Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1791 Accepted Submission(s): 686
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
Source 2014ACM/ICPC亞洲區北京站-重制賽（感謝北師和上交）
題意：求兩圓環相交的重疊部分面積。
因為兩圓環全等
圓環交=大大交-2*大小交+小小交。
#include<iostream>
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#define maxn 1000005
#include<math.h>
#define MS(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
using namespace std;
double cirinter(int x1,int y1,int r1,int x2,int y2,int r2)//圓交面積公式
{
double d,s,t,a1,a2,s1,s2,s3;
if(r1<r2){t=r1;r1=r2;r2=t;}
d=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));//兩圓心距離
if(d>=r1+r2)return 0;//兩圓相離
else if(d<=(r1-r2))//兩圓内含
s=pi*r2*r2;
else//兩圓相交
{
a1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));//大圓中扇形圓心角的一半
a2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));//小圓中扇形圓心角的一半
s1=a1*r1*r1;//大圓中的那個扇形面積
s2=a2*r2*r2;//小圓中的那個扇形面積
s3=r1*sin(a1)*d;//兩圓心與兩交點組成的四邊形面積
s=s1+s2-s3;//圓交面積
}
return s;
}
int main()
{
int n,i,r,R,x1,y1,x2,y2;
double area;
cin>>n;
for(i=1;i<=n;i++)
{
cin>>r>>R;
cin>>x1>>y1;
cin>>x2>>y2;
area=cirinter(x1,y1,R,x2,y2,R)-2*cirinter(x1,y1,R,x2,y2,r)+cirinter(x1,y1,r,x2,y2,r);
printf("Case #%d: %.6lf\n",i,area);
}
return 0;
}