POJ1064

#include <iostream>
#include <iomanip> 
#include <cmath>
using namespace std;

int N;
int K;
double num[10001];

int calc(double len)
{
    int numK= 0;
    for(int i = 0; i < N; ++i)
    {
        numK += (int)(num[i] / len);    
    }
    return numK;
}

double solve()
{
    double low = 0;
    double upper = 100000;
    int count  = 0;
    double mid;
    int numK;
    while(count < 100)
    {
        mid = (low + upper) / 2;
        numK = calc(mid);
        if(numK >= K)
            low = mid;    
        else
            upper = mid;
        ++count;    
    }
    return mid;
}

int main()
{
    //基于二分法的思想实现 
    cin >> N >> K;
    for(int i = 0; i < N; ++i)
        cin >> num[i];
        //这里floor很重要 
    cout << setiosflags(ios::fixed) << setprecision(2) << floor(solve()*100)/100 << endl;
    return 0;
}