HDU4033:Regular Polygon(二分+余弦定理)

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.

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.

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”.

2
3
3.0 4.0 5.0
3
1.0 2.0 3.0

Case 1: 6.766
Case 2: impossible



题意：知道正多边形内一个点到各个顶点的距离，求出这个多边形的边长

思路：首先，可以将这个点看做圆心，那么正多边形就是这个圆的内接正多边形，通过这个点与各个顶点的连线，形成n个三角形，其角度之和为360，可以用余弦定理去求

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define pi acos(-1.0)
#define exp 1e-8
int main()
{
    int t,n,i,j,flag,cas = 1;
    double a[105];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        flag = 0;
        for(i = 0; i<n; i++)
            scanf("%lf",&a[i]);
        a[n] = a[0];
        double l = 0,r = 20000,mid;
        for(i = 1; i<=n; i++)
        {
            r = min(r,a[i]+a[i-1]);
            l = max(l,fabs(a[i]-a[i-1]));
        }
        while(r-l>exp)
        {
            mid = (l+r)/2;
            double s,sum=0;
            for(i = 1; i<=n; i++)
            {
                s = (a[i]*a[i]+a[i-1]*a[i-1]-mid*mid)/(2.0*a[i]*a[i-1]);
                sum+=acos(s);
            }
            if(fabs(sum-2*pi)<exp)
            {
                flag = 1;
                break;
            }
            else if(sum>2*pi)
                r = mid;
            else
                l = mid;
        }
        printf("Case %d: ",cas++);
        if(flag)
            printf("%.3f\n",mid);
        else
            printf("impossible\n");
    }

    return 0;
}