UVALive 3635 Pie 二分查找

                                                       Problem C - Pie

Time limit: 1 second

 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 ≤ 10000: the number of pies and the number of friends.
• One line with N integers r_i with 1 ≤ r_i ≤ 10000: 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

The 2006 ACM Northwestern European Programming Contest

题意：n个不同半径的1个长度高的圆柱形pie，如何能分尽可能大的k份相同体积的pie给k个人。要求每个人只能要一份，不能多份。

思路：二分查找的水题不解释，不过要注意精度如果设为1e-8会TLE。

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define eps 1e-6using namespace std;const double PI=acos(-1);double a[10005];int n,m;double binarysearch(double l,double r){if(r-l<eps)return l;double mid=(l+r)/2.0;int sum=0;for(int i=1;i<=n;i++){sum+=(int)(a[i]/mid);}//cout<<mid<<' '<<sum<<endl;if(sum>=m)return binarysearch(mid,r);else return binarysearch(l,mid);}int main(){int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);m++;//自己也要一份double s=0;for(int i=1;i<=n;i++){scanf("%lf",a+i);a[i]=a[i]*a[i]*PI;s+=a[i];}printf("%.4f\n",binarysearch(0,s/m));}}