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
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
题目类型：简单字符处理
算法分析：将string排序后按照尽可能排在一行的贪心策略输出即可
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/************************************************** filename :m.cpp author :maksyuki created time :2018/1/3 21:23:29 last modified :2018/1/3 21:48:35 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; vector<string> ans; int main() { #ifdef LOCAL CFF; //CFO; #endif int n; while(cin >> n) { ans.clear(); string s; int maxval = -INF; for(int i = 1; i <= n; i++) { cin >> s; int tmp = s.length(); maxval = max(maxval, tmp); ans.push_back(s); } sort(ans.begin(), ans.end()); int col = (60 - maxval) / (maxval + 2) + 1, row; if(n % col) row = n / col + 1; else row = n / col; for(int i = 1; i <= 60; i++) cout << "-"; cout << endl; string tt; for(int i = 0; i < row; i++) { for(int j = 0; j < col; j++) { if(i + j * row > n - 1) continue; tt = ans[i+j*row]; int len = tt.length(); cout << tt; int tv; if(j < col - 1) tv = maxval + 2; else tv = maxval; for(int k = len + 1; k <= tv; k++) cout << " "; } cout << endl; } } return 0; } |
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