HDU

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.

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：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
int nu[maxn],cnt,minn;
const double pi = acos(-1.0);
const double eps = 1e-6;
inline double getd(double xa,double ya,double xb,double yb)
{
    return sqrt((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb));
}
double getarea(double xa,double ya,double xb,double yb,double r,double R)
{
    double d=getd(xa,ya,xb,yb);
   // cout<<d<<"d"<<endl;
    if(d+eps>r+R) return 0; //不相交
    if(d<R-r+eps)//圆R与圆r无相交面积
    {
        return pi*r*r;
    }
    double cosa=(R*R+d*d-r*r)/(2*R*d);
  //  cout<<cosa<<endl;
    double c=R*cosa;
   // cout<<c<<endl;
    double anga=acos(cosa);
    double e=d-c;
    double angb=acos(e/r);
    return R*R*anga+r*r*angb-d*R*sin(anga);
}

int main()
{
    cout<<setiosflags(ios::fixed)<<setprecision(6);
    int t;
    double rr,RR,xa,ya,xb,yb,r,R;
    scanf("%d",&t);
    for(int cas=1;cas<=t;++cas)
    {
        cout<<"Case #"<<cas<<": ";
        cin>>rr>>RR;
        r=min(rr,RR);
        R=max(RR,rr);
        cin>>xa>>ya>>xb>>yb;
        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);
        cout<<setprecision(6)<<ans<<endl;
    }
    return 0;
}