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[ACM] poj 1064 Cable master (二進制搜索)

系統 2507 0
Cable master
Time Limit: 1000MS ? Memory Limit: 10000K
Total Submissions: 21071 ? Accepted: 4542

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
    

Source

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解題思路:

N條繩子,長度為別為li,要截成長度相等的K段,問切成小段的最大長度是多少。有的繩子能夠不切。

也就是求一個x ,?? l1/ x +l2/x +l3/x +.....=K,求最大的x。

求的過程中中間值x 。假設>k也是符合題意的。要求最大的x。==k.

條件C(x)=能夠得到K條長度為x的繩子

區間l=0,r等于無窮大,二分。推斷是否符合c(x) C(x)=(floor(Li/x)的總和大于或等于K

代碼:

      #include <iostream>

#include <iomanip>

#include <stdio.h>

#include <cmath>

using namespace std;

const int maxn=10003;

const int inf=0x7fffffff;

double l[maxn];

int n,k;



bool ok(double x)//推斷x是否可行

{

    int num=0;

    for(int i=0;i<n;i++)

    {

        num+=(int)(l[i]/x);

    }

    return num>=k;//被分成的段數大于等于K才可行

}



int main()

{

    cin>>n>>k;

    for(int i=0;i<n;i++)

        scanf("%lf",&l[i]);

    double l=0,r=inf;

    for(int i=0;i<100;i++)//二分,直到解的范圍足夠小

    {

        double mid=(l+r)/2;

        if(ok(mid))

            l=mid;

        else

            r=mid;

    }

    cout<<setiosflags(ios::fixed)<<setprecision(2)<<floor(l*100)/100;//l和r最后相等

    return 0;

}


    


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[ACM] poj 1064 Cable master (二進制搜索)


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