Poj：1064 ： ：Cable master (假定一个解并判断是否可行）（二分搜索答案）

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
题目大意：有n条绳子，长度分别为L[i]。如果从他们中切割出k条长度相同的绳子的话，这k条绳子每条最长能有多长？（答案保留小数点后两位，规定1单位长度的绳子最多可以切割成100份）。
这是一道明显二分搜索题。现在先进行二分搜索题的思考步骤。
设条件C(x)=可以得到k条长度为x的绳子。
现在问题变成求满足C(x)条件的最大的x。在区间初始化的时候，只需使用INF做上界即可：
st=;
en=INF;
那么现在的问题就变为了如何高效的判断C（x）是否满足。
由于长度为L的绳子最多可以切割出floor(L/x)段长度为x的绳子，因子C（x）=floor(Li/x)的总和是否不小于k，他可以在O（n）的时间内判断出来。
AC代码：

#include<stdio.h>
#include<math.h>
#define INF 0x3f3f3f3f
int n,k;
double a[];
bool C(double mid)
{
    int num=;
    for(int i =  ; i < n ; i++)
        num+=(int)(a[i]/mid);
    if(num>=k)
        return ;
    else
        return ;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i =  ; i < n ; i++)
        scanf("%lf",&a[i]);
      double st=,en=INF;
      ///重复循环，直到解的范围够小，/100次循环可以达到10-30的精度
       for(int i =  ; i <  ; i++)
       {
           double mid=(st+en)/;
           if(C(mid)==)
            st=mid;
            else
            en=mid;
       }
       printf("%.2f\n",floor(en*)/);

}

AC代码另解，实际也是一样的：

#include<stdio.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-10
using namespace std;
int n,k;
double a[];
bool C(double mid)
{
    int num=;
    for(int i =  ; i < n ; i++)
        num+=(int)(a[i]/mid);
    if(num>=k)
        return ;
    else
        return ;
}
int main()
{
    scanf("%d%d",&n,&k);
    double en=-;
    for(int i =  ; i < n ; i++)
    {
        scanf("%lf",&a[i]);
        en=max(en,a[i]);
    }

      double st=;
      while(en-st>=eps)
      {
          double mid=(en+st)/;
          if(C(mid)==)
             st=mid;
          else
           en=mid;
      }
       printf("%.2f\n",floor(en*)/);

}

floor()函数：向下整取函数，头文件<math.h>

#include <stdio.h>
#include <math.h>

int main ()
{
  printf ("floor of 2.3 is %.1lf/n", floor (2.3) );
  printf ("floor of 2.6 is %.1lf/n", floor (2.6) );
  printf ("floor of -2.3 is %.1lf/n", floor (-2.3) );
  printf ("floor of -2.6 is %.1lf/n", floor (-2.6) );
  return ;
}

输出：

floor of 2.3 is 2.0
floor of 2.6 is 2.0
floor of -2.3 is -3.0
floor of -2.6 is -3.0