The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text. Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that it points…

题意 https://vjudge.net/problem/CodeForces-1263E 您要设计一个只有一行的打字机,这一行的长度是无限大,一开始可以认为每个字符都是空.您的打字机有一个光标只指向一个字符,一开始指向最左侧的字符. 使用者有三种操作: L 将光标向左移一格(当光标已经在最左侧时,忽略这次操作) R 将光标向右移一格 一个小写字符或者'(',')' 将当前字符替换为给定字符 您需要在每次操作后,判断这一行是否是合法括号序列(例如 (ahakioi) 就是合法的,(ige))(…

[Codeforces]1263A Sweet Problem [Codeforces]1263B PIN Code [Codeforces]1263C Everyone is a Winner! [Codeforces]1263D Secret Passwords [Codeforces]1263E Editor…

普通的栈大家都会写,STL的栈据说默认实现方式是deque,没关系反正deque跑得飞快. 这里收录的是一些奇怪的栈,当然双栈实现的队列收录在队列里面. 对顶栈 众所周知,栈可以维护一系列前缀和,包括前缀最值.但是怎么维护全局的呢?当每次都只会修改栈顶(换句话说是顺序移动,而不是随机移动),那么可以用两个反方向的栈来维护这一段序列. Codeforces - 1263E - Editor 题意:要求实现一种数据结构,可以查询全局的前缀最大值,全局的前缀最小值,可在任意位置修改,不过修改光标是顺序…

E. Correct Bracket Sequence Editor time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Recently Polycarp started to develop a text editor that works only with correct bracket sequences (abbrev…

E. Correct Bracket Sequence Editor 题目连接: http://www.codeforces.com/contest/670/problem/E Description Recently Polycarp started to develop a text editor that works only with correct bracket sequences (abbreviated as CBS). Note that a bracket sequence…

E. Editor The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text. Your editor consists of a line with infinite length and cursor, which points to the current character. Please note that…

链接: https://codeforces.com/contest/1263/problem/E 题意: The development of a text editor is a hard problem. You need to implement an extra module for brackets coloring in text. Your editor consists of a line with infinite length and cursor, which point…

E. Editor 题目链接: https://codeforces.com/contest/1263/problem/E 题目大意: 输入一个字符串S1含有‘(’ , ‘)’ , ‘R’ , ‘L’ 以及其他字符.根据这个字符串,得到相应的字符串S2.起始idx=1即S2的初始坐标,然后从左到右读取S1,当遇到'L',idx减小1(当无法左移的情况下idx不减小),当遇到'R',idx增加1,当读取到其他字符时,将idx这个位置上的字符更新为读取到的新的字符.然后输出每一步得到的字符串S2最大…

题目链接: http://codeforces.com/contest/670/problem/E 题解: 用STL的list和stack模拟的,没想到跑的还挺快. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<map> #include<list> #include<stack> using namespace std; + ; const in…