POJ 3122 Pie (二分查找）

Pie

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14107 Accepted: 4837 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

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

Sample Output

25.13273.141650.2655

Source

Northwestern Europe 2006

二分法查找，类似于猜数字游戏，不断地进行折半缩小范围

题意：生日家里来了F个朋友，他家里有好N个Pie，主人希望把Pie分出F+1份(自己也要一个)，要求体积相同，所有的Pie不需要都分完，问你每个人最大能分到多大体积的Pie。

#include<stdio.h>#include<string.h>#include<algorithm>#define PI 3.14159265359using namespace std;double pi[110000];int n,m;int main(){int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);double maxx=0;for(int i=0;i<n;i++){scanf("%lf",&pi[i]);//pie的半径 pi[i]*=pi[i];maxx=max(maxx,pi[i]);//取最大的pie}m=m+1;double l=0;double r=maxx;double mid;while(r-l>0.000001){mid=(r+l)/2;int sum=0;for(int i=0;i<n;i++){if((pi[i]-mid)>0.000001){sum+=(int)pi[i]/mid;}}if(sum>=m){l=mid;}else{r=mid;}}printf("%.4lf\n",PI*mid);}    return 0;}