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 }

 

 
posted @ 2012-09-04 08:19  加拿大小哥哥  阅读(331)  评论(0)  编辑  收藏  举报