垃圾键盘(bushi) (悲剧文本)

问题 B: Broken Keyboard
时间限制: 1 Sec 内存限制: 128 MB

题目描述
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.
输入
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). The size of input file does not exceed 5MB.
输出
For each case, print the Beiju text on the screen.
样例输入 Copy
This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University
样例输出 Copy
BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University

题意: 给你若干字符串, 即输入的文本, 要求模拟光标乱动后产生的悲剧, ‘[’ --> home , ‘]’ --> end …

简单又有趣的一道题, 代码如下:

#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define RE register int
#define INF 0x3f3f3f3f
#define EPS 1e-8
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> P;
const double Pi = acos(-1);
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const ll inf = 1e15;
string s;int main() {//    freopen("test.txt", "r", stdin);ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);while (cin >> s) {bool prev_is_home = false; // 默认刚开始光标是在末尾的bool prev_is_end = true; // 个人感觉这俩flag要一个就狗了, 当然这样更清楚点string ans = "", tmp = "";for (auto & i : s) {if (i == '[') {prev_is_home = true;prev_is_end = false;ans = tmp + ans;tmp = "";} else if (i == ']') {prev_is_end = true;prev_is_home = false;ans = tmp + ans;tmp = "";} else {if (prev_is_home) {tmp += i;} else if (prev_is_end) {ans += i;}}} cout << ans << endl;}return 0;
}

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