簡單，二分法，可是在oj上交的時候會有精度的限制，所以僅僅能把數(shù)擴得大一點，并且在擴的時候為防止尾數(shù)會自己主動生成一些非零數(shù)，所以還是自己吧擴到的位置設(shè)置為0，還有輸出時由于%.2lf會自己有4設(shè)5入，所以能夠自己算出小數(shù)點后兩位的數(shù)值，在輸出，保證要求的精度

      #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);

    }
}

    

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