简单，二分法，可是在oj上交的时候会有精度的限制，所以仅仅能把数扩得大一点，并且在扩的时候为防止尾数会自己主动生成一些非零数，所以还是自己吧扩到的位置设置为0，还有输出时由于%.2lf会自己有4设5入，所以能够自己算出小数点后两位的数值，在输出，保证要求的精度

#include <iostream>

#include <stdio.h>

#include <algorithm>

#define MAX 10010

using namespace std;

long long input[MAX];

int m,n;

bool cmp(int a,int b)

{

    return a<b;

}

bool judge(int mid)

{

    int x = 0;

    for(int i = 0;i < n;i++){

        x += input[i] / mid;

    }

    if(x >= m)

        return true;

    else

        return false;

}

int main()

{

//#ifndef ONLINE_JUDGE

  //  freopen("in.txt","r",stdin);

//#endif

    long long high,low,mid,res;

    double a;

    while(cin>>n>>m){

        high = 0;

        low = 0;

        for(int i = 0;i < n;i++){

            cin>>a;

            input[i] = (a+1e-8)*10000;

            high = high > input[i]?high : input[i];

        }

        sort(input,input+n,cmp);

        res = 0;

        low = 10000/100;

        while(low <= high){

            mid = (low+high)/2;

            if(judge(mid)){

                res = mid;

                low = mid+1;

            }

            else

                high = mid-1;

        }

        int th = res/10000;

        int hu = (res - th * 10000)/1000;

        int te = (res - th * 10000 - hu * 1000)/100;

      printf("%d.%d%d\n",th,hu,te);

    }

}