Regular Polygon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 682    Accepted Submission(s): 174


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
 

Recommend
lcy
 
 
二分求解：
 
#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
const double eps=1e-10;
double m[110][2];
int n;
const double PI=acos(-1.0);

int jug(double mid)
{
    double sum=0.0;
    int i;
    double temp;
    for(i=0;i<n;i++)
    {
        if(mid>(m[i][0]+m[i][1])-eps)  return 1;
        if(mid<fabs(m[i][0]-m[i][1])+eps)  return -1;
        temp=(m[i][0]*m[i][0]+m[i][1]*m[i][1]-mid*mid)/(2.0*m[i][0]*m[i][1]);
        sum+=acos(temp);
    }    
    if(sum>PI*2.0+eps)  return 1;
    if(sum<PI*2.0-eps) return -1;
    else  return 0;
}   
int main()
{
    int i,T;
    int iCase=0;
    scanf("%d",&T);
    double l,r,temp;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%lf",&m[i][1]);
            m[i+1][0]=m[i][1];
        }   
        m[0][0]=m[n][0];
        bool flag=false;
        double mid;
        int temp1;
        l=0;
        r=20000;
        while((r-l)>eps)
        {
            mid=(r+l)/2;
            temp1=jug(mid);
            if(temp1==0)
            {
                flag=true;
                break;
            }    
            if(temp1<0)  l=mid;
            else  r=mid;
        }  
        if(!flag) printf("Case %d: impossible\n",iCase);
        else  printf("Case %d: %.3lf\n",iCase,mid);  
    }  
    return 0;  
}


文章来源:http://www.cnblogs.com/kuangbin/archive/2011/09/11/2173825.html