POJ-3122 Pie

J -Pie

Time Limit:1000MSMemory Limit:65536KB64bit IO Format:%I64d & %I64u

SubmitStatus

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：有N块蛋糕，F+1个人，每人只能分一块（不能拿几块小的拼成大的），求每人最多拿多少体积（高相等，体积即面积）。

二分，代码：

#include<iostream>
#include<cmath>
using namespace std;
int n,k,i,j;
double s[10005],pi=acos(-1.0);//pi的定义 
bool search(double a)
{
	int count=0;
	for(i=1;i<=n;i++)
		count+=s[i]/a;//按每人a体积，共能分几块 
	if(count>=k+1)
		return true;//超过人数，返回真 
	else
		return false;
}
int main()
{
	int t,temp;
	scanf("%d",&t);
	while(t--)
	{
		double mid=0.0,max=0.0,min=0.0;
		scanf("%d%d",&n,&k);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&temp);
			s[i]=pi*temp*temp;
			max+=s[i];
		}
		max/=(k+1);
		while(max-min>1e-5)//精度 
		{
			mid=(max+min)/2.0;
			search(mid)?min=mid:max=mid;//如果search值为真，即体积mid够分，把mid赋值给min 
		}
		printf("%.4lf\n",mid);
	}
	return 0;
}