hdu 6354

Problem Description

Edward is a worker for Aluminum Cyclic Machinery. His work is operating mechanical arms to cut out designed models. Here is a brief introduction of his work.
Assume the operating plane as a two-dimensional coordinate system. At first, there is a disc with center coordinates

Input

The first line contains one integer

Output

For each test case, print the perimeter of the remaining area in one line. Your answer is considered correct if its absolute or relative error does not exceed

Sample Input

1 4 10 6 3 5 10 -4 3 -2 -4 4 0 9 1

Sample Output

81.62198908430238475376

Source

2018 Multi-University Training Contest 5

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 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define N 120
 4 #define pi acos(-1.0)
 5 struct point{
 6     double  x,y;
 7 };
 8 struct circle{
 9     point po;
10     double  r;
11 }cir[N];
12 double dist (point a,point b){
13     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
14 }
15 int t,m;
16 double R;
17 int main()
18 {
19     scanf("%d",&t);
20     while(t--)
21     {
22         scanf("%d%lf",&m,&R);
23         circle a;
24         a.po.x=0;a.po.y=0;
25         a.r=R;    
26         for(int i=0;i<m;i++)
27         {
28             scanf("%lf%lf%lf",&cir[i].po.x,&cir[i].po.y,&cir[i].r);
29         }
30         double ans=2*pi*R;
31         for(int i=0;i<m;i++)
32         {
33             double dis=dist(a.po,cir[i].po);
34             if(dis-cir[i].r<a.r&&dis+cir[i].r>=a.r){
35                 double d1=2*acos((dis*dis+a.r*a.r-cir[i].r*cir[i].r)/(2*dis*a.r));
36                 double d2=2*acos((dis*dis+cir[i].r*cir[i].r-a.r*a.r)/(2*dis*cir[i].r));
37                 double l1=d1*a.r;
38                 double l2=d2*cir[i].r;
39                 ans-=l1;
40                 ans+=l2;
41             }
42         }
43         printf("%.10f
",ans);
44     }
45     
46     return 0;
47 }
48 /*
49 //判段两个圆的位置关系：
50 相离  ：  dis(a,b)>a.r+b.r
51 外切  ：  dis(a.b)==a.r+b.r
52 相交  ：  dis(a,b)-min(a.r,b.r)<max(a.r,b.r)&&dis(a,b)+min(a.r,b.r)>max(a.r,b.r)
53 内切  ：  dis(a,b)+min(a.r,b.r)==max(a.r,b.r)
54 内含  ：  dis(a,b)+min(a,r)<max(a.r,b.r)
55 */

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