Codeforces 3 D 题意:有一个括号序列,其中一些位置是问号,把第\(i\)个问号改成(需要\(a_i\)的代价,把它改成)需要\(b_i\)的代价. 问使得这个括号序列成立所需要的最小代价. 思路1: 这个是正统的贪心. 首先我们假设所有的位置上都是),那么我们在从左向右扫描的途中会发现一些问题. 比如:我们原来的序列是(??),现在假设成了())),那么在第三个字符处就会发现我们的打开的左括号数量为\(-1\),这是肯定不行的,所以我们必须把第二个字符或者第三个字符改成左括号以把左…

题目链接 给一个字符串, 由( ) 以及? 组成, 将?换成( 或者 ) 组成合法的括号序列, 每一个?换成( 或者 ) 的代价都不相同, 问你最小代价是多少, 如果不能满足输出-1. 弄一个变量num, 如果是( 那么num++,如果是)那么num--. 如果碰到?, 那么先将这个地方弄成), 然后把make_pair(b-a, i)加到一个队列里面.a是换成左括号的值, b是换成右括号的值, i是这个位置的坐标. 如果遇到num小于0的情况, 那么就将队首元素取出, 将这个位置换成左括号.…

题目链接 D. Least Cost Bracket Sequence time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output This is yet another problem on regular bracket sequences. A bracket sequence is called regular, if by inse…

D. Least Cost Bracket Sequence time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output This is yet another problem on regular bracket sequences. A bracket sequence is called regular, if by inserting…

传送门 题意 从左到右有n个连续的组,每一组有Li个括号,要么全是左括号,要么全是右括号,以及该组的每一个左括号翻成右括号, 或者右括号翻成左括号的花费Di.可以对这n个组的括号进行翻转,每一个括号都可以选择翻或者不翻,使整个括号序列是一个合法括号序列. 分析 首先读入的时候将所有左括号变成右括号,并将费用变成负费用 遍历组,每次计算当前需要多少左括号,条件是如果当前有n个括号,则左括号的个数需>=(n+1)/2.并且将该组括号入优先队列.取优先队列队首进行处理即可 trick 1.XTU的读入…

Least Cost Bracket Sequence CodeForces - 3D 题目描述 This is yet another problem on regular bracket sequences. A bracket sequence is called regular, if by inserting "+" and "1" into it we get a correct mathematical expression. For example,…

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…

C. Heap Operations time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Petya has recently learned data structure named "Binary heap". The heap he is now operating with allows the following…

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…

http://codeforces.com/problemset/problem/346/A 观察了一下,猜测和他们的最大公因数有关,除以最大公因数前后结果是不会变的. 那么怎么证明一定是有n轮呢?我猜就是因为现在至少有几个是互质的,所以总是可以构造出1?具体怎么证明呢?还是看看别人的思路吧…… 首先最终停止的状态一定是一个等差数列,这个是毫无疑问的.设首项为d,那么肯定停止于d,2d,3d,...,n,那么很显然d就是他们的最大公因数啊……对哦?! #include<bits/stdc++.h…