UVA - 400 Unix ls

 Unix ls
Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu

Description

The computer company you work for is introducing a brand new computer line and is developing a new Unix-like operating system to be introduced along with the new computer. Your assignment is to write the formatter for the ls function.

Your program will eventually read input from a pipe (although for now your program will read from the input file). Input to your program will consist of a list of (F) filenames that you will sort (ascending based on the ASCII character values) and format into (C) columns based on the length (L) of the longest filename. Filenames will be between 1 and 60 (inclusive) characters in length and will be formatted into left-justified columns. The rightmost column will be the width of the longest filename and all other columns will be the width of the longest filename plus 2. There will be as many columns as will fit in 60 characters. Your program should use as few rows (R) as possible with rows being filled to capacity from left to right.
Input

The input file will contain an indefinite number of lists of filenames. Each list will begin with a line containing a single integer ( tex2html_wrap_inline41 ). There will then be N lines each containing one left-justified filename and the entire line's contents (between 1 and 60 characters) are considered to be part of the filename. Allowable characters are alphanumeric (a to z, A to Z, and 0 to 9) and from the following set { ._- } (not including the curly braces). There will be no illegal characters in any of the filenames and no line will be completely empty.

Immediately following the last filename will be the N for the next set or the end of file. You should read and format all sets in the input file.
Output

For each set of filenames you should print a line of exactly 60 dashes (-) followed by the formatted columns of filenames. The sorted filenames 1 to R will be listed down column 1; filenames R+1 to 2R listed down column 2; etc.

Sample Input

10tiny2short4mevery_long_file_nameshortersize-1size2size3much_longer_name12345678.123mid_size_name12WeaserAlfalfaStimeyBuckwheatPorkyJoeDarlaCottonButchFroggyMrs_CrabappleP.D.19Mr._FrenchJodyBuffySissyKeithDannyLoriChrisShirleyMarshaJanCindyCarolMikeGregPeterBobbyAliceRuben

Sample Input

------------------------------------------------------------12345678.123         size-1               2short4me            size2                mid_size_name        size3                much_longer_name     tiny                 shorter              very_long_file_name  ------------------------------------------------------------Alfalfa        Cotton         Joe            Porky          Buckwheat      Darla          Mrs_Crabapple  Stimey         Butch          Froggy         P.D.           Weaser         ------------------------------------------------------------Alice       Chris       Jan         Marsha      Ruben       Bobby       Cindy       Jody        Mike        Shirley     Buffy       Danny       Keith       Mr._French  Sissy       Carol       Greg        Lori        Peter

题意：把每组测试数据中的文件名排序（字典序），然后按照行数最少的格式输出，那么行最多输出的列就有col = 62 / (smax+2); 要考虑多余的两个空格

#include "iostream"#include "stdio.h"#include "algorithm"#include "cmath"using namespace std;string str[10000];int main(){   //freopen("1.in", "r", stdin);    //freopen("1.out", "w", stdout);    int T;    int i, j, k;    while(scanf("%d", &T) == 1)    {   cout<<"------------------------------------------------------------"<<endl;        int smax = 0;        int row, col;        for(i = 0; i < T; i++)        {            cin>>str[i];            smax = max(smax, (int)str[i].size());        }        col = 62 / (smax+2);//输出的列计算        row = ceil(1.0*T/col);//输出的行计算，由列控制        sort(str, str + T);        for(i = 0; i < row; i++)        {            for(j = 0; j < col; j++)            {                   //i+1+j*row 用于判断是否最后一列，如果是最后一列且没有元素的话就不什么都不输出，如果有元素输出，则只输出补全空格，不输出多余的两个空格。                if(i+1+j*row > T) continue;                else if(i+1+j*row <= T) {cout<<str[i+j*row]; for(k = 1; k <= smax-str[i+j*row].size(); k++) cout<<' '; if(j != col-1) cout<<"  ";}            }            cout<<endl;        }    }    return 0;}