hdu 4033Regular Polygon(二分+余弦定理)

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

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

 

  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。

思路：

       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。

 

代码：

 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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