POJ 3122 Pie（简单二分）

Pie

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 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 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output

25.13273.1416
50.2655

Source

Northwestern Europe 2006

          题意：主人过生日，m个人来庆生，所以n块蛋糕，m+1个人（还有主人自己）分，问每个人分到的最大体积的蛋糕是多大，PS每 个人所分的蛋糕必须是在同一个蛋糕上写下来的。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;

const double PI = 3.14159265359;
const double eps = 1e-7;
double pie[100010];
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",&pie[i]);
            pie[i] *= pie[i];
            if(maxx<pie[i])
            {
                maxx = pie[i];
            }
        }
        m  = m + 1;
        double l = 0;
        double r = maxx;
        double mid;
        while(r-l>eps)
        {
            mid = (r+l)/2;
            int sum = 0;
            for(int i=0;i<n;i++)
            {
                if(pie[i]-mid>eps)
                {
                    sum += (int)pie[i]/mid;
                }
            }
            if(sum>=m)
            {
                l = mid;
            }
            else
            {
                r = mid;
            }
        }
        printf("%.4lf\n",mid*PI);
    }
    return 0;
}