IntersectionHDU-5120（求两圆交集面积）

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
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

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

大致题意：就是告诉你两个圆环的圆心坐标，小圆和大圆的半径，让你求两个相同的圆环的交集的面积。如下图所示

思路：所求阴影部分的面积=两个大圆相交的面积+两个小圆相交的面积-2*大圆和小圆相交的面积。

注意精度！pi的值多取几位

代码如下

#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long 
using namespace std;
const double pi=3.141592654;
//网上找的一份求两圆交集面积的代码
double intersect(double x1,double y1,double r1,double x2,double y2,double r2){
    double s,temp,p,l,ans;
    l=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
    if(l>=r1+r2)    ans=0;
    else if(l<=abs(r1-r2)){
        if(r1<=r2)      ans=pi*r1*r1;
        else            ans=pi*r2*r2;
    }
    else{
        p=(l+r1+r2)/2;
        s=2*sqrt(p*(p-l)*(p-r1)*(p-r2));
        if(r1>r2){
            temp=x1;x1=x2;x2=temp;
            temp=y1;y1=y2;y2=temp;
            temp=r1;r1=r2;r2=temp;
        }
        ans=acos((r1*r1+l*l-r2*r2)/(2*r1*l))*r1*r1+acos((r2*r2+l*l-r1*r1)/(2*r2*l))*r2*r2-s;
    }
    return ans;
}

//六个参数，意思分别是第一个圆的圆心，半径，第二个圆的圆心，半径，返回相交部分的面积，如果不相交，则返回零。
int main()
{

    int T;
    int n;
    int r,R;
    int x1,y1,x2,y2;
    double ans;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        ans=0;
        scanf("%d%d",&r,&R);
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        ans+=intersect(x1,y1,R,x2,y2,R)+intersect(x1,y1,r,x2,y2,r);
        ans-=2*intersect(x1,y1,R,x2,y2,r);
        printf("Case #%d: %.6lf\n",cas,ans);
    }
    return 0;
}