Cable master

Cable master

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37481		Accepted: 7960

Description

Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestants using a "star" topology - i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it. 
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible. 
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled. 
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.

Input

The first line of the input file contains two integer numb ers N and K, separated by a space. N (1 = N = 10000) is the number of cables in the stock, and K (1 = K = 10000) is the number of requested pieces. The first line is followed by N lines with one number per line, that specify the length of each cable in the stock in meters. All cables are at least 1 meter and at most 100 kilometers in length. All lengths in the input file are written with a centimeter precision, with exactly two digits after a decimal point.

Output

Write to the output file the maximal length (in meters) of the pieces that Cable Master may cut from the cables in the stock to get the requested number of pieces. The number must be written with a centimeter precision, with exactly two digits after a decimal point. 
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output file must contain the single number "0.00" (without quotes).

Sample Input

4 11
8.02
7.43
4.57
5.39

Sample Output

2.00

题目的意思是有n条绳子，长度分别为L[i]，如果从它们中切割出K条长度相同的绳子的话，这K条绳子最长能有多长？结果保留两位小数。

对于此类问题，我们可以套二分搜索的模型去解决。二分搜索不仅仅是在有序表中查找某一值，在求最优解的问题上也非常有用，例如求“满足某个条件C(x)的最小的x”这一问题就可以用二分搜索。每次取中点mid=(lb+ub)/2,判断是否满足直到(lb,ub]足够小时，此时的ub就是所求的结果。这就是二分搜索算法。

这一题套这个模板，条件C(x)：可以得到k条长度为x的绳子。区间初始化时，用充分大的数INF(INF>maxL),lb=0,ub=INF.由于长度为L[i]的绳子最多可以切出L[i]/x条长度为x的绳子，因此C(x)=∑L[i]/x>=k。可以在O(n)时间内求出

下面是AC代码(G++超时，C++直接AC)

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define maxn 10000+5
#define INF 0x7fffffff
int n,k;
double L[maxn];
//判断是否满足条件
bool C(double x)
{
    int num=0;
    for(int i=0;i<n;i++){
        num+=(int)(L[i]/x);
    }
    return num>=k;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++){
        cin>>L[i];
    }
    double lb=0,ub=INF;
    for(int i=0;i<100;i++){
        double mid=(lb+ub)/2;
        if(C(mid)) lb=mid;
        else ub=mid;
    }
    printf("%.2f\n",floor(ub*100)/100);
    return 0;
}

注意:二分搜索的结束判定

在输出小数的问题中，一般都会允许有指定的误差范围或者指定输出小数点后几位，因

此在使用二分搜索法时，有必要设置合理的结束条件来满足精度要求。在上面的程序中，

指定循环次数作为终止条件:每次区间都会减小到一半，100次后可以达到精度为10^(-30)

的精度范围，基本没有问题。也可以设置成(ub-lb)>EPS这样，指定一个区间的大小，

EPS不能取太小，否则会因为浮点小数精度的原因导致陷入死循环。。。