Uva 派 (Pie,NWERC 2006,LA 3635)

依然是一道二分查找

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 using namespace std;
 5 
 6 const double PI=acos(-1.0);
 7 int N,F;
 8 double r[10001];
 9 
10 bool ok(double area)
11 {
12     int sum=0;
13     for(int i=0;i<N;i++)
14         sum+=floor(r[i]/area);
15     return sum>=F+1;
16 }
17 
18 int main()
19 {
20     int T;
21     cin>>T;
22     while(T--)
23     {
24         cin>>N>>F;
25         double maxn=-1;
26         for(int i=0;i<N;i++)
27         {
28             int a;
29             cin>>a;
30             r[i]=PI*a*a;
31             maxn=max(maxn,r[i]);
32         }
33         double L=0,R=maxn;
34         while(R-L>1e-4)
35         {
36             double M=(L+R)/2;
37             if(ok(M)) L=M;else R=M;
38         }
39         printf("%.4lf
",L);
40     }
41     return 0;
42 }
原文地址:https://www.cnblogs.com/InWILL/p/5560801.html