HDU 4033

Regular Polygon

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2515 Accepted Submission(s): 782

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

 1 //给定一个正num边型内一点距离所有点的距离，求边长。如果无解输出impossible

 2 //所有角度和是否为2*PI，大于则缩小边，否则扩大边 

 3 #include <cstdio>

 4 #include <iostream>

 5 #include <cstring>

 6 #include <cmath>

 7 using namespace std;

 8 

 9 double ch[200];

10 const double PI = 4 * atan( double( 1 ) );

11 //const double PI = 3.1415926;//这个wa 

12 int num;

13 

14 double is_triangle( double a, double b, double c )

15 {

16     if( c > a + b )

17         return 100;

18     else if( c < fabs( a - b ) )    

19         return -100;

20     else

21         return acos( ( a * a + b * b - c * c ) / ( 2 * a * b ) );

22 }

23 

24 double Acos( double a, double b, double c )

25 {

26     return is_triangle( a, b, c );

27 }

28 

29 double total_angle( double mid )

30 {

31     double ans = 0;

32     for( int i = 2; i <= num; ++i )

33     {

34         double t = Acos( ch[i], ch[i-1], mid );

35         if( fabs( t ) == 100 ) //  如果mid的这个取值使得某些相邻的两条边不            

36             return t;        //  能够形成三角形，则及时退出

37         else

38             ans += Acos( ch[i], ch[i-1], mid );

39     }

40     return ans;

41     // 最后返回总的角度大小

42 }

43 

44 double Bsearch( double ss, double ee )

45 {

46     double left = ss, right = ee, mid, sum_angle;

47     while( right - left >= 1e-9 )

48     { 

49         mid = ( left + right ) / 2.0;

50         double t = total_angle( mid ); // 计算出一部分的角度总和

51         if( fabs( t ) == 100 )

52         {

53             if( t > 0 )

54                 right = mid;//缩小边 

55             else

56                 left = mid;//扩大边 

57         }

58         else

59         {

60             sum_angle = t + Acos( ch[1] , ch[num], mid ); 

61             if( sum_angle - 2 * PI > 1e-9 )

62                 right = mid;

63             else if( sum_angle - 2 * PI < -1e-9 )

64                 left = mid;

65             else 

66                 return mid;

67         }

68     }

69     return 0;

70 }

71 

72 int main(  )

73 {

74     int i,j,k,T;

75     cin>>T;

76     for(i = 1; i <= T; ++i )

77     {

78         cin>>num;

79         for(j = 1; j <= num; ++j )

80             cin>>ch[j];

81         //  接下来选择第num条边以及第1条边来二分答案

82         //  根据三角形中有 C < A+B && C > A-B;   

83         double ans = Bsearch( fabs( ch[num] - ch[1] ), ch[1] + ch[num] );

84         printf( "Case %d: ", i );

85         if( ans == 0 )

86             puts( "impossible" );

87         else

88             printf( "%.3lf\n", ans );

89     }

90     return 0;

91 }