Pie _HDU1969

题目

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 

 n个派，f+1个人。

f+1个人均分n个派

要求每个人获得的派必须是完整的，不能使拼接起来的。

采用二分来猜测每个人获得的派的大小

相对来说比较简单，需要注意以下几点

1 pi的大小精度要设置好 pi=acos(-1.0);

2 我一开始是设置high为所有半径和均分，这样得到的结果与题目给出的样例不符合，可能是精度不够，后来改成体积之和就ok了

3 有一个点要想到  一张派可以切成好几份，这样照样可能是最大值 也就是

for(int j=0;j<n;j++) cnt+=int(r[j]/mid);

4 题目中的f是好友数量，要加上自己，一共f+1个人均分派

代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
double r[10010],pi=acos(-1.0);
int main()
{
	int t,n,f;
	cin>>t;
	while(t--)
	{
		cin>>n>>f;
		f++;
		double sum=0;
		for(int i=0;i<n;i++) {cin>>r[i]; r[i]=r[i]*r[i]*pi; sum+=r[i];}
		double low=0,high=sum/f,mid;
		while(high>low+1e-6)
		{
			int cnt=0;
			mid=(low+high)/2;
			for(int j=0;j<n;j++) cnt+=int(r[j]/mid);
			if(cnt>=f) low=mid;
			else high=mid;
		}
		printf("%.4lf\n",mid);
	}
	
	return 0;
}