• UVa400


    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 << "
    ";
    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 << "
    ";
    35         }
    36     }
    37     return 0;
    38 }
    View Code
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  • 原文地址:https://www.cnblogs.com/cyb123456/p/5834257.html
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