這題容斥原理求覆蓋面積就行

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#define maxn 2200005
#define LL long long
#define lson i<<1
#define rson (i<<1)+1
using namespace std;
int r,R,x11,y11,x22,y22;
//這個函式返回兩個圓的相交面積
double cal(int
 
 r,int x1,int y1,int R,int x2,int y2)
{
       if((x11-x22)*(x11-x22)+(y11-y22)*(y11-y22)>=(r+R)*(r+R))return 0.0;
       if(r>=R)
       {
           swap(r,R);
           swap(x11,x22);
           swap(y11,y22);
       }
       if(sqrt((x11-x22)*(x11-x22)+(y11-y22)*(y11-y22))+r<=R)return acos(-1.0
 
)*r*r;
       if((x11-x22)*(x11-x22)+(y11-y22)*(y11-y22)>=R*R)
       {
           double area = 0.0;
           int ed = (x11-x22)*(x11-x22)+(y11-y22)*(y11-y22);
           double jiao = ((double)R*R+ed-r*r)/(2.0*R*sqrt((double)ed));
           jiao = acos(jiao);
           jiao*=2.0;
           area+=R*R
 
*jiao/2;
           jiao = sin(jiao);
           area-=R*R*jiao/2;
           jiao = ((double)r*r+ed-R*R)/(2.0*r*sqrt((double)ed));
           jiao = acos(jiao);
           jiao*=2;
           area+=r*r*jiao/2;
           jiao = sin(jiao);
           area-=r*r*jiao/2;
           return area;
       }
       double area = 0.0;
       int ed = (x11-x22)*(x11-x22)+(y11-y22)*(y11-y22);
       double jiao = ((double)r*r+ed-R*R)/(2.0*r*sqrt(ed));
       jiao = acos(jiao);
       area+=r*r*jiao;
       jiao = ((double)R*R+ed-r*r)/(2.0*R*sqrt(ed));
       jiao =acos(jiao);
       area+= R*R*jiao-R*sqrt(ed)*sin(jiao);
       return area;
}
int main()
{
    int t,i1 = 1,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d%d%d%d",&r,&R,&x11,&y11,&x22,&y22);
        double sum = 0.0;
        sum+=cal(R,x11,y11,R,x22,y22);
        sum-=cal(R,x11,y11,r,x22,y22);
        sum-=cal(r,x11,y11,R,x22,y22);
        sum+=cal(r,x11,y11,r,x22,y22);
        printf("Case #%d: ",i1);
        i1++;
        printf("%.6f\n",sum);
    }
    return 0;
}