Problem 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

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

题目大意是要办生日Party，有n个馅饼，有f个朋友，接下来是n个馅饼的半径。然后是分馅饼了，

注意咯自己也要，大家都要一样大，形状没什么要求，但都要是一整块的那种，也就是说不能从两个饼中

各割一小块来凑一块，像面积为10的和6的两块饼（饼的厚度是1，所以面积和体积相等），

如果每人分到面积为5，则10分两块，6切成5，够分3个人，如果每人6，则只能分两个了！

题目要求我们分到的饼尽可能的大！

只要注意精度问题就可以了,一般WA 都是精度问题

运用2分搜索：

首先用总饼的体积除以总人数，得到每个人最大可以得到的V，但是每个人手中不能有两片或多片拼成的一块饼，

最多只能有一片分割过得饼。用2分搜索时，把0设为left，把V 设为right。mid=(left+right)/2;

搜索条件是：以mid为标志，如果每块饼都可以分割出一个mid，那么返回true,说明每个人可以得到的饼的体积可以

大于等于mid;如果不能分出这么多的mid,那么返回false,说明每个人可以得到饼的体积小于等于mid。

（1）精度为：0.000001

(2)   pi 用反余弦求出，精度更高。

 

 1 #include <iostream>

 2 #include <stdio.h>

 3 #include <math.h>

 4 using namespace std;

 5 double pi = acos(-1.0);

 6 int F,N;

 7 double V[10001];

 8 bool test(double x)

 9 {

10     int num=0;

11     for(int i = 0; i < N;i++)

12     {

13         num += int(V[i]/x);

14     }

15     if(num>=F)

16     return true;

17     else return false;

18 }

19 int main()

20 {

21     int t,r;

22     double v,max,left,right,mid;

23     scanf("%d",&t);

24     while(t--)

25     {

26         scanf("%d%d",&N,&F);

27         F = F+1;

28         for(int i = 0; i < N; i++)

29         {

30             scanf("%d",&r);

31             V[i] = pi*r*r;

32             v += V[i];

33         }

34         max = v/F;

35         left = 0.0;

36         right = max;

37         while((right-left)>1e-6)//注意这里的精度问题。

38         {

39             mid = (left+right)/2;

40             if(test(mid))

41             left = mid;

42             else right = mid;

43         }

44         printf("%.4f\n",mid);

45     }

46     return 0;

47 }