UVa400

Posted 2020-08-06 Yan_Bin

400 Unix ls
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 (1  N  100). 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
10
tiny
2short4me
very_long_file_name
shorter
size-1
size2
size3
much_longer_name
12345678.123
mid_size_name
12
Weaser
Alfalfa
Stimey
Universidad de Valladolid OJ: 400 – Unix ls 2/2
Buckwheat
Porky
Joe
Darla
Cotton
Butch
Froggy
Mrs_Crabapple
P.D.
19
Mr._French
Jody
Buffy
Sissy
Keith
Danny
Lori
Chris
Shirley
Marsha
Jan
Cindy
Carol
Mike
Greg
Peter
Bobby
Alice
Ruben
Sample Output
------------------------------------------------------------
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：

       输入正整数n以及n个文件名，排序后按照列优先的方式左对齐输出，假设最长文件名有M个字符，则最右列宽至少M字符，其它列宽皆为M+2字符，而输出的总宽为60个字符，输出文件名前要先输出60个’-’字符。

分析：

       找出M值，并计算出输出的行数和列数。设maxcol=60，则列数cols = (maxcol - M) / (M + 2) + 1，行数rows = (n - 1) / cols + 1。然后排序输出文件名。

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxcol = 60;
 6 const int maxn = 100;
 7 string filenames[maxn + 1];
 8 // 输出字符串s，长度不足则补字符extra
 9 void print(const string& s,int len,char extra){
10     cout << s;
11     for(int i = 0 ; i < len - s.length() ; i++)
12         cout << extra;
13 }
14 int main(){
15     int n;
16     while(cin >> n){
17         int M = 0;
18         for(int i = 0 ; i < n ; i++){
19             cin >> filenames[i];
20             M = max(M,(int)filenames[i].length());
21         }
22         // 计算行数和列数
23         int cols = (maxcol - M) / (M + 2) + 1;
24         int rows = (n - 1) / cols + 1;
25         print("",60,\'-\');
26         cout << "\\n";
27         sort(filenames,filenames + n);
28         for(int r = 0 ; r < rows ; r++){
29             for(int c = 0 ; c < cols ; c++){
30                 int idx = c * rows + r;
31                 if(idx < n)
32                     print(filenames[idx],c == cols - 1 ? M : M + 2,\' \');
33             }
34             cout << "\\n";
35         }
36     }
37     return 0;
38 }

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