HDU 1969 PIE

题目：

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

33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2 

 

Sample Output

25.13273.141650.2655 

这个题目其实和求方程的零点差不多，只是判断条件不一样而已。

对每个实数mid，判断能不能每个人都分到大小为mid的pie（忽略了pi，最后再算）

然而我其实是败在了题意上面，题目应该是要求四舍五入保留四位小数，然而Output里面写的明明不是这样，被坑了。

总之，用cout << fixed << setprecision(4) << mid*acos(-1.0) << endl;保留4位小数，它会自动四舍五入，题目就解决了。

代码：

#include<iostream>#include<math.h>#include<iomanip>using namespace std;int main(){int t;cin >> t;int n, f, ni;double low, high, mid;while (t--){cin >> n >> f;int *list = new int[n];low = 0;high = 0;for (int i = 0; i < n; i++){cin >> ni;list[i] = ni*ni;if (list[i]>high)high = list[i];}while (low + 0.00000001 < high){mid = (low + high) / 2;int sum = 0;for (int i = 0; i < n; i++)sum += int(list[i] / mid);if (sum>f)low = mid;else high = mid;}cout << fixed << setprecision(4) << mid*acos(-1.0) << endl;}return 0;}