CSU1019: Simple Line Editor

1019: Simple Line Editor

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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 ‘#’.

there are no more than 1000 letters for each test case.

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

2
ab#a
ab@aa

Sample Output

aa
aa

题意：模拟水题，碰到#，前面一个字符无效，碰到@前面所有的字符无效，用栈模拟一下就可以了，碰到#pop，碰到@清空。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
using namespace std;
const int maxn=1005;
int T;
char s[maxn];
char ans[maxn];
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        stack<char>q;
        scanf("%s",s);
        int len=strlen(s);
        for(int i=0;i<len;i++)
        {
            if(s[i]=='#')
                q.pop();
            else if(s[i]=='@')
            {
                while(!q.empty())
                    q.pop();
            }
            else
                q.push(s[i]);
        }
        int j=0;
        while(!q.empty())
        {
            ans[j++]=q.top();
            q.pop();
        }
        for(int i=0;i<j;i++)
            printf("%c",ans[i]);
        printf("
");
    }
    return 0;
}


/**********************************************************************
    Problem: 1019
    User: therang
    Language: C++
    Result: AC
    Time:0 ms
    Memory:0 kb
**********************************************************************/