HDU 1969 Pie(二分查找)

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

Source

NWERC2006

 #include<iostream>

 #include<stdio.h>

 #include<cmath>

 using namespace std;

 int pi,pe,num;//pi是派的数量，pe是人的数量,num是可以分到的人数

 double pie[];//pie的大小

 double mi,ma,mid;//最少能分到的和最多能分到的,mid是二分法的中间变量

 double pai=acos(-1.0);//pi的定义

 int main()

 {

     int T;

     cin>>T;

     while(T--)

     {

         cin>>pi>>pe;

         pe++;//pe个朋友加上自己

         ma=0.0;

         mi=0.0;

         for(int i=;i<pi;i++)

         {

             cin>>pie[i];

             pie[i]=pai*pie[i]*pie[i];

             ma+=pie[i];

             if(pie[i]>mi)

                 mi=pie[i];

         }

         ma/=pe;

         mi/=pe;

         while(mi+0.00001<ma)//因为两个都是double型无法相等所以+0.0001控制

         {

             mid=(ma+mi)/;

             num=;

             for(int i=;i<pi;i++)

             {

                 num+=(int)(pie[i]/mid);

             }

             if(num>=pe)mi=mid;//足够，往大继续二分

             else ma=mid;//不够，往小继续二分

         }

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

     }

     return ;

 }

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