ACM Course Practice 2--1003

Title Description

My birthday is coming up and traditionally I ' m serving pie. Not just one pie, no, I had a number N of them, of various tastes and of various sizes. F of my friends is coming to my party and each of the them gets a piece of pie. This should is one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My Friends is 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 (and 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 is of the same size.

What's the largest possible piece size all of us can get? All the Pies is cylindrical in shape and they all has the same height 1, but the radii of the Pies can is different.

Input
One line with a positive integer:the number of test cases. Then to each test case:
-one line with a integers n and F with 1 <= N, F <= 000:the number of pies and the number of friends.
-one line with N integers ri with 1 <= ri <= 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 is given as a floating point number with an absolute error of in 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

Main topic

There are f+1 individuals divided into n pieces of cake, each person can only be divided into one piece, and the size of each must be equal

Ideas

As the size of the cake area increases, can be divided into the number of blocks (note that this is not a linear function relationship, because the cake can not be re-assembled, so there will be a piece of cake cut out of the same area after a few pieces, because the remaining area is not larger than the previous block, can only discard the remaining area of the situation. That's why it's not easy to divide the total area by the number of people.
Because of the decline in reverse order, we can use dichotomy to find out the solution.
This is a two-way water problem.

AC Code

  
 

   
  

  
#include<iostream>
   
  

  
#include<cmath>
   
  

  
#include<iomanip>
   
  

  
#include<stdio.h>
   
  

  
#define max(a,b) (((a)>(b))?(a):(b))
   
  

  
using namespace std;
   
  

  
const double pi=acos(-1.0);
   
  

  
int main(){
   
  

  
 //freopen("date.in","r",stdin);
   
  

  
 //freopen("date.out","w",stdout);
   
  

  
 int N,T,renshu,tem1,sum;
   
  

  
 cin>>T;
   
  

  
 double maxMian,tem2,low,up;
   
  

  
 double mianji[10001];
   
  

  
 while(T--){
   
  

  
 up=0;
   
  

  
 cin>>N>>renshu;
   
  

  
 renshu++;
   
  

  
 for(int i=0;i<N;i++){
   
  

  
 cin>>tem1;
   
  

  
 mianji[i]=pi*tem1*tem1;
   
  

  
 up=max(mianji[i],up);
   
  

  
 }
   
  

  
 low=0;
   
  

  
 sum=0;
   
  

  
 while(up-low>1e-6){
   
  

  
 sum=0;
   
  

  
 tem2=(up+low)/2;
   
  

  
 for(int j=0;j<N;j++){
   
  

  
 sum+=((int)(mianji[j]/tem2));
   
  

  
 }
   
  

  
 if(sum>=renshu) low=tem2;
   
  

  
 else up=tem2;
   
  

  
 }
   
  

  
 cout<<fixed<<setprecision(4)<<tem2<<endl;
   
  

  
 }
   
  

  
}
  
 

From for notes (Wiz)

ACM Course Practice 2--1003