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：
1 #include<cstdio>
2 #include<cstring>
3 #include<cmath>
4 #include<algorithm>
5 #define pi acos(-1.0)
6 #define esp 1e-8
7 using namespace std;
8 double aa[105];
9 int main()
10 {
11 int cas,n;
12 double rr,ll;
13 scanf("%d",&cas);
14 for(int i=1;i<=cas;i++)
15 {
16 scanf("%d",&n);
17 for(int j=0;j<n;j++)
18 scanf("%lf",aa+j);
19 //确定上下边界
20 ll=20001,rr=0;
21 for(int j=0;j<n;j++)
22 {
23 rr=max(rr,aa[j]+aa[(j+1)%n]);
24 ll=min(ll,fabs(aa[j]-aa[(j+1)%n]));
25 }
26 double mid,sum,cosa;
27 printf("Case %d: ",i);
28 bool tag=0;
29 while(rr>esp+ll)
30 {
31 mid=ll+(rr-ll)/2;
32 sum=0;
33 for(int j=0;j<n;j++){
34 //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
35 cosa=(aa[j]*aa[j]+aa[(j+1)%n]*aa[(j+1)%n]-mid*mid)/(2.0*aa[j]*aa[(j+1)%n]);
36 sum+=acos(cosa);
37 }
38 if(fabs(sum-2*pi)<esp){
39 tag=1;
40 printf("%.3lf\n",mid);
41 break;
42 }
43 else
44 if(sum<2*pi) ll=mid;
45 else
46 rr=mid;
47 }
48 if(tag==0)
49 printf("impossible\n");
50 }
51 return 0;
52 }