hdu-1969-Pie(二分，精度)

题目链接---hdu-1969

                                               Pie

                                                Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                Total Submission(s): 14720    Accepted Submission(s): 5214

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

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

 

Sample Output

25.1327

3.1416

50.2655

 

Source

NWERC2006  

题意：本题主要讲述有m相同高度不同半径的圆柱什么什么东西(反正知道是个圆柱的就行)分给n个人，要求每个人所分的体积相同，求每个人所能分得的最大的体积；

题解：本题主要是用二分的思想解决问题，但是要注意精度；

代码实现如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define PI acos(-1.0) 
using namespace std;
int a[10005];
int main()
{
	int m,n,b,max,t;
	double head,tail,mid,ans;
	cin>>t; 
	while(t--)
	{
		cin>>m>>n;
		max=0;
		for(int i=0;i<m;i++)
		{
			cin>>a[i];
			a[i]=a[i]*a[i];
			if(a[i]>max) 
			max=a[i];
		}
		head=0;
		tail=max;
		mid=(head+tail)/2;
		while(fabs(tail-head)>1e-7)  //控制精度 
		{
			int x=0;
			for(int i=0;i<m;i++)
			{
				x=x+(int)(a[i]/mid); 
			}
			if(x>=(n+1))
			{
				ans=mid;
				head=mid;
			}
			else
			{
				tail=mid;
			}
			mid=(head+tail)/2;
		}
		printf("%.4lf\n",ans*PI);
	}
return 0;	
}