
{
double th1=acos((R*R+dis*dis-r*r)/(2*R*dis));
double s=R*dis*sin(th1);
double th2=acos((r*r+dis*dis-R*R)/(2*r*dis));
double area=th1*R*R+th2*r*r-s;
return area;
}
}
int main()
{
int t,num=0;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&r,&R);
scanf("%lf%lf%lf%lf",&xr,&yr,&xR,&yR);
double ans=ins(R,R)+ins(r,r)-ins(r,R)*2;
printf("Case #%d: %.6lf\n",++num,ans);
}
return 0;
}