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

HINT

题目大意:输入给定数量的文件名,按照字典顺序排序,按照列优先,输出。保证行数最小。

题目难点:

  1. 如何计算行数和列数?

    这个直接看代码里面的公式就好了,一看就懂。

  2. 如何输出?

    对于文件名数组来说,每一次输出都要计算好输出的坐标,应当采用二层循环来实现。另外,每一个文件名达不到最大长度的使用预先初始化好的字符数组,输出前M-len位就好。

注意点:memset()头文件在csting中;sort()在algorithm中

Aceepted

#include<iostream>
#include<vector>
#include<cstring>
#include <algorithm> using namespace std; int main()
{
int sum; //文件名总数
char arr[60];
memset(arr, ' ', 60); //输出空格
while (cin >> sum)
{
string s;
vector<string>filenames; //存储文件名
int M = 0; //记录最长文件名长度
for (int i = 0;i < sum;i++)
{
cin >> s;
filenames.push_back(s); //录入文件名
if (M < s.length())M =s.length();//记录最长文件名长度
}
sort(filenames.begin(), filenames.end());//排序
cout << "------------------------------------------------------------" << endl;
int c = (60 - M) / (M + 2) + 1;//计算列数
int r = (sum - 1) / c + 1; //计算行数
for (int i = 0;i < r;i++) //行号
{
for (int j = 0 ;j <c;j++) //列号
{
int k = j * r + i; //对应的数组内部的编号
if (k >= sum)break;
if (j)cout << " "; //补全两个空格
cout << filenames[k]; //输出文件名
arr[M - filenames[k].length()] = '\0';//输出和最长文件差的字符数
cout << arr;
arr[M - filenames[k].length()] = ' ';
}
cout << endl; //输出空格
}
}
}

Unix ls UVA - 400的更多相关文章

  1. 【紫书】 Unix ls UVA - 400 模拟

    题意:中文版https://vjudge.net/problem/UVA-400#author=Zsc1615925460 题解:首先读取字符,维护一个最长字符串长度M,再排序. 对于输出,写一个pr ...

  2. UVA 400 - Unix ls (Unixls命令)

    csdn : https://blog.csdn.net/su_cicada/article/details/86773007 例题5-8 Unixls命令(Unix ls,UVa400) 输入正整数 ...

  3. UVa400.Unix ls

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

  4. UVA 400 (13.08.05)

     Unix ls  The computer company you work for is introducing a brand new computer line and is developi ...

  5. UVA 400 Unix ls by sixleaves

    题目其实很简单,答题意思就是从管道读取一组文件名,并且按照字典序排列,但是输入的时候按列先输出,再输出行.而且每一行最多60个字符.而每个文件名所占的宽度为最大文件名的长度加2,除了输出在最右边的文件 ...

  6. UVa 400 (水题) Unix ls

    题意: 有n个文件名,排序后按列优先左对齐输出.设最长的文件名的长度为M,则最后一列长度为M,其他列长度为M+2. 分析: 这道题很简单,但要把代码写的精炼,还是要好好考虑一下的.lrj的代码中有两个 ...

  7. Uva - 400 - Unix ls

    先计算出最长文件的长度M,然后计算列数和行数,最后输出即可. AC代码: #include <iostream> #include <cstdio> #include < ...

  8. uva 400 Unix ls 文件输出排版 排序题

    这题的需要注意的地方就是计算行数与列数,以及输出的控制. 题目要求每一列都要有能够容纳最长文件名的空间,两列之间要留两个空格,每一行不能超过60. 简单计算下即可. 输出时我用循环输出空格来解决对齐的 ...

  9. 【例题5-8 UVA - 400】Unix ls

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 设n个字符串中出现的最长的为len; 最后一列能容纳len个字符,然后前面的列能容纳len+2个字符. 每行最多60个字符. 按照这 ...

随机推荐

  1. Mac上的Redis安装和使用

    redis简介 REmote DIctionary Server(Redis) 是一个由 Salvatore Sanfilippo 写的 key-value 存储系统,是跨平台的非关系型数据库. Re ...

  2. 微信小程序(七)-项目实例(原生框架 MINA转云开发)==02-云开发-配置

    云开发:1.就是用云函数的型式来使用云存储和云数据库完成各种操作!     2.只关注调什么函数,完成什么功能即可,无需关心HTTP请求哪一套!     3.此模式不代表没有服务器,只是部署在云环境中 ...

  3. C语言柔性数组和动态数组

    [前言]经常看到C语言里的两个数组,总结一下. 一.柔性数组 参考:https://www.cnblogs.com/veis/p/7073076.html #include<stdio.h> ...

  4. SpringBoot(九):SpringBoot集成Mybatis

    (1)新建一个SpringBoot工程,在pom.xml中配置相关jar依赖 贴代码: <!--加载mybatis整合springboot--> <dependency> &l ...

  5. Kubernetes-6.Service

    docker version:20.10.2 kubernetes version:1.20.1 本文概述Kubernetes Service的基本原理和使用. 服务 Service是将运行在一组Po ...

  6. GDB调试:从入门到入土

    GDB是类Unix操作糸统下使用命令行调试的调试软件,全名GNU Debugger,在NOI系列竞赛使用的NOI Linux系统中起很大作用(如果不想用毒瘤Guide或直接输出)(XXX为文件名) 1 ...

  7. Typora For Markdown 语法

    数学表达式 要启用这个功能,首先到Preference->Editor中启用.然后使用$符号包裹Tex命令,例如:$lim_{x \to \infty} \ exp(-x)=0$将产生如下的数学 ...

  8. POJ-2236(并查集)

    Wireless NetWork POJ-2236 需要注意这里的树的深度需要初始化为0. 而且,find函数需要使用路径压缩,这里的unint合并函数也使用了优化(用一开始简单的合并过不了). #i ...

  9. Go语言GC实现原理及源码分析

    转载请声明出处哦~,本篇文章发布于luozhiyun的博客:https://www.luozhiyun.com/archives/475 本文使用的 Go 的源码1.15.7 介绍 三色标记法 三色标 ...

  10. [set]JZOJ 5821 手机信号

    Description