Serval and Parenthesis Sequence CodeForces - 1153C

题目大意:一个字符串只含有? ( ),?可以变成 ) 或者 ( ,将字符串中所有的?变成) 或者 ( 使得字符串合法. 合法就是让括号配对,并且不可以提前结束比如:()()这样是不合法的. 题解:既然不能提前结束,那第一个字符必须和最后一个匹配,所以我们只要关注从2~n-1就可以了. 我们可以正着扫一遍,在任何一个位置")"的数目必须小于等于"("+"?". 然后再倒着扫一遍,同样地,在任何位置的"("的数目必须小于等于&qu…

C. Serval and Parenthesis Sequence 【括号匹配】 Codeforces Round #551 (Div. 2)

冲鸭,去刷题:http://codeforces.com/contest/1153/problem/C C. Serval and Parenthesis Sequence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Serval soon said goodbye to Japari kindergarten, and b…

CF1153C Serval and Parenthesis Sequence

题目地址:CF1153C Serval and Parenthesis Sequence 思路:贪心如果有解,那么 $s_0 = ($ && $s_{n-1} = )$ && $n % 2 = 0$ . 如果有解,那么 $s_1$ ~ $s_{n-2}$ 为一个合法的括号序列. 那么已知的 $($ 和 $)$ 个数不能超过一半接下来贪心:如果有解,那么一定有一组解是把 $?$ 中左边一部分填成 $($ ,右边一部分填成 $)$ ,…

Serval and Parenthesis Sequence【思维】

Serval and Parenthesis Sequence 题目链接(点击) Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School. In his favorite math class, the teacher taught him the following interesting definitions. A parenthesis sequence is…

cf-Round551-Div2-C. Serval and Parenthesis Sequence（贪心）

题目链接:http://codeforces.com/contest/1153/problem/C 题意:给定由'(',')','?'组成的字符串,问是否能将其中的?全部换成'(‘,’)'使得字符串的任意非空真字串不构成正确的括号表达式,而整个字符串构成括号表达式,其中正确的括号表达式是指通过插入'1','+'能构成算术式. 思路:我们记'('为-1,')'为1,显然所有字串应满足前面的和<0,字串等于0的话就不满足字串不构成正确的括号表达式了,且整个字符串的和=0(题目可能出现'((((??'…

cf——C. Serval and Parenthesis Sequence

括号正确匹配问题,应该不难 #include <iostream> #include <cstring> #include <string> #include <map> #include <set> #include <algorithm> #include <fstream> #include <cstdio> #include <cmath> #include <stack> #i…

Increasing Sequence CodeForces - 11A

Increasing Sequence CodeForces - 11A 很简单的贪心.由于不能减少元素,只能增加,过程只能是从左到右一个个看过去,看到一个小于等于左边的数的数就把它加到比左边大,并记录加的次数. 错误记录: 但是很容易错...以前错了4次..过几个月来再做还是不能1A... 比如下面这个有很明显错误的程序 #include<cstdio> int n,d,last,now,ans; int main() { int i; scanf("%d%d",&…

Almost Regular Bracket Sequence CodeForces - 1095E （线段树，单点更新，区间查询维护括号序列）

Almost Regular Bracket Sequence CodeForces - 1095E You are given a bracket sequence ss consisting of nn opening '(' and closing ')' brackets. A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expressio…

AC日记——The Child and Sequence codeforces 250D

D - The Child and Sequence 思路: 因为有区间取模操作所以没法用标记下传: 我们发现,当一个数小于要取模的值时就可以放弃: 凭借这个来减少更新线段树的次数: 来,上代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 100005 #define ll lo…

C. Vasily the Bear and Sequence Codeforces 336C(枚举，思维)

C. Vasily the Bear and Sequence time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasily the bear has got a sequence of positive integers a1, a2, ..., *a**n*. Vasily the Bear wants to write o…

Prefix Product Sequence CodeForces - 487C (数论,构造)

大意: 构造一个[1,2,...n]的排列, 使得前缀积模n为[0,1,...,n-1]的排列这种构造都好巧妙啊, 大概翻一下官方题解好了对于所有>=6的合数$n$, 有$(n-1)! \equiv 0 \space (mod \space n)$, 一定不成立对于素数可以构造$[1,\frac{2}{1},\frac{3}{2},\frac{4}{3},...,\frac{n-1}{n-2},n]$, 或者构造$[1,g,g^{-2},g^3,g^{-4},...,n]$, $g$为$n…

【Codeforces】Codeforces Round #551 (Div. 2)

Codeforces Round #551 (Div. 2) 算是放弃颓废决定好好打比赛好好刷题的开始吧 A. Serval and Bus 处理每个巴士最早到站且大于t的时间 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ')…

codeforces 1153 D

cf-551-div2-D C. Serval and Parenthesis Sequence 题意:给定由'(',')','?'组成的字符串,问是否能将其中的?全部换成'(‘,’)'使得字符串的任意非空真字串不构成正确的括号表达式,而整个字符串构成括号表达式,其中正确的括号表达式是指通过插入'1','+'能构成算术式. 思路:我们记'('为-1,')'为1,显然所有字串应满足前面的和<0,字串等于0的话就不满足字串不构成正确的括号表达式了,且整个字符串的和=0(题目可能出现'((((??'这…

Codeforces Round #551 (Div. 2) A-E

A. Serval and Bus 算出每辆车会在什么时候上车, 取min即可 #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #include<iostream> #define ll long long #define M 101 #define mmp make_pair using namespace std; int read() { int…

Codeforces Round #551 (Div. 2) A~E题解

突然发现上一场没有写,那就补补吧本来这场应该5题的,结果一念之差E fail了 A. Serval and Bus 基本数学不解释,假如你没有+1 -1真的不好意思见人了 #include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> using namespace std; con…

CF-551：部分题目总结

题目链接:http://codeforces.com/contest/1153 A .Serval and Bus pro:给出n种公交车的首班车时间和两班车之间的时间间隔,找t时间以后的第一辆车是第几种车 sol:对于每种车,找到t时间以后的第一班车的时间,计算这个时间和t的差距,然后找离t最近的暴力 #include "bits/stdc++.h" using namespace std; ; const int INF = 0x3f3f3f3f; int ans, k = IN…

2016年湖南省第十二届大学生计算机程序设计竞赛---Parenthesis（线段树求区间最值）

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…

湖南省第十二届大学生计算机程序设计竞赛 G Parenthesis

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…

2016年省赛G题， Parenthesis

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…

CSU 1809 Parenthesis（线段树+前缀和）

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…