（POJ3122）Pie <二分法>

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 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

Northwestern Europe 2006

题意：
有n个半径为ri的高为1的圆柱形派，现在要分给f+1个人，每个人的大小要是相同的（每个派剩下的要丢掉），问每人最大得到多少派？

分析：
我们就是要找到一个【0，sum】中的最大值满足可以分f+1份。所以我们直接用二分法查找即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define PI 3.1415926535898
#define esp 1e-5
using namespace std;

const int maxn = 10010;
double a[maxn];

int main()
{
    int t,n,f,x;
    double sum;
    scanf("%d",&t);
    while(t--)
    {
        sum = 0;
        scanf("%d%d",&n,&f);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            a[i] = x*x;
            sum += a[i];
        }
        double low,high,mid;
        low = 0; high = sum;
        while(high - low > esp)
        {
            mid = (low + high) / 2;
            int cnt = 0;
            for(int i=0;i<n;i++)
            {
                cnt += a[i]/mid;
            }
            if(cnt < f + 1)
                high = mid;
            else low = mid;
        }
        printf("%.4f\n",high*PI);
    }
    return 0;
}