A - WUB(*?-?*)WUB
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
208A
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 200characters. 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
Input
WUBWUBABCWUB
Output
ABC
Input
WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB
Output
WE ARE THE CHAMPIONS MY FRIEND
Hint
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB".
That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE"
— between them Vasya added two "WUB".
题意就是WUB变成空格，如果在首不不输出空格，中间连续多个只算一个空格。
上代码
#include <stdio.h>
#include <queue>
#include <algorithm>
using namespace std;
#include <string.h>
#include <stdlib.h>
int main()
{
char s[2222];
int i,j;
while(scanf("%s",s)!=EOF)
{
int l=strlen(s);
for(i=0;i<l;i++)
{
if(s[i]=='W' &&s[i+1]=='U'&&s[i+2]=='B')
{
s[i]=1;
s[i+1]=1;
s[i+2]=1; //全变为一
}
}
j=0;
for(i=0;i<l;i++)
{
if(j==1)
{
if(s[i]==1 && s[i+1]==1&& s[i+2]==1)
{
printf(" ");
j=0; //保证连续的空格只输出一个。
}
}
if(s[i]!=1) //不等于1才输出
{
printf("%c",s[i]);
j=1; //只有先输出首字母 上面的J==1才有输出空格的权利
}
}
printf("\n");
}
return 0;
}