100722C

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 sev-

eral 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

r

i

with 1

≤

r

i

≤

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

.做过啊 二分半径

#include <set>
#include <map>
#include <queue>
#include <deque>
#include <cstdio>
#include <string>
#include <vector>
#include <math.h>
#include <time.h>
#include <utility>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
const double PI=3.141592653589793;
int T;
double a[];
double l,r,n,f;
bool C(double val)
{
    int temp=;
    for(int i=;i<=n;i++)
    {
        temp+=a[i]/val;
    }
    if(temp>=f)return true;
    else return false;
}
int main()
{
    cin>>T;
    while(T--)
    {
        cin>>n>>f;
        memset(a,,sizeof(a));
        for(int i=;i<=n;i++)
        {
            cin>>a[i];
            a[i]=a[i]*a[i]*PI;
            r=max(a[i],r);
        }
        r++;
        l=;
        f++;
        double mid=;
        while(r-l>0.000001)
        {
            mid=(l+r)/;
            if(C(mid))l=mid;
            else r=mid;
        }
        if(C(r))printf("%.4lf\n",r);
        else printf("%.4lf\n",mid);
    }
    return ;
}