题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=341

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 (  ). 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 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

题解：其实该归为水题类的，但是自己缺WA了两次，不能开set，题目里面没有说不相同。

 1 /****************************************/ 2 /**              Desgard_Duan          **/ 3 /****************************************/ 4 //#pragma comment(linker, "/STACK:102400000,102400000") 5 #define _CRT_SECURE_NO_WARNINGS 6 #include <iostream> 7 #include <iomanip> 8 #include <cstdio> 9 #include <cstdlib>10 #include <cstring>11 #include <string>12 #include <algorithm>13 #include <stack>14 #include <map>15 #include <queue>16 #include <vector>17 #include <set>18 #include <functional>19 #include <cmath>20 #include <numeric>21 22 using namespace std;23 24 inline void get_val(int &a) {25     int value = http://www.mamicode.com/0, s = 1;26     char c;27     while ((c = getchar()) == ‘ ‘ || c == ‘\n‘);28     if (c == ‘-‘) s = -s; else value = http://www.mamicode.com/c - 48;29     while ((c = getchar()) >= ‘0‘ && c <= ‘9‘)30         value = http://www.mamicode.com/value * 10 + c - 48;31     a = s * value;32 }33 34 vector<string> S;35 map<int, vector<string> > ans;36 37 int main () {38     int n, len = 0;39     string str;40     //freopen ("test.in", "r", stdin);41     while (cin >> n) {42         S.clear();43         ans.clear();44         len = 0;45         for (int i = 0 ; i < n; ++ i)  {46             cin >> str;47             S.push_back (str);48             len = max (len, (int)str.size());49         }50         sort (S.begin(), S.end());51         int cols = (60 - len) / (len + 2) + 1;  //列数52         int rows = (n - 1) / cols + 1;          //行数53         int rows_cnt = 1;54         int cols_cnt = 1;55         //cout << "c: " << cols << endl56         //     << "r: " << rows << endl;57         vector<string> :: iterator it = S.begin();58         while (it != S.end()) {59             if (rows_cnt > rows) {60                 rows_cnt = 1;61                 cols_cnt ++;62             }63             ans[cols_cnt].push_back (*it);64             rows_cnt ++;65             it ++;66         }67         cout.setf (ios :: left);68 69         puts("------------------------------------------------------------");70         for (int i = 0; i < rows; ++ i) {71             for (int j = 1; j <= cols; ++ j) {72                 if (i >= ans[j].size()) break;73                 cout << setw(len + 2) << ans[j][i];74             }75             puts("");76         }77     }78     return 0;79 }

UVa400.Unix ls