二分_D

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

33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2

Sample Output

25.13273.141650.2655

生日分pie，每个人都应该分得相同大小，可以是一块中的小分布或者整一块，但不可以拼凑

/*这题的关键是check函数用来判断能否均分F+1个人*/#include<iostream>#include"algorithm"#include<cstdio>#include<cmath>#include<cstring>#define pi acos(-1.0)#define EPS 1e-6static double a[10005];int N,F;using namespace std;double binary(double);bool check(double);int main(){  int T;  cin>>T;  double sum,rad;  while(T--){    sum=0.0;    double rad=0.0;    cin>>N>>F;    for(int i=0;i<N;i++){      cin>>rad;      a[i]=rad*rad*pi;      sum+=a[i];    }    double Max = sum/(F+1);    double ans=binary(Max);    printf("%.4lf\n",ans);    memset(a,0,sizeof(a));  }  return 0;}double binary(double high){  double low=0.0;  double mid=0.0;  //注意次二分与数组二分不同  while(high-low>EPS){    mid=((high-low)>>1)+low;    if(check(mid)) low=mid;    else  high=mid;  }  return mid;}bool check(double x){  int Count=0;  for(int i=0;i<N;i++)  Count+=int(a[i]/x);  if(Count<F+1)  return false;  else  return true;}