Pie 【二分】

//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

题意 ： 有n个蛋糕，f+1个人，分蛋糕，可以切开蛋糕，但是不可以合并，问每个人最大得到的蛋糕的大小。

我们可以二分 每个人得到的蛋糕大小 ，直到它可以满足所有人，并且是最大的 ；
代码

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 ;
const double pi=acos(-1.0);
const int inf =0x3f3f3f3f;
const double eps =1e-5;// 精度问题,精度不能够太高【脑残的me精度一直开1e-8】。其实一般比题目要求高一个就可以
double arr[MAXN];
bool cmp(double a,double b ){
    return a>b;
}
int n,m;
bool f(double x){
    int cnt=0;
    for(int i=0;i<n;i++){
        int k=int(arr[i]/x);
        cnt+=k;
        if(cnt>=m) return 1;
    }
    return 0;
}
int main(){
    int t;scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);m++;
        double ri=0.0;double le=0.0;
        for(int i=0;i<n;i++){
                int r;
            scanf("%d",&r);
            arr[i]=pi*r*r;
            ri=max(ri,arr[i]);
        }
        sort(arr,arr+n,cmp);
        while(ri-le>eps){
            double mid=(le+ri)/2;
            if(f(mid)) le=mid;
            else ri=mid;
        }
        printf("%.4lf\n",le);
    }
    return 0;
}