# Regular Polygon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 3274    Accepted Submission(s): 996

Problem Description

In a 2_D plane, there is a point strictly in a regular polygon with N sides. If you are given the distances between it and N vertexes of the regular polygon, can you calculate the length of reguler polygon's side? The distance is defined as dist(A, B) = sqrt( (Ax-Bx)*(Ax-Bx) + (Ay-By)*(Ay-By) ). And the distances are given counterclockwise.

Input

First a integer T (T≤ 50), indicates the number of test cases. Every test case begins with a integer N (3 ≤ N ≤ 100), which is the number of regular polygon's sides. In the second line are N float numbers, indicate the distance between the point and N vertexes of the regular polygon. All the distances are between (0, 10000), not inclusive.

Output

For the ith case, output one line “Case k: ” at first. Then for every test case, if there is such a regular polygon exist, output the side's length rounded to three digits after the decimal point, otherwise output “impossible”.

Sample Input

2 3 3.0 4.0 5.0 3 1.0 2.0 3.0

Sample Output

Case 1: 6.766 Case 2: impossible

Source

已知一个点到正n边形的n个顶点的距离，求正n边形的边长。

在已知的表达式中，求不出n边形的边长。但是依据两边之和大于第三边，两边之差小鱼第三边。可以得到这个边的范围.

然后由于n边形的以任意一个点，连接到所有顶点，所有的夹角之和为360,所以只需要采取二分依次来判断，是否满足。

``` 1 #include<cstdio>
2 #include<cstring>
3 #include<cmath>
4 #include<algorithm>
5 #define pi acos(-1.0)
6 #define esp 1e-8
7 using namespace std;
8 double aa[105];
9 int main()
10 {
11   int cas,n;
12   double rr,ll;
13   scanf("%d",&cas);
14   for(int i=1;i<=cas;i++)
15   {
16       scanf("%d",&n);
17       for(int j=0;j<n;j++)
18         scanf("%lf",aa+j);
19     //确定上下边界
20      ll=20001,rr=0;
21     for(int j=0;j<n;j++)
22     {
23       rr=max(rr,aa[j]+aa[(j+1)%n]);
24       ll=min(ll,fabs(aa[j]-aa[(j+1)%n]));
25     }
26     double mid,sum,cosa;
27     printf("Case %d: ",i);
28     bool tag=0;
29     while(rr>esp+ll)
30     {
31       mid=ll+(rr-ll)/2;
32       sum=0;
33       for(int j=0;j<n;j++){
34           //oosr=a*a+b*b-mid*mid; 余弦定理求夹角，然后判断所有的夹角之和是否为360
35           cosa=(aa[j]*aa[j]+aa[(j+1)%n]*aa[(j+1)%n]-mid*mid)/(2.0*aa[j]*aa[(j+1)%n]);
36         sum+=acos(cosa);
37       }
38        if(fabs(sum-2*pi)<esp){
39               tag=1;
40            printf("%.3lf\n",mid);
41            break;
42        }
43        else
44          if(sum<2*pi)  ll=mid;
45          else
46              rr=mid;
47     }
48     if(tag==0)
49         printf("impossible\n");
50   }
51   return 0;
52 }```

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