Cable master (POJ No.1064)

//#define LOCAL
#include<stdio.h>
#include<math.h>
int const MAX_N=10005;
int const MAX_M=100;
double const INF=100000000;
int N,K;
double d[MAX_N],lb,ub;
//判断是否满足条件
bool C(double x)//假设截得的每段长度为x,则所有绳子中能不能截成K段 
{    
    int sum=0;
    for(int i=0;i<N;i++)
    { 
        sum=sum+(int)(d[i]/x);
    } 
    return sum>=K;
} 
void solve()
{
    for(int i=0;i<N;i++)
    {
        scanf("%lf",&d[i]);
    }
    lb=0.0,ub=INF;
    for(int i=0;i<MAX_M;i++)
    {
        double mid=(lb+ub)/2;
        if(C(mid))//如果能截，往上搜让mid变大一些，看行不行 
            lb=mid; 
        else ub=mid;//如果不能截往下搜(找最近能截的点) 
    }
    printf("%.2f\n",floor(ub*100)/100);
}
int main()
{
#ifdef LOCAL
    freopen("Cable master.in","r",stdin);    
    freopen("Cable master.out","w",stdout);
#endif
    while(~scanf("%d%d",&N,&K))
    {
        solve();
    }
    return 0;
}