BZOJ4355: Play with sequence https://lydsy.com/JudgeOnline/problem.php?id=4355 分析: 模板题. 把\(2\)操作看成先区间加再区间取\(max\). 查询转化成求最小值个数. 需要维护\(3\)个标记. 这里我比较naive地维护了三个标记,多维护了一个最小值需要加的值. 然后下传的时候对左右儿子判断是否应该下传. 实际上不需要,直接修改最小值即可. 代码: #include <cstdio> #include &…

传送门 维护区间覆盖成非负数,区间变成max(xi+a,0)" role="presentation" style="position: relative;">max(xi+a,0)max(xi+a,0),询问区间中0" role="presentation" style="position: relative;">00的个数. 由于每次操作之后区间中都是非负数,因此相当于只用维护最小值和最小值…

题意 题目链接 Sol 传说中的吉司机线段树??感觉和BZOJ冒险那题差不多,就是强行剪枝... 这题最坑的地方在于对于操作1,$C >= 0$, 操作2中需要对0取max,$a[i] >= 0$,这不就是统计最小值出现的次数么?? 按照套路 维护好区间赋值标记 / 区间加法标记 / 区间max标记 / 区间最小值 / 区间最小值出现的次数 / 区间次小值 对于第二个操作就拆成区间加 和 区间max 区间max是一个很神奇的操作 设当前加入的数为val 若val>=mn,那该操作对该区间…

算导: 核算法 给每种操作一个摊还代价(是手工定义的),给数据结构中某些东西一个“信用”值(不是手动定义的,是被动产生的),摊还代价等于实际代价+信用变化量. 当实际代价小于摊还代价时,增加等于差额的信用: 当实际代价大于摊还代价时,减少等于差额的信用. 显然总摊还代价等于总实际代价+总信用变化量. 如果信用变化量始终>=0,那么总摊还代价给出总实际代价的一个上界:设法证明信用变化量始终>=0 势能法 对整个数据结构定义一个“势”函数$\Phi$ 定义一个操作的摊还代价为实际代价加上势的变化量…

题意: 已知\(n\)个数字,进行以下操作: \(1.\)区间\([L,R]\) 赋值为\(x\) \(2.\)区间\([L,R]\) 赋值为\(max(a[i] + x, 0)\) \(3.\)区间\([L,R]\) 询问\(0\)个数 已知初始值\(\geq 0\),\(x\geq0\). 思路: 吉司机线段树. 操作\(1\)可以直接打覆盖标记. 操作\(2\)可以分为两步:区间加\(x\),然后取区间\(max(a[i],0)\). 操作\(3\)只要维护最小值的个数,因为不管怎么操作最…

Segment Tree Beats 区间最值问题 线段树一类特殊技巧! 引出:CF671C Ultimate Weirdness of an Array 其实是考试题,改题的时候并不会区间取最值,区间求和,之后秉承着好好学习的态度,学习了Segment tree Beats 套路是维护出区间最小值和次小值,以及区间最小值数量.之后再维护出题目中需要的东西就好了.之后怎么处理呢,如果我们需要维护出区间和x取max,那么,如果x<=minn[rt],那么直接return;如果x<minx[rt]…

[BZOJ4355]Play with sequence Description 维护一个长度为N的序列a,现在有三种操作: 1)给出参数U,V,C,将a[U],a[U+1],...,a[V-1],a[V]都赋值为C. 2)给出参数U,V,C,对于区间[U,V]里的每个数i,将a[i]赋值为max(a[i]+C,0). 3)给出参数U,V,输出a[U],a[U+1],...,a[V-1],a[V]里值为0的数字个数. Input 第一行包含两个正整数N,M(1<=N,M<=300000),分别…

题目描述 维护一个长度为N的序列a,现在有三种操作: 1)给出参数U,V,C,将a[U],a[U+1],...,a[V-1],a[V]都赋值为C. 2)给出参数U,V,C,对于区间[U,V]里的每个数i,将a[i]赋值为max(a[i]+C,0). 3)给出参数U,V,输出a[U],a[U+1],...,a[V-1],a[V]里值为0的数字个数. 输入 第一行包含两个正整数N,M(1<=N,M<=300000),分别表示序列长度和操作个数. 第二行包含N个整数,其中第i个数表示a[i](0&l…

https://www.lydsy.com/JudgeOnline/problem.php?id=4355 维护一个长度为N的序列a,现在有三种操作: 1)给出参数U,V,C,将a[U],a[U+1],...,a[V-1],a[V]都赋值为C. 2)给出参数U,V,C,对于区间[U,V]里的每个数i,将a[i]赋值为max(a[i]+C,0). 3)给出参数U,V,输出a[U],a[U+1],...,a[V-1],a[V]里值为0的数字个数. 吉如一论文板子题(当然所有的操作都来自论文拼起来的那…

oracle创建序列化: CREATE SEQUENCE seq_itv_collection            INCREMENT BY 1  -- 每次加几个              START WITH 1399       -- 从1开始计数              NOMAXVALUE        -- 不设置最大值              NOCYCLE               -- 一直累加,不循环              CACHE 10; oracle修改序列…

