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100722C

2024-08-17 17:10:10   215人阅读

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[1000010];
double l,r,n,f;
bool C(double val)    
{
    int temp=0;
    for(int i=1;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,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            a[i]=a[i]*a[i]*PI;
            r=max(a[i],r);
        }
        r++;
        l=0;
        f++;
        double mid=0;
        while(r-l>0.000001)
        {
            mid=(l+r)/2;
            if(C(mid))l=mid;
            else r=mid;
        }
        if(C(r))printf("%.4lf\n",r);
        else printf("%.4lf\n",mid);
    }
    return 0;
}

 

100722C


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