hdu1969之二分查找

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2735    Accepted Submission(s): 1052

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

25.13273.141650.2655

题意:有n块饼分给f个人,一块饼可以分成多块,每个人得到的饼只能是来自某一块饼,每个人分到的饼都是一样多,问最多每个人能分到多少

分析:求出每块饼的面积s[i],并记录面积最大的s[i]为maxs,然后在0~maxs之间进行二分得到mid,判断是否每个人都能得到面积为mid的饼,如果能则mid=left,否则right=mid

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<cmath>#include<iomanip>#define INF 99999999using namespace std;const int MAX=10000+10;const double PT=3.14159265358979323846;double s[MAX];int t,n,m;bool search(double v){int sum=0;for(int i=0;i<n;++i){sum+=(int)(s[i]/v);}if(sum>=m)return true;else return false;} int main(){cin>>t;while(t--){cin>>n>>m;++m;double maxs=-INF,mins=0,mid=0;for(int i=0;i<n;++i){cin>>s[i];s[i]=PT*s[i]*s[i];maxs=max(maxs,s[i]*1.0);}while(fabs(mins-maxs)>1e-6){mid=(mins+maxs)/2;if(search(mid))mins=mid;else maxs=mid;}cout<<fixed<<setprecision(4)<<mid<<endl;}return 0;}