You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is basically a text editor with bells and whistles. You are working on a module that takes a piece of code containing some definitions or other tabular information and aligns each column on a fixed vertical position, while keeping the resulting code as short as possible, making sure that only whitespaces that are absolutely required stay in the code. So, that the first words on each line are printed at position p1 = 1; the second words on each line are printed at the minimal possible position p2, such that all first words end at or before position p2−2; the third words on each line are printed at the minimal possible position p3, such that all second words end at or before position p3− 2, etc.

  For the purpose of this problem, the code consists of multiple lines. Each line consists of one or more words separated by spaces. Each word can contain uppercase and lowercase Latin letters, all ASCII punctuation marks, separators, and other non-whitespace ASCII characters (ASCII codes 33 to 126 inclusive). Whitespace consists of space characters (ASCII code 32).

Input

  The input file contains one or more lines of the code up to the end of file. All lines (including the last one) are terminated by a standard end-of-line sequence in the file. Each line contains at least one word, each word is 1 to 80 characters long (inclusive). Words are separated by one or more spaces. Lines of the code can have both leading and trailing spaces. Each line in the input file is at most 180 characters long. There are at most 1000 lines in the input file.

Output

  Write to the output file the reformatted, aligned code that consists of the same number of lines, with the same words in the same order, without trailing and leading spaces, separated by one or more spaces such that i-th word on each line starts at the same position pi.

Note for the Sample:

  The ‘⊔’ character in the example below denotes a space character in the actual files (ASCII code 32).

Sample Input

␣␣start:␣␣integer;␣␣␣␣//␣begins␣here
stop:␣integer;␣//␣␣ends␣here
␣s:␣␣string;
c:␣␣␣char;␣//␣temp

Sample Output

start:␣integer;␣//␣begins␣here
stop:␣␣integer;␣//␣ends␣␣␣here
s:␣␣␣␣␣string;
c:␣␣␣␣␣char;␣␣␣␣//␣temp

HINT

题目的解决思路很清晰,就是录入和输出。录入的方式采用的是读取整行然后赋给stringstream对单词进行分段。分段的该过程中要对每一列的最大长度进行计算。输出就按照正常输出就行,然后后面计算空格的个数。

注意:每一行最后一个单词后面直接换行,不要输出空格!!!

Accepted

#include<iostream>
#include<vector>
#include<cstring>
#include <algorithm>
#include<queue>
#include<sstream>
using namespace std;
int main()
{
queue<string>str[1010];
int maxlen[200] = { 0 };
char arr[200] = { 0 };
memset(arr, ' ', 200);
int num = 0;
string s;
char ss[200];
while (cin.getline(ss,200))
{
int i = 0;
stringstream sss(ss);
while (sss >> s)
{
str[num].push(s);
if (s.length() > maxlen[i++])maxlen[i-1] = s.length();
}
num++;
}
for (int i = 0;i < num;i++)
{
int j = 0;
while (!str[i].empty())
{
cout << str[i].front();
if (str[i].size() != 1)
{
arr[maxlen[j] - str[i].front().length()] = '\0';
cout << arr;
arr[maxlen[j++] - str[i].front().length()] = ' ';
}
str[i].pop();
if (str[i].size())cout << " ";
else {
cout << endl;break;
}
}
}
}

Alignment of Code UVA - 1593的更多相关文章

  1. [刷题]算法竞赛入门经典(第2版) 5-1/UVa1593 - Alignment of Code

    书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,0 ms) //UVa1593 - Alignment of Code #include&l ...

  2. UVA 1593 Alignment of Code(紫书习题5-1 字符串流)

    You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is ...

  3. UVa 1593 (水题 STL) Alignment of Code

    话说STL的I/O流用的还真不多,就着这道题熟练一下. 用了两个新函数: cout << std::setw(width[j]);    这个是设置输出宽度的,但是默认是在右侧补充空格 所 ...

  4. Uva - 1593 - Alignment of Code

    直接用<iomanip>的格式输出,setw设置输出宽度,setiosflags(ios::left)进行左对齐. AC代码: #include <iostream> #inc ...

  5. UVA 1593: Alignment of Code(模拟 Grade D)

    题意: 格式化代码.每个单词对齐,至少隔开一个空格. 思路: 模拟.求出每个单词最大长度,然后按行输出. 代码: #include <cstdio> #include <cstdli ...

  6. 【习题5-1 UVA - 1593】Alignment of Code

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 模拟题,每一列都选最长的那个字符串,然后后面加一个空格就好. 这个作为场宽. 模拟输出就好. [代码] #include <b ...

  7. UVa 1593代码对齐

    原题链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem ...

  8. poj 3959 Alignment of Code <vector>“字符串”

    Description You are working in a team that writes Incredibly Customizable Programming Codewriter (IC ...

  9. A - Alignment of Code(推荐)

    You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is ...

随机推荐

  1. 配置JDK环境及其相关问题

    1.首先找到JDK的安装目录 如果忘记了安装目录在那个地方,可以通过dos命令java -verbose,进行查看 配置jdk环境 新建系统变量JAVA_HOME: 编辑系统变量Path: 新建系统变 ...

  2. Docker 一次性进程与对话进程

    目录 一次性进程 对话进程 退出的方法 参考 Docker在运行程序的时候,需要区分运行的程序是一次性进程还是对话进程,不同的进程操作方式有差异. 一次性进程 一些简单进程是不需要交互的,比如hell ...

  3. 关于Laravel框架中Guard的底层实现

    1. 什么是Guard 在Laravel/Lumen框架中,用户的登录/注册的认证基本都已经封装好了,开箱即用.而登录/注册认证的核心就是: 用户的注册信息存入数据库(登记) 从数据库中读取数据和用户 ...

  4. free命令查看内存

    [root@jojo ~]# free -h total used free shared buff/cache available Mem: 991M 273M 64M 1.1M 653M 535M ...

  5. PAT-1132(Cut Integer )数的拆分+简单题

    Cut Integer PAT-1132 #include<iostream> #include<cstring> #include<string> #includ ...

  6. 记录自己第一次搭建本地fabric框架

    写在前,第一次搭建fabric框架,对于小白的我很是艰辛,参考了很多博主的博客才最终完成,在此记录一下搭建过程. 参考的网站 https://blog.csdn.net/smallone233/art ...

  7. [源码分析] 消息队列 Kombu 之 Consumer

    [源码分析] 消息队列 Kombu 之 Consumer 目录 [源码分析] 消息队列 Kombu 之 Consumer 0x00 摘要 0x01 综述功能 0x02 示例代码 0x03 定义 3.1 ...

  8. Java Swing 自定义Dialog确认对话框

    Java Swing 自定义Dialog 需求:当点击JFrame窗口的关闭按钮时,弹框询问是否确定关闭窗口,如果是则关闭程序,否就让弹框消失什么也不做(使用Dialog). 分析:虽然Java提供了 ...

  9. go中sync.Cond源码解读

    sync.Cond 前言 什么是sync.Cond 看下源码 Wait Signal Broadcast 总结 sync.Cond 前言 本次的代码是基于go version go1.13.15 da ...

  10. 危害api收集

    以下每一条代码,无论其通过什么方式被调用,在哪个类里被调用,传入什么参数,都具有唯一不变性(在逆向出来的的smali文件里),故可以作为匹配的凭证.     网络操作相关: Ljava/net/URL ...