//二分搜索//1e-8是什么神奇的东西？//Pie------三F

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

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

解题思路：

现在好像有点明白二分怎么用了。首先得确定一个答案所在的范围，比如这道题就是大于0小于最大的饼的面积。还必须有一个能确定mid是新的上界还是下界的条件，在这道题里就是分的份数和人数的对比，如果份数多了答案就应该更小，份数少了答案就应该更大。

然后因为这道题上界和下界都是double型，两个double型绝对值差在10的负8次方以内就视作相等。e就是10的意思？1e-8就是10的负8次方？但是这道题里用10的负8次方会超时，因为精度要求没那么高，10的负4次方就可以了，不会超时。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;

const double pi=3.14159265358979323846;
double area[10010];
int n,f;

bool cmp(double a,double b)
{
return a>b;
}

int check(double m)
{
int sum=0;
for(int i=1;i<=n;i++)
{
sum+=(int)(area[i]/m);
if(sum>=f+1) return 1;
}
return 0;
}

int main()
{
int t;
double l,r,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&f);
for(int i=1;i<=n;i++)
{
scanf("%lf",&area[i]);
area[i]=area[i]*area[i]*pi;
}
sort(area+1,area+n+1,cmp);
l=0;
r=area[1];
while(fabs(l-r)>1e-5)
{
m=(l+r)/2;
if(check(m)==1)
l=m;
else
r=m;
}
printf("%.4lf\n",l);
}
}