E. Correct Bracket Sequence Editor   Recently Polycarp started to develop a text editor that works only with correct bracket sequences (abbreviated as CBS). Note that a bracket sequence is correct if it is possible to get a correct mathematical expre…

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…

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…

题目链接:http://codeforces.com/contest/670/problem/E 给你n长度的括号字符,m个操作,光标初始位置是p,'D'操作表示删除当前光标所在的字符对应的括号字符以内的所有字符(比如'(()())'),'R'操作表示右移光标,'L'操作表示左移光标.删除操作后光标向右移,要是再向右移没有字符的话,那就停在最后的字符上了.问你最后的括号字符是怎么样的. 这题用链表写简单多了,直接模拟操作就行了,我用stl里的list. #include <bits/stdc++…

题目链接:https://codeforces.com/contest/670/problem/E 题意: 给出一个已经匹配的括号串,给出起始的光标位置(光标总是指向某个括号). 有如下操作: 1.往左移动一下光标: 2.往左移动一下光标: 3.删除当前光标指向的括号,以及和它匹配的那个括号,以及这两个括号之间的所有括号. 要求你给出在做完所有操作后的括号串. 题解: 用线段树维护,每个括号是否存在,存在记为 $1$,被删掉了记为 $0$. 然后我们只需要实现:①区间求和.②区间赋值.③根据 $…

D. Babaei and Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/D Description As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake. Simple cake is a cylinder of som…

题目链接:http://codeforces.com/contest/474/problem/F 题意简而言之就是问你区间l到r之间有多少个数能整除区间内除了这个数的其他的数,然后区间长度减去数的个数就是答案. 要是符合条件的话,那这个数的大小一定是等于gcd(a[l]...a[r]). 我们求区间gcd的话,既可以利用线段树性质区间递归下去然后返回求解,但是每次查询是log的,所以还可以用RMQ,查询就变成O(1)了. 然后求解区间内有多少个数的大小等于gcd的话,也是利用线段树的性质,区间递…

C. Day at the Beach Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/599/problem/C Description One day Squidward, Spongebob and Patrick decided to go to the beach. Unfortunately, the weather was bad, so the friends were unab…

题目地址:http://codeforces.com/contest/474/problem/F 由题意可知,最后能够留下来的一定是区间最小gcd. 那就转化成了该区间内与区间最小gcd数相等的个数.区间最小gcd一定小于等于区间最小值.所以仅仅要先推断最小值是否是最小gcd.若是的话,就求出最小值的个数.然后用r-l+1-个数就可以. 对于以上信息.能够用线段树来维护.分别维护区间gcd,区间最小值以及区间最小值的个数. 代码例如以下: #include <iostream> #includ…

B. "Or" Game Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/578/problem/B Description You are given n numbers a1, a2, ..., an. You can perform at most k operations. For each operation you can multiply one of the number…