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

//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
//所有角度和是否为2*PI，大于则缩小边，否则扩大边 
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;

double ch[200];
const double PI = 4 * atan( double( 1 ) );
//const double PI = 3.1415926;//这个wa 
int num;

double is_triangle( double a, double b, double c )
{
    if( c > a + b )
        return 100;
    else if( c < fabs( a - b ) )    
        return -100;
    else
        return acos( ( a * a + b * b - c * c ) / ( 2 * a * b ) );
}

double Acos( double a, double b, double c )
{
    return is_triangle( a, b, c );
}

double total_angle( double mid )
{
    double ans = 0;
    for( int i = 2; i <= num; ++i )
    {
        double t = Acos( ch[i], ch[i-1], mid );
        if( fabs( t ) == 100 ) //  如果mid的这个取值使得某些相邻的两条边不            
            return t;        //  能够形成三角形，则及时退出
        else
            ans += Acos( ch[i], ch[i-1], mid );
    }
    return ans;
    // 最后返回总的角度大小
}

double Bsearch( double ss, double ee )
{
    double left = ss, right = ee, mid, sum_angle;
    while( right - left >= 1e-9 )
    { 
        mid = ( left + right ) / 2.0;
        double t = total_angle( mid ); // 计算出一部分的角度总和
        if( fabs( t ) == 100 )
        {
            if( t > 0 )
                right = mid;//缩小边 
            else
                left = mid;//扩大边 
        }
        else
        {
            sum_angle = t + Acos( ch[1] , ch[num], mid ); 
            if( sum_angle - 2 * PI > 1e-9 )
                right = mid;
            else if( sum_angle - 2 * PI < -1e-9 )
                left = mid;
            else 
                return mid;
        }
    }
    return 0;
}

int main(  )
{
    int i,j,k,T;
    cin>>T;
    for(i = 1; i <= T; ++i )
    {
        cin>>num;
        for(j = 1; j <= num; ++j )
            cin>>ch[j];
        //  接下来选择第num条边以及第1条边来二分答案
        //  根据三角形中有 C < A+B && C > A-B;   
        double ans = Bsearch( fabs( ch[num] - ch[1] ), ch[1] + ch[num] );
        printf( "Case %d: ", i );
        if( ans == 0 )
            puts( "impossible" );
        else
            printf( "%.3lf\n", ans );
    }
    return 0;
}