pie 饼子问题

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.<br><br>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. <br><br>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.<br>

 

Input

One line with a positive integer: the number of test cases. Then for each test case:<br>---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.<br>---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.<br>

 

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<br>3 3<br>4 3 3<br>1 24<br>5<br>10 5<br>1 4 2 3 4 5 6 5 4 2<br>

 

Sample Output

25.1327<br>3.1416<br>50.2655<br>

 

Source

NWERC2006

这个题

薇薇说直接选最大的开始二分

我选的是平均开始的

防止超时啦

嘿嘿

并且

我没有

算加的那个人

还

a了

一直不知道为什me  可能水体小王子

#include <cstdio>#include<iostream>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<map>#include<set>#include<vector>#include<iomanip>using namespace std;double b[10000];double a;double pai=acos(-1.0);int main(){    int n;    scanf("%d",&n);    int i;    for(i=1;i<=n;i++)    {        int x1,x2;      scanf("%d",&x1);      scanf("%d",&x2);        int j;        double sum=0;        sum=0;        for(j=1;j<=x1;j++)       {            cin>>a;;           b[j]=a*a*pai;           sum=sum+b[j];       }       sum=sum/x2;       double sup;       double mid;       double down;        int pao;         sup=sum;         down=0;         int x=0;         int paopao=0;          while(1)       {   paopao=0;           mid=(sup+down)/2;           for(x=1;x<=x1;x++)           {   pao=b[x]/mid;               paopao=paopao+pao;           }           if(paopao>x2)           {               down=mid;           }          else          {              sup=mid;          }           if((sup-down)<0.00001)           {               cout<<fixed<<setprecision(4)<<(sup+down)/2<<endl;               break;           }       }    }       return 0;}