综合-学长杂题讲义

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 int N;
 8 double R[1005];
 9 const double _pi = asin(1)*2;
10 
11 bool Check(double v)
12 {
13     double tmp = 0; int i;
14     for (i = 1; i <= N; ++i)
15         if ((tmp += asin(R[i]/2/v)*2) > 2*_pi) break;
16     return i == N+1;
17 }
18 
19 int main()
20 {
21     freopen("zeriba.in", "r", stdin);
22     int T;
23     scanf("%d", &T);
24     for (int ca = 1; ca <= T; ++ca) {
25         scanf("%d", &N);
26         double sum = 0, mx = 0; int i;
27         for (i = 1; i <= N; ++i)
28             scanf("%lf", &R[i]), sum += R[i], mx = max(mx, R[i]);
29         for (i = 1; i <= N; ++i)
30             if (sum-R[i] <= R[i]) break;
31         if (i != N+1) { printf("Case %d: can‘t form a convex polygon\n", ca); continue; }
32         double l = mx/2, r = sum/4;
33         while (fabs(l-r) >= 0.000001) {
34             double mid = (l+r)/2;
35             if (Check(mid)) r = mid;
36             else l = mid;
37         }
38         printf("Case %d: %.4lf\n", ca, l);
39     }
40     return 0;
41 }