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.
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.
InputThe 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.OutputFor 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
这题本是水题，无奈我思维太僵硬。愣是问了人家才ac。发现自己不会总结归纳，很多规律我明明已经找到了，但就是没法总结成一个式子，得练啊。
题意：给两个环，求其相交面积。
解题思路：可以轻易推出一般情况下的公式是ans=areaRR-2*arearR+arearr。在其他情况下，其实也一样，只不过有的area为0了而已，只需要把这两个环变成两种圆R和r相交情况就行了。具体看代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <iomanip> 6 #include <cmath> 7 using namespace std; 8 typedef long long ll; 9 const int maxn = 1e6+10; 10 int nu[maxn],cnt,minn; 11 const double pi = acos(-1.0); 12 const double eps = 1e-6; 13 inline double getd(double xa,double ya,double xb,double yb) 14 { 15 return sqrt((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb)); 16 } 17 double getarea(double xa,double ya,double xb,double yb,double r,double R) 18 { 19 double d=getd(xa,ya,xb,yb); 20 // cout<<d<<"d"<<endl; 21 if(d+eps>r+R) return 0; //不相交 22 if(d<R-r+eps)//圆R与圆r无相交面积 23 { 24 return pi*r*r; 25 } 26 double cosa=(R*R+d*d-r*r)/(2*R*d); 27 // cout<<cosa<<endl; 28 double c=R*cosa; 29 // cout<<c<<endl; 30 double anga=acos(cosa); 31 double e=d-c; 32 double angb=acos(e/r); 33 return R*R*anga+r*r*angb-d*R*sin(anga); 34 } 35 36 int main() 37 { 38 cout<<setiosflags(ios::fixed)<<setprecision(6); 39 int t; 40 double rr,RR,xa,ya,xb,yb,r,R; 41 scanf("%d",&t); 42 for(int cas=1;cas<=t;++cas) 43 { 44 cout<<"Case #"<<cas<<": "; 45 cin>>rr>>RR; 46 r=min(rr,RR); 47 R=max(RR,rr); 48 cin>>xa>>ya>>xb>>yb; 49 double ans=getarea(xa,ya,xb,yb,R,R)-2.0*getarea(xa,ya,xb,yb,r,R)+getarea(xa,ya,xb,yb,r,r); 50 cout<<setprecision(6)<<ans<<endl; 51 } 52 return 0; 53 }