HDU 1969（二分法）

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 ≤ 10000: the number of pies and the number of friends. • One line with N integers ri with 1 ≤ ri ≤ 10000: 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 oating 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

题目意思：有N个馅饼，要分给F+1个人。要求每个人分到的面积相同，求最大的面积是多少！（分的要求，每个人手上只能有一个馅饼.....馅饼可以分割）

解题思路：

               1.题目的最终目的不外乎就是确定 一个最大的面积值。这个面积值得范围是0至所有馅饼面积之和sum...然后再想想，它要分给F+1个人，那么它的范围又缩小到了

0至sum/(F+1)。

 　　　　　2.然后就想办法二分缩小范围，直到确定最大面积值。通过来判断分的实际个数t与F+1比较来二分。如果t>=F+1，说明要求的值在右边，否则在左边。（注意要有等于，不然输出相差太大）

　　　　　3.输出

程序代码：

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
const double pi = 4.0 * atan(1.0);
int n, f, r[10010];
double Left, Right;
void solve();
void input()
{
    int T;
    cin >> T;
    while (T--)
    {
        cin >> n >> f;
        f++;
        for (int i = 1; i <= n; i++)
            cin >> r[i];
        solve();
    }
}
void solve()
{
    Left = Right = 0;
    for (int i = 1; i <= n; i++)
    {
        r[i] *= r[i];
        if (r[i] > Right)
            Right = r[i];
    }
    while (Right - Left > 1e-6)
    {
        int tmp = 0;
        double mid = (Left + Right) / 2;
        for (int i = 1; i <= n; i++)
            tmp += r[i] / mid;
        if (tmp >= f)
            Left = mid;
        else
            Right = mid;
    }
    printf("%.4lf
", Left * pi);
}
int main()
{
    input();
    return 0;
}