poj3122 pie 二分

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

题面：给你n个蛋糕 ，m个朋友 算上自己一共m+1个人 使每个人分的蛋糕一样大 形状可以不一样。

比如第一个sample 就是把第一个4的蛋糕分成两半 每块就是面积(4*4*pai)/2  然后把3蛋糕按照半径为2倍根号2分 剩下的扔掉 就是答案。

思路 ：二分，二分的下限是让最大的那块的面积/总人数 上限就是所有蛋糕面积/总人数 这分别是最坏情况和最好情况。 其实你也可以把

这个限制定的送一点也没什么 然后每次二分比较的内容就是 分别在所有蛋糕中计算该蛋糕可以满足多少人 一个for循环搞定。

#include<iostream>#include<cmath>#include<algorithm>using namespace std;double pi=acos(-1.0);double pie[10010],maxn,sum;double eps=1e-6;int main(){    int n,fri,t,r,cnt;    scanf("%d",&t);    while(t--)    {      scanf("%d%d",&n,&fri);      fri++;      maxn=0,sum=0;      for(int i=0;i<n;i++)      {         scanf("%lf",&pie[i]);         pie[i]=pie[i]*pie[i]*pi;         maxn=max(pie[i],maxn);         sum+=pie[i];      }      double l=maxn*1.0/fri,r=sum*1.0/fri,mid;      while(l+eps<r)      {        mid=(l+r)/2;        cnt=0;        for(int i=0;i<n;i++)          cnt+=(int)(pie[i]/mid);        if(cnt<fri) r=mid;        else l=mid;      }      printf("%.4lf\n",l);    }    return 0;}