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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
233.0 4.0 5.031.0 2.0 3.0
Sample Output
Case 1: 6.766Case 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36035 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 else44 if(sum<2*pi) ll=mid;45 else46 rr=mid;47 }48 if(tag==0)49 printf("impossible\n");50 }51 return 0;52 }
hdu 4033Regular Polygon(二分+余弦定理)
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