hdu 4033 Regular Polygon 计算几何 二分+余弦定理
给一个n个顶点的正多边形, 给出多边形内部一个点到n个顶点的距离, 让你求出这个多边形的边长。
二分边长, 然后用余弦定理求出给出的相邻的两个边之间的夹角, 看所有的加起来是不是2Pi。
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> #include <stack> #include <bitset> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, n, a) for(int i = a; i<n; i++) #define fi first #define se second typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; double a[104]; int main() { int t, n; cin>>t; for(int casee = 1; casee<=t; casee++) { printf("Case %d: ", casee); cin>>n; for(int i = 0; i<n; i++) { scanf("%lf", &a[i]); } double l = -inf, r = inf; for(int i = 0; i<n; i++) { l = max(l, fabs(a[i]-a[(i+1)%n])); r = min(r, a[i]+a[(i+1)%n]); } int flag = 0; while(fabs(r-l)>eps) { double mid = (l+r)/2, ans = 0; for(int i = 0; i<n; i++) { int tmp = (i+1)%n; ans += acos((a[i]*a[i]+a[tmp]*a[tmp]-mid*mid)/(2*a[i]*a[tmp])); } if(fabs(ans-2*PI)<eps) { flag = 1; break; } if(ans>2*PI) r = mid; else l = mid; } if(flag) { printf("%.3f\n", l); } else { puts("impossible"); } } return 0; }