原题链接 http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1809 Description Bobo has a balanced parenthesis sequence P=p1 p2…pn of length n and q questions. The i-th question is whether P remains balanced after pai and pbi swapped. Note that questions ar…
1809: Parenthesis Description Bobo has a balanced parenthesis sequence P=p1 p2…pn of length n and q questions. The i-th question is whether P remains balanced after pai and pbi swapped. Note that questions are individual so that they have no affect…
Problem G: Parenthesis Time Limit: 5 Sec Memory Limit: 128 MBSubmit: 398 Solved: 75[Submit][Status][Web Board] Description Bobo has a balanced parenthesis sequence P=p1 p2…pn of length n and q questions. The i-th question is whether P remains balan…
31 Rikka with Parenthesis II (六花与括号II) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to…
Parenthesis Problem Description: Bobo has a balanced parenthesis sequence P=p1 p2-pn of length n and q questions. The i-th question is whether P remains balanced after pai and pbi swapped. Note that questions are individual so that they have no affec…
Rikka with Parenthesis II 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5831 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Correct parenthe…
Traceback (most recent call last): File "androidmarket82.py", line 108, in <module> main() File "androidmarket82.py", line 54, in main pattern=re.compile('<label id="ctl00_AndroidMaster_Content_Apk_SoftVersion…
3300: [USACO2011 Feb]Best Parenthesis Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 89 Solved: 42[Submit][Status] Description Recently, the cows have been competing with strings of balanced parentheses and comparing them with each other to see who…
题目链接:hdu_5831_Rikka with Parenthesis II 题意: 给你一些括号的排列,必须交换一次,能不能将全部的括号匹配 题解: 模拟一下括号的匹配就行了,注意要特判只有一对括号是NO,因为必须要交换一次 #include <cstdio> #include <cstring> ; char s[N]; int st[N],top; int main() { int t; scanf("%d", &t); while (t--)…
Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules: Any left parenthesis '(' must have a corresponding right parenthesi…
