CSU1019-Simple Line Editor-模拟

最新推荐文章于 2018-07-09 22:26:20 发布

Zickshen

最新推荐文章于 2018-07-09 22:26:20 发布

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分类专栏： ACM训练

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本文介绍了一道关于早期行编辑器的模拟题，该编辑器只能处理英文字符，并支持撤销和清除命令。文章提供了完整的C++代码实现，通过样例输入输出展示了如何处理这些命令来最终形成文档。

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D: Simple Line Editor

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

2
ab#a
ab@aa

Sample Output

aa
aa

裸的基础模拟题

#include <bits/stdc++.h>
#define N 10100
#define INF 0x3f3f3f3f
#define LL long long
#define mem(a,n) memset(a,n,sizeof(a))
#define fread freopen("in.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    int n,cnt;
    cin>>n;
    string temp,str;
    while(n--){
        cin>>temp;
        str.resize(temp.size());
        cnt=0;
        for(int i=0;i<temp.size();++i){
            if(temp[i]=='#'){
                if(cnt){
                    --cnt;
                }
            }else if(temp[i]=='@'){
                cnt=0;
            }else{
                str[cnt++]=temp[i];
            }
        }
        for(int i=0;i<cnt;++i){
            cout<<str[i];
        }
        cout<<endl;
    }
    return 0;
}


/**********************************************************************
    Problem: 1019
    User: CSUzick
    Language: C++
    Result: AC
    Time:0 ms
    Memory:1688 kb
**********************************************************************/