hdu 4033Regular Polygon(二分+余弦定理)

Regular Polygon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3274    Accepted Submission(s): 996

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

 

  已知一个点到正n边形的n个顶点的距离，求正n边形的边长。

思路：

       在已知的表达式中，求不出n边形的边长。但是依据两边之和大于第三边，两边之差小鱼第三边。可以得到这个边的范围.

   然后由于n边形的以任意一个点，连接到所有顶点，所有的夹角之和为360,所以只需要采取二分依次来判断，是否满足。

 

代码：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #define pi acos(-1.0)
 #define esp 1e-8
 using namespace std;
 double aa[];
 int main()
 {
   int cas,n;
   double rr,ll;
   scanf("%d",&cas);
   for(int i=;i<=cas;i++)
   {
       scanf("%d",&n);
       for(int j=;j<n;j++)
         scanf("%lf",aa+j);
     //确定上下边界
      ll=,rr=;
     for(int j=;j<n;j++)
     {
       rr=max(rr,aa[j]+aa[(j+)%n]);
       ll=min(ll,fabs(aa[j]-aa[(j+)%n]));
     }
     double mid,sum,cosa;
     printf("Case %d: ",i);
     bool tag=;
     while(rr>esp+ll)
     {
       mid=ll+(rr-ll)/;
       sum=;
       for(int j=;j<n;j++){
           //oosr=a*a+b*b-mid*mid; 余弦定理求夹角，然后判断所有的夹角之和是否为360
           cosa=(aa[j]*aa[j]+aa[(j+)%n]*aa[(j+)%n]-mid*mid)/(2.0*aa[j]*aa[(j+)%n]);
         sum+=acos(cosa);
       }
        if(fabs(sum-*pi)<esp){
               tag=;
            printf("%.3lf\n",mid);
            break;
        }
        else
          if(sum<*pi)  ll=mid;
          else
              rr=mid;
     }
     if(tag==)
         printf("impossible\n");
   }
   return ;
 }