/*
题意:作者要开一个生日party,他现在拥有n块高度都为1的圆柱形奶酪,已知每块奶酪的底面半径为r不等,
作者邀请了f个朋友参加了他的party,他要把这些奶酪平均分给所有的朋友和他自己(f+1人),
每个人分得奶酪的体积必须相等(这个值是确定的),形状就没有要求。
现在要你求出所有人都能够得到的最大块奶酪的体积是多少?
要求是分出来的每一份必须出自同一个pie,也就是说当pie大小为3,2,1,只能分出两个大小为2的,
剩下两个要扔掉。
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.
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
*/
思路:
分出来的每一份只能出自一个蛋糕,那么分的蛋糕大小一定不能超过最大那个蛋糕的体积。所以二分在(0,max);
#include<stdio.h>
#include<string.h>
#define PI 3.1415926535897932
double s[1500],max;
int n,f;
double bsearch()
{
double l,r,mid;
int ans,i;
l=0;
r=max;
while(r-l>0.000001)
{
ans=0;
mid=(l+r)/2;
for(i=0;i<n;i++)
{
ans=ans+(int)(s[i]/mid);
}
if(ans>=f+1)
{
l=mid;
}
else
{
r=mid;
}
}
return r;
}
int main()
{
int t,r,i;
scanf("%d",&t);
while(t--)
{
max=0;
memset(s,0,sizeof(s));
scanf("%d%d",&n,&f);
for(i=0;i<n;i++)
{
scanf("%d",&r);
s[i]=PI*r*r;
if(s[i]>max)
{
max=s[i];
}
}
printf("%.4lf\n",bsearch());
}
return 0;
}