题目链接: 传送门

Dubstep

Time Limit: 1000MS     Memory Limit: 32768 KB

Description

Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.

Input

The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.

Output

Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.

Sample Input

WUBWUBABCWUB
WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB

Sample Output

ABC
WE ARE THE CHAMPIONS MY FRIEND 

思路:

简单字符串处理,删除"WUB"输出剩下的字符。

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;

int main()
{
    string str,tmp;
    cin >> str;
    int len = str.size();
    for (int i = 0;i < len;)
    {
        string s = str.substr(i,3);
        if (s == "WUB")
        {
            i += 3;
        }
        else
        {
            tmp += str[i];
            i++;
            string s = str.substr(i,3);
            if (s == "WUB")
            {
                tmp += ' ';
            }
        }
    }
    if (tmp != "")
        cout << tmp << endl;
    return 0;
} 

CF 208A Dubstep(简单字符串处理)的更多相关文章

  1. 第一部分之简单字符串SDS(第二章)

    一,什么是SDS? 1.引出SDSC字符串:c语言中,用空字符结尾的字符数组表示字符串简单动态字符串(SDS):Redis中,用SDS来表示字符串.在Redis中,包含字符串值的键值对在底层都是由SD ...

  2. hdu 5059 简单字符串处理

    http://acm.hdu.edu.cn/showproblem.php?pid=5059 确定输入的数是否在(a,b)内 简单字符串处理 #include <cstdio> #incl ...

  3. C 封装一个通用链表 和 一个简单字符串开发库

    引言 这里需要分享的是一个 简单字符串库和 链表的基库,代码也许用到特定技巧.有时候回想一下, 如果我读书的时候有人告诉我这些关于C开发的积淀, 那么会走的多直啊.刚参加工作的时候做桌面开发, 服务是 ...

  4. 1442: Neo 的简单字符串(字符串)

    1442: Neo 的简单字符串 时间限制: 10 Sec 内存限制: 128 MB 提交: 9 解决: 3 统计 题目描述 Neo 给你一系列字符串,请你输出字符串中的不同单词个数以及总单词个数. ...

  5. Codeforces 208A:Dubstep(字符串)

    题目链接:http://codeforces.com/problemset/problem/208/A 题意 给出一个字符串,将字符串中的WUB给删去,如果两个字母间有WUB,则这两个字母用空格隔开 ...

  6. Java实验--关于简单字符串回文的递归判断实验

    首先题目要求写的是递归的实验,一开始没注意要求,写了非递归的方法.浪费了一些时间,所谓吃一堑长一智.我学习到了以后看实验的时候要认真看实验中的要求,防止再看错. 以下是对此次的实验进行的分析: 1)递 ...

  7. python 简单字符串字典加密

    1 def crypt(source,key): from itertools import cycle result='' temp=cycle(key) for ch in source: res ...

  8. 洛谷 简单字符串 'P1055ISBN号码' 问题

    题目描述如下: 知识点①:char数组与int型数字进行运算时,需要将 char[i]-'0' .比如 char c[5]; int i; for(i=0;i<5;i++) scanf(&quo ...

  9. c# 进程间的通信实现之一简单字符串收发

       使用Windows API实现两个进程间(含窗体)的通信在Windows下的两个进程之间通信通常有多种实现方式,在.NET中,有如命名管道.消息队列.共享内存等实现方式,这篇文章要讲的是使用Wi ...

随机推荐

  1. Datatable删除行的Delete和Remove方法

    在C#中,如果要删除DataTable中的某一行,大约有以下几种办法: 1,使用DataTable.Rows.Remove(DataRow),或者DataTable.Rows.RemoveAt(ind ...

  2. 如何优化 FineUI 控件库的性能,减少 80% 的数据上传量!

    在开始正文之前,请帮忙为当前排名前 10 唯一的 .Net 开源软件 FineUI 投一票: 投票地址: https://code.csdn.net/2013OSSurvey/gitop/codevo ...

  3. 45个JavaScript小技巧

    原文地址 http://modernweb.com/2013/12/23/45-useful-javascript-tips-tricks-and-best-practices/ 这篇文章的质量个人感 ...

  4. RabbitMQ集群环境搭建-4

    确保成功安装好JDK,erlang,RabbitMQ等,并且RabbitMQ能正常启动,多台电脑之间能互相ping得通. 1. 安装 erlang.rabbitmq 如: 192.168.1.1.19 ...

  5. poj-1384 Piggy-Bank

    poj-1384 Piggy-Bank 地址:http://poj.org/problem?id=1384 题意: 知道盒子里面的物体的总重量,得到每一种硬币的价格和重量,求最少钱构成盒子物体总重量的 ...

  6. Yii 字段验证

    关于验证的属性: $enableClientValidation:是否在客户端验证,也即是否生成前端js验证脚本(如果在form中设置了ajax验证,也会生成这个js脚本). $enableAjaxV ...

  7. RabbitMQ集群、镜像部署配置

    1   RABBITMQ简介及安装 RabbitMQ是一个开源的AMQP实现,服务器端用Erlang语言编写,支持多种客户端,如:Python.Ruby..NET.Java.JMS.C.PHP.Act ...

  8. js 基础(一)

    <!--最近需要用到js相关的知识 就把在W3cSchool 下学到的东西做个笔记,方便以后再看 --><!DOCTYPE html> <html> <hea ...

  9. URL(待整合到HTTP书中哦)

    一:scheme://host.domain:port/path/filename scheme - 定义因特网服务的类型.最常见的类型是 http host - 定义域主机(http 的默认主机是 ...

  10. [转]SpringMVC+Hibernate+Spring 简单的一个整合实例

    原文地址:http://langgufu.iteye.com/blog/2088355 下面开始实例,这个实例的需求是对用户信息进行增删改查.首先创建一个web项目test_ssh,目录结构及需要的J ...