Codeforces Round #130 (Div. 2) A. Dubstep

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.

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.

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

sample input

WUBWUBABCWUB

sample output

ABC

sample input

WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB

sample output

WE ARE THE CHAMPIONS MY FRIEND

给你一串字母，求去掉分割标记"WUB"之后的句子。

开一个缓存，直接模拟，匹配一个"WUB"之后把缓存里的值取出来，清空缓存然后跳过这个“WUB"继续做。

注意事项：退出循环之后要检查一下缓存里面是不是有东西。

（数据很小，用c++的string类做比较方便。）

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

const int maxn=1e6+10;
typedef long long LL;

string str;
vector<string> ans;

int n;

int main(){
cin>>str;
string tmp="";
for(int i=0;i<str.length();i++){
if(i+2<str.length()&&str[i]=='W'&&str[i+1]=='U'&&str[i+2]=='B'){
if(tmp!=""){
ans.push_back(tmp);
}
tmp="";
i+=2;
}else{
tmp+=str[i];
}
}
if(tmp!="") ans.push_back(tmp);
for(int i=0;i<ans.size()-1;i++) cout<<ans[i]<<' ';
cout<<ans[ans.size()-1]<<endl;
return 0;
}