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( ).There will then be N lines each containing one left-justifiedfilename 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 followingset { ._- } (not including the curly braces). There will be no illegalcharacters 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

题意:

制表输出给出的一堆单词;

制表规则:

根据所有单词中, 长度最长的一个, 决定了有几列

一行最多输出60个字符

做法及注意点:

网上挺多份代码说60会WA, 所以我听信了, 根据别人题解的建议, 用62;

然后输出的话我们可以先qsort一下, 接着, 看我AC代码吧, 怎么按列输出

AC代码:

#include<stdio.h>
#include<string.h>
#include<stdlib.h> char str[100][65]; int cmp(const void *a, const void *b) {
return strcmp((char *)a, (char *)b);
} int main() {
int N;
int row, col;
while(scanf("%d", &N) != EOF) {
getchar();
int max_l = 0;
for(int i = 0; i < N; i++) {
gets(str[i]);
if(strlen(str[i]) > max_l)
max_l = strlen(str[i]);
}
qsort(str, N, sizeof(str[0]), cmp);
printf("------------------------------------------------------------\n");
col = 62 / (max_l+2);
row = (N-1) / col+1;
for(int i = 0 ; i < row; i++) {
for(int j = 0; j < col; j++) {
int pos = j * row + i;
int len, blank;
if(pos >= N)
break;
printf("%s", str[pos]);
len = strlen(str[pos]);
blank = max_l - len;
for(int k = 0; k < blank; k++)
printf(" ");
printf(" ");
}
printf("\n");
}
}
return 0;
}

UVA 400 (13.08.05)的更多相关文章

  1. UVA 10194 (13.08.05)

    :W Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (ameri ...

  2. UVA 253 (13.08.06)

     Cube painting  We have a machine for painting cubes. It is supplied withthree different colors: blu ...

  3. UVA 573 (13.08.06)

     The Snail  A snail is at the bottom of a 6-foot well and wants to climb to the top.The snail can cl ...

  4. UVA 10499 (13.08.06)

    Problem H The Land of Justice Input: standard input Output: standard output Time Limit: 4 seconds In ...

  5. UVA 10025 (13.08.06)

     The ? 1 ? 2 ? ... ? n = k problem  Theproblem Given the following formula, one can set operators '+ ...

  6. UVA 465 (13.08.02)

     Overflow  Write a program that reads an expression consisting of twonon-negative integer and an ope ...

  7. UVA 10494 (13.08.02)

    点此连接到UVA10494 思路: 采取一种, 边取余边取整的方法, 让这题变的简单许多~ AC代码: #include<stdio.h> #include<string.h> ...

  8. UVA 424 (13.08.02)

     Integer Inquiry  One of the first users of BIT's new supercomputer was Chip Diller. Heextended his ...

  9. UVA 10106 (13.08.02)

     Product  The Problem The problem is to multiply two integers X, Y. (0<=X,Y<10250) The Input T ...

随机推荐

  1. Redis keys命令

    序号 命令及描述 1 DEL key该命令用于在 key 存在时删除 key. 2 DUMP key 序列化给定 key ,并返回被序列化的值. 3 EXISTS key 检查给定 key 是否存在. ...

  2. HDU Today hdu 2112

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=2112 文章末有一些相应的测试数据供参考. 此题就是一个求最短路的问题,只不过现在的顶点名称变成了字符串而不 ...

  3. python 与 mongodb的交互

  4. CodeDom生成类文件

    仅供个人学习 需要先引入System.CodeDom nuget包 using CodeGenerate.Entities; using System; using System.CodeDom; u ...

  5. iOS 11开发教程(八)定制iOS11应用程序图标

    iOS 11开发教程(八)定制iOS11应用程序图标 在图1.9中可以看到应用程序的图标是网状白色图像,它是iOS模拟器上的应用程序默认的图标.这个图标是可以进行改变的.以下就来实现在iOS模拟器上将 ...

  6. python 模式之工厂模式

    转自:https://www.cnblogs.com/lizhitai/p/4471952.html 工厂模式是一个在软件开发中用来创建对象的设计模式. 工厂模式包涵一个超类.这个超类提供一个抽象化的 ...

  7. Ubuntu下修改为永久DNS的方法

    安装好Ubuntu之后设置了静态IP地址,再重启后就无法解析域名.想重新设置一下DNS,打开/etc/resolv.conf cat /etc/resolv.conf # Dynamic resolv ...

  8. Curl 及 Curl的使用介绍

    Curl 简介 Curl是Linux下一个很强大的http命令行工具,其功能十分强大. 1) 二话不说,先从这里开始吧! $ curl http://www.linuxidc.com 回车之后,www ...

  9. redis的搜索组件 redis-search4j

    redis-search4j是一款基于redis的搜索组件. 特点 1.基于redis,性能高效 2.实时更新索引 3.支持Suggest前缀.拼音查找(AutoComplete功能) 4.支持单个或 ...

  10. Asky极简教程:零基础1小时学编程,已更新前8节

    Asky极简架构 开源Asky极简架构.超轻量级.高并发.水平扩展.微服务架构 <Asky极简教程:零基础1小时学编程>开源教程 零基础入门,从零开始全程演示,如何开发一个大型互联网系统, ...