HDU 1551 Cable master 二分

题意：把N条电缆截成相同的M段，问截成的最大长度是多少？

思路：二分截的长度即可。（我不会告诉你我的程序样例都过不了）

/*********************************************
Problem : HDU 1551
Author  : NMfloat
InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};
double A[10005] ;
double rights;
double sum;

void input() {
double tmp;
rights = 0;
rep(i,1,n) {
scanf("%lf",&A[i]);
rights += A[i];
}
}

int calc(double mid) {
int count = 0;
rep(i,1,n) {
count += (int)( A[i] / mid );
}
return count;
}

void solve() {
double left = 0;
double right = rights;
double mid ;
while(fabs(right-left)>eps) {
mid = (left + right) / 2;
if(calc(mid) < m) right = mid;
else left = mid;
}
printf("%.2lf\n",mid);
}

int main(void) {
//freopen("a.in","r",stdin);
while(scanf("%d %d",&n,&m),n+m) {
input();
solve();
}
return 0;
}