Harbin Polytechnic Oj--simple Line Editor

Simple line Editor

Time limit:1000 MS Memory limit:65536 K Total submit:573 (151 users) Total accepted:183 (139 users) Rating: Special Judge:no

Description

Early Computer used line editor, which allowed text to is created and changed only within one line at a time. However, in line editor programs, typing, editing, and document display does not occur simultaneously (unlike the modern Tex T editor like Microsoft Word). Typically, typing does not enter text directly into the document. Instead, users modify the document text by entering-commands on a text-only terminal. Here's an example of a simple line editor which can only process 中文版. In addition, it has both commands. ' @ ' and ' # '. ' # ' means to cancel the previous letter, and ' @ ' was a command which invalidates all the letters typed before. That's to say, if you want Type "AA", but has mistakenly entered "AB", then you should enter ' #a ' or ' @aa ' to correct it . Note that if there are no letter in the current document, ' @ ' or ' # ' command would do nothing.

Input

The first line contains a integer T, which is the number of the test cases. Each test case was a typing sequence of a line editor, which contains only lower case letters, ' @ ' and ' # '.

Output

For each test case, print one line which represents the final document of the user. There would is no empty line in the test data.

Sample Input

2 Ab#a [Email protected]

Sample Output

Aa Aa

Think of two ways to do, with the STL to do, a lot of container details, do not understand, these details to understand, do up to a duck.

Vector container to do:

#include <iostream> #include <vector> #include <string>using namespace Std;int main () {string Str;int T ; Cin>>t;while (t--) {cin>>str;vector<char>ls;for (int i=0;i<str.size (); i++) {if (((str[i]== ' # ') || (str[i]== ' @ ')) && (Ls.size () ==0)) Continue;else if (str[i]== ' @ ') ls.clear (); else if (str[i]== ' # ') ls.pop_back (); else {Ls.push_ Back (Str[i]);}} For (Vector<char>::iterator Iter=ls.begin (); Iter!=ls.end (); iter++) cout<<*iter;cout<<endl; Ls.clear ();} return 0;}

Do it with the stacks you just learned!

More smoothly! Be aware that the stack is checked for empty before each pop ().

AC:

#include <iostream> #include <stack> #include <string>using namespace Std;int main () {int t;cin> >t;string Str;while (t--) {stack<char>gq;cin>>str;for (int i=0;i<str.size (); i++) {if (str[i]!= ' # ') &&str[i]!= ' @ ') Gq.push (Str[i]), else if (str[i]== ' # ' &&!gq.empty ()) Gq.pop (); else if (str[i]== ' @ ' & &!gq.empty ()) {while (!gq.empty ()) {Gq.pop ();}}} int Dic=0;char cnt[100000];while (!gq.empty ()) {cnt[dic++]=gq.top (); Gq.pop ();} for (int t=dic-1;t>=0;t--) Cout<<cnt[t];cout<<endl;} return 0;}

Harbin Polytechnic Oj--simple Line Editor