HDU 1969 Pie（二分法）

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 ri with 1 ≤ ri ≤ 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 oating 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

题目大意：主人家里来了F个他的朋友，他家里有n个Pie，主人希望把Pie分出F+1份，体积相同（包括主人），所有的Pie不需要都分完，问你每个人最大能分到多大体积的Pie。

主要是猜那个二分最小值x，对，就是猜。

#include <iostream>
#include <cstdio>
#include<cmath>
using namespace std;
double pi=acos(-1.0);             // 圆周率的表示。。。
int T,n,f;
double b[10003];
int juge(double x)
{
    int total =0;
    for(int i=1; i<=n; i++)
    {
        total+=int(b[i]/x);
    }
    if(total>f)
        return 1;
    else
        return 0;
}
int main()
{
    double l,r,mid,rad;
    cin>>T;
    while(T--)
    {
        cin>>n>>f;
        double sum=0;
        for(int i =1; i<=n; i ++)
        {
            cin>>rad;
            b[i]=rad*rad*pi;
            sum+=b[i];
        }
        l=0;
        double r=sum/(f+1);
        while(r-l>0.0001)           // 此题精度小数点后四位
        {
            mid=(l+r)/2;
            if(juge(mid))
                l=mid;
            else
                r=mid;
        }
        printf("%.4lf\n",mid);
    }
    return 0;
}