• poj 1064 高精度 二分


      
    Cable master
    Time Limit: 1000MS
    Memory Limit: 10000K
    Total Submissions: 32191
    Accepted: 6888

    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

    Northeastern Europe 2001

    解法一:浮点数二分           直接在double上二分

    错因:没有向下取整 

    解答:二分+高精度

    <span style="color:#3333ff;">#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    double a[10005];
    int main()
    {
        int n,k;
        while(~scanf("%d %d",&n,&k))
        {
           for(int i=1;i<=n;i++)
             scanf("%lf",&a[i]);
           int t=100;double r=100000,l=0,mid;
           while(t--)
           {
               int cnt=0;mid=(r+l)/2.0;
               for(int i=1;i<=n;i++)
                 cnt+=int(a[i]/mid);    //记得取整
               if(cnt>=k)
                 l=mid;      //尽量向大的取
               else
                 r=mid;
           }
           </span><span style="color:#ff0000;">printf("%0.2f
    ",(floor(r*100))/100);  //注意利用floor函数向下取整,所以要预先*100!!!!!!!!!!向下取整是关键</span><span style="color:#3333ff;">
        }
        return 0;
    }</span>
    解法二:整数二分           先乘以100再运算,最后/100         

    错因:l需要初始化为1而不能是0,否则会re;!!!!!!!!!!!!

    解答:二分

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    int  a[10005];
    int main()
    {
        int n,k,maxn;double temp;
        while(~scanf("%d %d",&n,&k))
        {
           maxn=0;
           for(int i=1;i<=n;i++)
             {
                 scanf("%lf",&temp);
                 a[i]=int(temp*100);
                 if(a[i]>maxn)
                    maxn=a[i];
             }
           int mid,r=maxn,l=1;//l初始化为0会re,,因为l==0时若r==1或0则mid==0,下面的除以mid就会re!!!!!!!!!
           while(l<=r)
           {
               int cnt=0;mid=(r+l)/2;
               for(int i=1;i<=n;i++)
                 cnt+=int(a[i]/mid);
               if(cnt>=k)
                 l=mid+1;
               else
                 r=mid-1;
           }
           printf("%0.2lf
    ",r*0.01);//注意是*0.01,不能为/100,因为r是整型
        }
        return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/smilesundream/p/6642536.html
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