hdu 1966 Pie

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14709    Accepted Submission(s):
5209

Problem Description

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 <= 10 000: the number of pies and the number of friends.
---One
line with N integers ri with 1 <= ri <= 10 000: 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

 

#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
double a[];
    double pi=3.1415926535897932;
int main()
{
    int T,n,f,i;

    cin>>T;
    while(T--)
    {
        cin>>n>>f;
        f++;//包括我自己
        double sum=;
        for(i=;i<=n;i++)
        {
            cin>>a[i];
            a[i]=a[i]*a[i];
            sum=sum+a[i];
        }
        double l=0.0,r=sum/f;
        double mid=;
        while(r-l>0.000001)
        {
             mid=(r+l)/;
            int num=;
            for(i=;i<=n;i++)
            {
                num=num+(int)(a[i]/mid);
            }
            if(num>=f)  l=mid;//等于号很重要
            else r=mid;
        }
         printf("%.4lf\n",pi*mid);
    }
    return ;
}