[poj 3122] Pie 二分

Pie
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14126 Accepted: 4844 Special Judge

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

Source
Northwestern Europe 2006

题目链接：http://poj.org/problem?id=3122

题意：给出n个pie的直径，有f+1个人，如果给每人分的大小相同（形状可以不同），每个人可以分多少。要求是分出来的每一份必须出自同一个pie，也就是说当pie大小为3，2，1，只能分出两个大小为2的份，剩下两个要扔掉。

思路：double 类型的二分 ，注意double的输出为.lf；
二分姿势while（rx-lx>1e-n） lx=mid||rx=mid;

代码：

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;#define PI  3.1415926535897932double a[10005];const double e=1e-6;  double rx,lx;int n,m;int main(){      int T;      scanf("%d",&T);      while(T--)      {        lx=0;rx=0;        scanf("%d%d",&n,&m);         m+=1;         for(int i=1;i<=n;i++)        {            scanf("%lf",&a[i]);            a[i]*=a[i];            rx=max(rx,a[i]);        }         double mid;          while(rx-lx>e)        {            mid=(rx+lx)/2;            int cnt=0;           for(int i=1;i<=n;i++)           {             cnt+=(int)(a[i]/mid);           }           if(cnt<m) rx=mid;           else lx=mid;       }       printf("%.4f\n",mid*PI);      }}