Simple Line Editor

Simple Line Editor Time Limit:1000MS    Memory Limit:131072KB    64bit IO Format:%lld & %llu

Submit Status Practice CSU 1019

Description

Early computer used line editor, which allowed text to be created and changed only within one line at a time. However, in line editor programs, typing, editing, and document display do not occur simultaneously (unlike the modern text editor like Microsoft Word). Typically, typing does not enter text directly into the document. Instead, users modify the document text by entering simple commands on a text-only terminal. 

Here is an example of a simple line editor which can only process English. In addition, it has two commands. ‘@’ and ‘#’. ‘#’ means to cancel the previous letter, and ‘@’ is a command which invalidates all the letters typed before. That is to say, if you want type “aa”, but have mistakenly entered “ab”, then you should enter ‘#a’ or ‘@aa’ to correct it. Note that if there is no letter in the current document, ‘@’ or ‘#’ command will do nothing.

Input

The first line contains an integer T, which is the number of test cases. Each test case is 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 be no empty line in the test data.

Sample Input

2ab#aab@aa

Sample Output

aaaa解题报告：这类字符匹配题目都可以用栈来做。比如括号匹配。code：
#include<iostream>#include<algorithm>#include<stdio.h>#include<queue>#include<stack>#include<math.h>#include<string.h>using namespace std;typedef long long ll;int main(){   // freopen("input.txt","r",stdin);    int t;    char s[1000];    scanf("%d",&t);    getchar();    while(t--){        scanf("%s",s);        stack<char> st;        int len=strlen(s);        for(int i=0;i<len;i++){            if(s[i]=='#'&&!st.empty())                st.pop();            else if(s[i]=='@')                while(!st.empty())                    st.pop();            else                st.push(s[i]);        }        char ch[1000];        int i=0;        while(!st.empty()){            ch[i++]=st.top();           st.pop();        }        ch[i]='\0';        for(int j=i-1;j>=0;j--)            printf("%c",ch[j]);        printf("\n");    }    return 0;}