二分 Problem C-CFANZ编程社区

Problem C

Time Limit : 5000/1000ms (Java/Other) Memory Limit :65536/32768K (Java/Other)

Total Submission(s) : 16 Accepted Submission(s) : 6

Problem Description

My birthday is coming up andtraditionally 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 partyand each of them gets a piece of pie. This should be one piece of one pie, notseveral small pieces since that looks messy. This piece can be one whole piethough. My friends are very annoying and if one of them getsa bigger piece than the others, they start complaining. Therefore all of themshould get equally sized (but not necessarily equally shaped) pieces, even ifthis leads to some pie getting spoiled (which is better than spoiling theparty). Of course, I want a piece of pie for myself too, and that piece shouldalso be of the same size. What is the largest possiblepiece size all of us can get? All the pies are cylindrical in shape and theyall have the same height 1, but the radii of the pies can bedifferent.

Input

One line with a positiveinteger: the number of test cases. Then for each test case: ---Oneline with two integers N and F with 1 <= N, F <= 10 000: the number ofpies 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 oneline with the largest possible volume V such that me and my friends can all geta pie piece of size V. The answer should be given as a floating point numberwith an absolute error of at most 10^(-3).

Sample Input

3 3

4 3 3

1 24

10 5

1 4 2 3 4 5 6 5 4 2

Sample Output

25.1327

3.1416

50.2655

代码实现：

#include<bits/stdc++.h>

#define pi acos(-1.0)//百度看到

using namespace std;

double a[10002];

int n, f;

int main()

{

 

    int i, j, T, t;

    double  high, mid, low;

    cin>>T;

    while(T--)

    {

        high = 0;

        cin >> n >> f;

        f++;

        for(i=0;i<n;i++)

        {

            cin>>a[i];

            a[i]*=a[i]*pi;

            high=max(high,a[i]);

        }

        low=0;

        while(high-low>=1e-7)

        {

            mid=(high+low)/2.0;

            t=0;

            for(i=0;i<n;i++)

            t+=(int)(a[i]/mid); //蛋糕分几分

            if (t>=f)

                low=mid;

            else

                high=mid;

        }

         printf("%.4lf\n",mid);

    }

    return 0;

}