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 ≤ 10000: the number
of pies and the number of friends. • One line with N integers ri with 1 ≤ ri ≤ 10000: 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 oating 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
25.1327 3.1416 50.2655
题目大意：我有个生日宴会邀请了f个朋友，宴会有n块饼，饼的高度都为1，分别给出n块饼的面积，要使得我和f个朋友每个人所分得的饼的体积都一样，
求每人分得的最大体积。
注意：每个人只能得到一块饼且不能由两块或两块以上不同的饼组合而成。
分析：首先用求出各个饼的体积，再将他们相加求出总体积（V），用V除以总人数（f+1）每人就可以得到最大的饼，由于每人不能由两块及两块以上不同的饼组合而成，
所以需要将V/(f+1)二分，直到找到最大且最合适的值。
代码如下：
#include <iostream>
#include <cstdio>
#include <cmath>
double pi=acos(-1);
const int maxn=10005;
double a[maxn];
int n,f;
using namespace std;
bool test(double x)
{
int num=0;
for(int i=0;i<n;i++)
{
num+=int(a[i]/x);
}
if(num>=f)
return true;
else
return false;
}
int main()
{
int t,r;
double max,v,left,right,mid;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&f);
f=f+1;
for(int i=0;i<n;i++)
{
scanf("%d",&r);
a[i]=pi*r*r;
v+=a[i];
}
max=v/f;
left=0.0;
right=max;
while((right-left)>1e-6)
{
mid=(right+left)/2;
if(test(mid))
left=mid;
else
right=mid;
}
printf("%.4f
",mid);
}
}