功能:备份存储过程,视图,函数触发器,Sequence序列号等准备工作:--1.创建文件夹 :'E:/OracleBackUp/ProcBack';--文本存放的路径--2.执行:create or replace directory MyProcBakPath as 'E:/OracleBackUp/ProcBack';--3.赋权限:sqlplus /nologconn user/pswd as sysdbagrant select on DBA_OBJECTS to user;--4.创建…

环境:Oracle 11.2.0.4 DG 故障现象: 客户在备库告警日志中发现GAP sequence提示信息: Mon Nov 21 09:53:29 2016 Media Recovery Waiting for thread 1 sequence 12034 Fetching gap sequence in thread 1, gap sequence 12034-12078 Mon Nov 21 09:55:20 2016 FAL[client]: Failed to request…

The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

Check whether the original sequence org can be uniquely reconstructed from the sequences in seqs. The org sequence is a permutation of the integers from 1 to n, with 1 ≤ n ≤ 104. Reconstruction means building a shortest common supersequence of the se…

Given a binary tree, find the length of the longest consecutive sequence path. The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from p…

Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree. You may assume each number in the sequence is unique. Follow up: Could you do it using only constant space complexity? 这道题让给了我们一个一维数组,让我们…

Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example, Given [100, 4, 200, 1, 3, 2], The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. Your algorithm should run i…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

传送门 Description Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad. Let's consider the following algorithm of generating a…

传送门 NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 1585    Accepted Submission(s): 688 Description NanoApe, the Retired Dog, has returned back to prepare for for the…

传送门 NanoApe Loves Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others) Total Submission(s): 1323    Accepted Submission(s): 521 Description NanoApe, the Retired Dog, has returned back to prepare for the Natio…

系列文章 [UML]UML系列——用例图Use Case [UML]UML系列——用例图中的各种关系(include.extend) [UML]UML系列——类图Class [UML]UML系列——类图class的关联关系(聚合.组合) [UML]UML系列——类图class的依赖关系 [UML]UML系列——类图class的泛化关系 [UML]UML系列——类图class的实现关系Realization [UML]UML系列——包图Package [UML]UML系列——活动图activity…

                                    Bracket Sequence Time Limit: 3000MS   Memory Limit: 65536KB   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description There is a sequence of brackets, which supports two kinds of operations.1. we c…

转载链接:http://www.cnblogs.com/zhangyoushugz/archive/2012/11/09/2762720.html 众所周知,在之前的sqlserver版本中,一般采用GUID或者identity来作为标示符,但是identity是一个表对象,只能保证在一张表里面的序列,当我们遇到以下情况时, 表1 ID 订单类型 价格 1 火车票 200 4 飞机票 2000 5 船票 600 表2 ID 订单类型 价格 2 酒店 400 3 公园门票 170 6 租车费用 3…

原题连接:https://www.patest.cn/contests/pat-a-practise/1051 题目: Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop seq…

Check whether the original sequence org can be uniquely reconstructed from the sequences in seqs. The org sequence is a permutation of the integers from 1 to n, with 1 ≤ n ≤ 104. Reconstruction means building a shortest common supersequence of the se…

<plugins> <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-war-plugin</artifactId> <version>2.1-beta-1</version> <configuration> <warName>admingift</warName> <webRe…

Sequence Project Showplan Operator 序列映射运算符 序列映射运算符会从一个已经排序的集合里通过不停添加集合里的列执行计算. 运算符根据一个或多个列的值把输入集合分为多个片段.然后运算符一次输出一个片段. 这些列在序列映射运算符里会被显示为参数. SQL Server支持四种类型函数:RANK, DENSE_RANK, ROW_NUMBER, NTILE 序列映射会生成有(一个序列映射)和(两个片段)的执行计划 The Sequence Project opera…

Farey Sequence 题意:给定一个数n,求在[1,n]这个范围内两两互质的数的个数.(转化为给定一个数n,比n小且与n互质的数的个数) 知识点: 欧拉函数: 普通求法: int Euler(int n) { int ans=n; for(int i=0;i<cnt&&prime[i]<=n;i++) { if(n%prime[i]==0) { ans=ans-ans/prime[i]; while(n%prime[i]==0) n/=prime[i]; } } if(…