uva 12097 - Pie（二分，4级）

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解决如何将多个不同大小的圆形蛋糕等分成若干份，确保每个人都能获得相同体积的一份，通过二分查找算法找到最大的可能体积。

Problem C - Pie

Time limit: 1 second

My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input

One line with a positive integer: the number of test cases. Then for each test case:

One line with two integers N and F with 1 ≤ N, F ≤ 10000: the number of pies and the number of friends.
One line with N integers r_i with 1 ≤ r_i ≤ 10000: the radii of the pies.

Output

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^-3.

Sample Input

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327
3.1416
50.2655

The 2006 ACM Northwestern European Programming Contest

思路：二分结果，判断是否能切出符合的块数

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const double pi=acos(-1.0);
const int mm=1e4+9;
const double mx=1e-6;
double f[mm];
int n,m;
bool judge(double mid)
{
  int ans=0;
  for(int i=0;i<n;++i)
    ans+=int(f[i]/mid);
  if(ans>m)return 1;
  else return 0;
}
double fabs(double x)
{
  if(x<0)return -x;
  return x;
}
int main()
{
 int cas;
 while(~scanf("%d",&cas))
 {
   while(cas--)
   {
     scanf("%d%d",&n,&m);
     for(int i=0;i<n;++i)
      {scanf("%lf",&f[i]);
       f[i]=f[i]*f[i]*pi;
      }
      double l=0,r=1e10,mid;
      while(fabs(r-l)>mx)
      {
        mid=(l+r)/2;
        if(judge(mid))l=mid;
        else r=mid;
      }
      ///cout<<(l+r)/2<<endl;
      printf("%.4lf\n",(l+r)/2);
   }
 }
}