Broken Keyboard (a.k.a. Beiju Text)

最新推荐文章于 2021-03-04 14:56:10 发布

jiruqianlong123

最新推荐文章于 2021-03-04 14:56:10 发布

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分类专栏：表，栈，队列

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本文介绍了一种使用C++ STL链表处理特殊键盘输入的方法，解决因Home和End键自动触发导致的文本错乱问题。通过示例说明如何正确处理这些特殊字符以获取正确的屏幕显示文本。

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网上看到说用链表，自己实现时发觉遇到’[‘和‘]’时都比较麻烦，很难维持结构，只好用STL的链表了(●ˇ∀ˇ●)

You’re typing a long text with a broken keyboard. Well it’s not so badly broken. The only problem
with the keyboard is that sometimes the “home” key or the “end” key gets automatically pressed
(internally).
You’re not aware of this issue, since you’re focusing on the text and did not even turn on the
monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).
In Chinese, we can call it Beiju. Your task is to find the Beiju text.
Input
There are several test cases. Each test case is a single line containing at least one and at most 100,000
letters, underscores and two special characters ‘[’ and ‘]’. ‘[’ means the “Home” key is pressed
internally, and ‘]’ means the “End” key is pressed internally. The input is terminated by end-of-file
(EOF).
Output
For each case, print the Beiju text on the screen.
Sample Input
This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University
Sample Output
BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<list>
#include<iterator>
using namespace std;
#define ll long long int 
#define ull unsigned long long int 
#define e 2.718281828459
#define INF 0x7fffffff
#pragma warning(disable:4996)
#define pf printf
#define sf scanf
#define max(a,b) (a)>(b)?(a):(b);
#define min(a,b) (a)<(b)?(a):(b);
#define pi  acos(-1.0);
#define  eps 1e-9;


int main(void) {
    char ch[1000005];
    list<char> lis;
    list<char>::iterator ite=lis.begin();//?
    char *c;
    while (sf("%s", ch)!=EOF) {
        c = ch;
        while (*c) {
            if (*c == '[')
                ite = lis.begin();
            else if (*c == ']')
                ite = lis.end();

            else {
                ite=lis.insert(ite, *c);//insert在指定迭代器前插入，返回插入元素的迭代器，所以为了向后继续插入，Ite要++
                ite++;

            }

         c++;//c后移

        }

        for (ite = lis.begin(); ite != lis.end(); ite++)//??
            putchar(*ite);
        putchar('\n');
        lis.clear();//元素清空
    }


    return 0;
}