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
Case 1: 6.766 Case 2: impossible 題意:知道正多邊形內一個點到各個頂點的距離,求出這個多邊形的邊長 思路:首先,可以將這個點看做圓心,那麼正多邊形就是這個圓的內接正多邊形,通過這個點與各個頂點的連線,形成n個三角形,其角度之和為360,可以用余弦定理去求#include#include #include #include 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 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) 2*pi) r = mid; else l = mid; } printf("Case %d: ",cas++); if(flag) printf("%.3f\n",mid); else printf("impossible\n"); } return 0; }