HDU 1969 精度二分

Pie

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12394    Accepted Submission(s): 4371

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

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

NWERC2006

 

二分每块pie的大小即可，主要是精度问题，由于精确到1e-4所以while(r-l>0.0001)即可，但是里面对于r的操作不写精度+-得话就能A，如果写必须注意精度了开到1e-7才A   (⊙﹏⊙)b

#include<bits/stdc++.h>
using namespace std;
#define PI acos(-1.0)
double a[10005],F,N;
bool can(double s)
{
  int sum=0,i;
  for(i=1;i<=N;++i) sum+=floor(a[i]/s);
  return sum>=F;
}
int main()
{
    int i,j,k,t;
    cin>>t;
    while(t--){
        cin>>N>>F;
        F++;
        for(i=1;i<=N;++i) {
                cin>>a[i];
                a[i]=a[i]*a[i]*PI;
        }
        double l=0,r=10000*10000*PI,mid;
        while(r-l>0.00001){
            mid=r-(r-l)/2;
            if(can(mid)){
                l=mid;
            }
            else{
                r=mid-0.0000001;
            }
        }
        printf("%.4f\n",l);
    }
    return 0;
}