Andy, 8, has a dream - he wants to produce his very own dictionary. This is not an easy task for him, as the number of words that he knows is, well, not quite enough. Instead of thinking up all the words himself, he has a briliant idea. From his bookshelf
he would pick one of his favourite story books, from which he would copy out all the distinct words. By arranging the words in alphabetical order, he is done! Of course, it is a really time-consuming job, and this is where a computer program is helpful. You
are asked to write a program that lists all the different words in the input text. In this problem, a word is defined as a consecutive sequence of alphabets, in upper and/or lower case. Words with only one letter are also to be considered. Furthermore, your
program must be CaSe InSeNsItIvE. For example, words like “Apple”, “apple” or “APPLE” must be considered the same.

Input

The input file is a text with no more than 5000 lines. An input line has at most 200 characters. Input is terminated by EOF.

Output

Your output should give a list of different words that appears in the input text, one in a line. The words should all be in lower case, sorted in alphabetical order. You can be sure that he number of distinct words in the text does not exceed 5000.

Sample Input

Adventures in Disneyland

Two blondes were going to Disneyland when they came to a fork in the road. The sign read: "Disneyland Left."

So they went home.

Sample Output

a

adventures

blondes

came

disneyland

fork

going

home

in

left

read

road

sign

so

the

they

to

two

went

were

when

题意:将文章中所有单词转换成小写,并按字典序排序

解法:可使用STL的vector容器,储存每个单词,然后使用sort函数,将vector中的每个单词按照字典序排,最后输出

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<ctype.h>
#include<vector>
using namespace std;
char c,s[10010];
vector<string> vec;
int main()
{ int i,m=0;
while(~(c=getchar()))
{ if(isupper(c)||islower(c)) //可使用isalpha 小写为2 大写为1
s[m++]=tolower(c);
else if(m!=0)
{ s[m]='\0';
vec.push_back(s); //在尾部加入字符串
m=0;
}
}
sort(vec.begin(),vec.end());
for(i=0;i<vec.size();i++)
if(i==0 || vec[i]!=vec[i-1] )
cout<<vec[i]<<endl;
}

UVA - 10815 - Andy's First Dictionary STL的更多相关文章

  1. UVa 10815 Andy's First Dictionary

    感觉这道题要比之前几个字符串处理的题目难度要大了一些. 题目大意:给若干行字符串,提取出所有单词并去掉重复的,最后按字典顺序输出. 对于输入大致有两种思路,一种是逐个读入字符,遇到字母的话就放到wor ...

  2. UVA 10815 Andy's First Dictionary (C++ STL map && set )

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

  3. UVA 10815 Andy's First Dictionary ---set

    题目链接 题意:输入一个文本,找出所有不同的单词(连续的字母序列),按字典序从小到大输出.单词不区分大小写. 刘汝佳算法竞赛入门经典(第二版)P112 #include <iostream> ...

  4. UVA 10815 Andy's First Dictionary【set】

    题目链接:https://vjudge.net/contest/211547#problem/C 题目大意: 输入一个文本,找出所有不同的单词(连续的字母序列),按字典序从小到大输出,单词不区分大小写 ...

  5. Uva 10815 Andy's First Dictionary(字符串)

    题目链接:https://vjudge.net/problem/UVA-10815 题意 找出一段文本中的所有单词,以小写形式按照字典序输出. 思路 用空白符替换文本中所有非字母字符后再次读入. 代码 ...

  6. UVA 10815 Andy&#39;s First Dictionary(字符处理)

    Andy, 8, has a dream - he wants to produce his very own dictionary. This is not an easy task for him ...

  7. UVa10815.Andy's First Dictionary

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

  8. UVA-10815 Andy's First Dictionary (非原创)

    10815 - Andy's First Dictionary Time limit: 3.000 seconds Problem B: Andy's First DictionaryTime lim ...

  9. 【UVA - 10815】Andy's First Dictionary (set)

    Andy's First Dictionary Description 不提英文了 直接上中文大意吧 XY学长刚刚立下了再不过CET就直播xx的flag,为了不真的开启直播模式,XY学长决定好好学习英 ...

随机推荐

  1. 简单的多对一传输ns2仿真

    实验名称:简单的多对一传输仿真 实验目的:1.研究怎么实现多对一传输. 实验步骤: 1.写c++代码并注册报文头. 先说一下多对一传输的方式.最开始,接收端发送控制报文给所有的发送端,告诉他们要发送多 ...

  2. 20155331 2016-2017-2 《Java程序设计》第6周学习总结

    20155331 2016-2017-2 <Java程序设计>第6周学习总结 教材学习内容总结 输入/输出基础 很多实际的Java应用程序不是基于文本的控制台程序.尽管基于文本的程序作为教 ...

  3. 阿里云mysql数据库设置让公网访问客户端访问

    第一步 首先使用root登入你的mysql ./mysql -u root -p 你的密码 第二步 备注:也可以添加一个用户名为yuancheng,密码为123456,权限为%(表示任意ip都能连接) ...

  4. 38、使用IO流进行文件拷贝

    使用IO流进行文件拷贝 需求:在项目的根目录里面创建一个java.txt的文件,然后将这个文件拷贝到file文件夹里面并且重命名为good.txt文件先以流的方式将java.txt文件读取到内存中,然 ...

  5. 【译】第八篇 Replication:合并复制-How it works

    本篇文章是SQL Server Replication系列的第八篇,详细内容请参考原文. 在这一系列的前几篇你已经学习了如何在多服务器环境中配置合并复制.这一篇将介绍合并代理并解释它在复制过程中扮演的 ...

  6. python作业ATM(第五周)

    作业需求: 额度 15000或自定义. 实现购物商城,买东西加入 购物车,调用信用卡接口结账. 可以提现,手续费5%. 支持多账户登录. 支持账户间转账. 记录每月日常消费流水. 提供还款接口. AT ...

  7. UNIX环境高级编程 第10章 信号

    SIGSTOP和SIGKILL区别是:前者是使进程暂时停止,即中止,也就是说使进程暂停,将进程挂起,比如你在终端里面执行一个脚本或者程序,执行到一半,你想暂停一下,你按下ctrl+z,就会导致终端发送 ...

  8. AopProxyUtils.getSingletonTarget(Ljava/lang/Object;)Ljava/lang/Object;大坑

    这个问题太坑了,试了好多个版本,都是依赖冲突导致的, https://blog.csdn.net/qq_15003505/article/details/78430595 最后找到这一篇博客解决了,就 ...

  9. 洛谷 P4093: bzoj 4553: [HEOI2016/TJOI2016]序列

    题目传送门:洛谷P4093. 题意简述: 给定一个长度为 \(n\) 的序列 \(a\). 同时这个序列还可能发生变化,每一种变化 \((x_i,y_i)\) 对应着 \(a_{x_i}\) 可能变成 ...

  10. JDK1.8源码TreeMap

    基于红黑树(Red-Black tree)的 NavigableMap 实现:键的排序由构造方法决定:自然排序,Comparator排序:非线程安全(仅改变与现有键关联的值不是结构上的修改):线程安全 ...