Pie

Pie

Time Limit : 5000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 18 Accepted Submission(s) : 7

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

题目大意：给定饼的数量n，及人数为M+1

                    输入n个饼的半径r

                   求在一个人得到的份额只来自一个饼的前提下所能获得的最大份额

解析：由于饼都为圆形，且厚度一样，所以只需计算面积r*r*Pi，二分求解

#include<cstdio>
#include<cmath>
const double Pi=acos(-1.0);//Pi的精确求法
using namespace std;
int main()
{

    int cases,n,peo,i,counter;
    double min,max,mid,sum,tmp,arr[10010];
    scanf("%d",&cases);
    while(cases--)
    {

        scanf("%d%d",&n,&peo);
        peo++,sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&tmp);
            arr[i]=tmp*tmp*Pi;
            sum+=arr[i];
        }
        max=sum/peo,min=0.0;//理论上可得到的最大面积
        while(max-min>=1e-5)
        {
            mid=(max+min)/2.0,counter=0;
            for(i=1;i<=n;i++)
                counter+=int(arr[i]/mid);//输出是可能为浮点数，强制转换
            if(counter>=peo)
                min=mid;
            else
                max=mid;

        }
        printf("%.4lf\n",mid);
    }
    return 0;
}