C. Replace To Make Regular Bracket Sequence 题目连接: http://www.codeforces.com/contest/612/problem/C Description You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and…

题目链接:http://codeforces.com/contest/612/problem/C 解题思路: 题意就是要求判断这个序列是否为RBS,每个开都要有一个和它对应的关,如:<()>满足条件,但<(>)就不满足条件,反正直接就是用栈做就行了,完美符合题目要求. #include <bits/stdc++.h> using namespace std; stack<char>st; /*struct node{ int num,id; }a[20000…

Codeforces Round #529 (Div. 3) 题目传送门 题意: 给你由左右括号组成的字符串,问你有多少处括号翻转过来是合法的序列 思路: 这么考虑: 如果是左括号 1)整个序列左括号个数比右括号多 2 2)在这个位置之前,所有位置的前缀左括号个数都不少于前缀右括号个数 3)在这个位置和这个位置之后,在修改后所有位置的前缀左括号个数减去前缀右括号个数大于2 (这里这么想,把左变成右,左-1,右+1) 右括号也是这样 代码: #include<bits/stdc++.h> usi…

传送门 题意: 给你一个只包含 '(' 和 ')' 的长度为 n 字符序列s: 给出一个操作:将第 i 个位置的字符反转('(' ')' 互换): 问有多少位置反转后,可以使得字符串 s 变为"Regular Bracket Sequence": 输出满足条件的位置的个数: 题解: 令 '(' = 1 , ')' = -1: 定义 sum[i]:括号序列的前缀和: 一个合法的括号匹配串的充要条件是: [1] 对于任何 i,sum[i] ≥ 0: [2] sum[n]=0: int n;…

题意:给你一串括号,每次仅可以修改一个位置,问有多少位置仅修改一次后所有括号合法. 题解:我们用栈来将这串括号进行匹配,每成功匹配一对就将它们消去,因为题目要求仅修改一处使得所有括号合法,所以栈中最后一定会有两个括号剩余,并且这两个括号要么是\(((\)要么是\())\),\()(\)是无论如何都不合法的,对于\())\),我们去找它左边的\()\)的个数贡献给答案(因为每次修改可以使[\((++\),\()--\)],所以栈中剩余的\())\)就没有了),对于\(((\)的情况也是一样的,我们…

C. Replace To Make Regular Bracket Sequence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). Th…

Replace To Make Regular Bracket Sequence You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For…

C. Longest Regular Bracket Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/5/C Description This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence…

You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bra…

[链接]:CF [题意]:给你一个只含有括号的字符串,你可以将一种类型的左括号改成另外一种类型,右括号改成另外一种右括号 问你最少修改多少次,才能使得这个字符串匹配,输出次数 [分析]: 本题用到了栈.如果遇上左括号,就加进栈里.如果遇上右括号,就判断栈里的左括号是否和它匹配,不匹配就加一.不论匹不匹配,判断后都要让左括号出栈. 如果最后栈不为空,或者栈在循环结束前就为空,那么不论怎么改变,左右括号都不可能刚好匹配. [代码]: #include<cstdio> #include<cst…