hdu1969

                                                Pie

                          Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                        Total Submission(s): 4008    Accepted Submission(s): 1623

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

33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2

 

Sample Output

25.13273.141650.2655解题思想：  大意为f个人分n块馅饼，接下来输入n块馅饼的半径，要求每个人的饼都是一整块的，不可以是多块饼拼起来的。首先用sumV求出饼的总体积，然后用总体积除以总人数，即sumV / n ,求出每人最大可以得到的馅饼体积maxV。然后用二分搜索，left = 0 ,right = maxV,mid = (left + right) / 2 。如果每块饼都可以分出一个mid，check函数返回true,说明每个人得到的饼的体积可以大于等于mid，于是把left更新为mid；反之，若check函数返回false,说明每个人得到的饼的体积小于等于mid，于是把right更新为mid.亮点：PI用反余弦求出，精度更高。AC代码：#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define N 10001double v[N];int n , f;double PI = acos(-1.0);double sumV , mid;bool check(double key){    int num = 0;    for(int i = 0 ; i < n ; i ++)        num += (int) (v[i] / key);    if(num >= f)        return true;    else        return false;}int main(){//freopen("in.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n, &f);        int r;        f += 1;        for(int i = 0 ; i < n; i ++)        {            scanf("%d",&r);            v[i] = PI * r * r;            sumV += v[i];        }        double maxV = sumV / n;        double left = 0.0;        double right = maxV;        while((right-left)>1e-6)        {            mid = (left + right) / 2;            if(check(mid))                left = mid;            else                right = mid;        }        printf("%.4lf\n",mid);    }}