split

题解 P1503 【鬼子进村】

天涯浪子 提交于 2020-02-28 09:12:02
提供两种做法,fhq_treap和set 思路: 首先在平衡树中加入0节点和n+1节点,是左右边界 'D x':在平衡树中加入x节点(为什么是加入而不是删除,我后面会详细讲到) 'R':既然是上一个点恢复了,而样例的最后一个询问为我们贴心地考虑了连续恢复的情况,那么就是要维护一个后进先出的 栈 来保存您删除的点啦~ 'Q x'查询x点的前驱pre上一个被删除的点),和x的后继suc下一个被删除的点,那么x能到达的点的个数就是suc-pre-1个(题目给的是一条编号连续的链,那么根据编号来查找个数就很方便啦^ _ ^) fhq_treap #include<bits/stdc++.h> using namespace std; #define rep(i,a,b) for(int i=a;i<=b;++i) #define dwn(i,a,b) for(int i=a;i>=b;--i) template <typename T> inline void rd(T &x){x=0;char c=getchar();int f=0;while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}x=f?-x:x;} const int N=5e4+10;

codewars:Split Strings

故事扮演 提交于 2020-02-26 17:45:36
codewars:Split Strings Instructions Complete the solution so that it splits the string into pairs of two characters. If the string contains an odd number of characters then it should replace the missing second character of the final pair with an underscore (’_’). Examples: solution(‘abc’) # should return [‘ab’, ‘c_’] solution(‘abcdef’) # should return [‘ab’, ‘cd’, ‘ef’] solution(’’) # should return [] 我的代码 这题我是用了递归的解法,递归就在于缩小问题的规模,同时有结束的最小条件。我个人觉得一遇到递归的问题脑子会先短路一下,这题最后是挺简单的,但我还是花了时间去捋的。 def solution ( s ) : if len ( s ) <= 1 : if len ( s ) == 0 : return [ ] else : return [ s + '_' ] return [ s

stratified splitting the data

一个人想着一个人 提交于 2020-02-26 08:24:33
问题 I have a large data set and like to fit different logistic regression for each City, one of the column in my data. The following 70/30 split works without considering City group. indexes <- sample(1:nrow(data), size = 0.7*nrow(data)) train <- data[indexes,] test <- data[-indexes,] But this does not guarantee the 70/30 split for each city. lets say that I have City A and City B, where City A has 100 rows, and City B has 900 rows, totaling 1000 rows. Splitting the data with above code will give

stratified splitting the data

我们两清 提交于 2020-02-26 08:23:45
问题 I have a large data set and like to fit different logistic regression for each City, one of the column in my data. The following 70/30 split works without considering City group. indexes <- sample(1:nrow(data), size = 0.7*nrow(data)) train <- data[indexes,] test <- data[-indexes,] But this does not guarantee the 70/30 split for each city. lets say that I have City A and City B, where City A has 100 rows, and City B has 900 rows, totaling 1000 rows. Splitting the data with above code will give

How to remove WindowsPath and parantheses from a string [duplicate]

妖精的绣舞 提交于 2020-02-26 04:28:45
问题 This question already has an answer here : Reference - What does this regex mean? (1 answer) Closed 4 days ago . I need to remove WindowsPath( and some of the closing parentheses ) from a directory string. x= (WindowsPath('D:/test/1_birds_bp.png'),WindowsPath('D:/test/1_eagle_mp.png')) What I need to have is x= ('D:/test/1_birds_bp.png', 'D:/test/1_eagle_mp.png') I don't know how to do multiple strings at once. by following @MonkeyZeus I tried; y = [re.sub("(?<=WindowsPath\(').*?(?='\))",'',a

How to remove WindowsPath and parantheses from a string [duplicate]

假装没事ソ 提交于 2020-02-26 04:27:48
问题 This question already has an answer here : Reference - What does this regex mean? (1 answer) Closed 4 days ago . I need to remove WindowsPath( and some of the closing parentheses ) from a directory string. x= (WindowsPath('D:/test/1_birds_bp.png'),WindowsPath('D:/test/1_eagle_mp.png')) What I need to have is x= ('D:/test/1_birds_bp.png', 'D:/test/1_eagle_mp.png') I don't know how to do multiple strings at once. by following @MonkeyZeus I tried; y = [re.sub("(?<=WindowsPath\(').*?(?='\))",'',a

Hbase的region合并与拆分详解

不羁的心 提交于 2020-02-25 23:32:05
1、region 拆分机制 region中存储的是大量的rowkey数据 ,当region中的数据条数过多的时候,直接影响查询效率.当region过大的时候.hbase会拆分region , 这也是Hbase的一个优点 . HBase的region split策略一共有以下几种: 1、 ConstantSizeRegionSplitPolicy 0.94版本前默认切分策略 当region大小大于某个阈值(hbase.hregion.max.filesize=10G)之后就会触发切分,一个region等分为2个region。 但是在生产线上这种切分策略却有相当大的弊端:切分策略对于大表和小表没有明显的区分。阈值(hbase.hregion.max.filesize)设置较大对大表比较友好,但是小表就有可能不会触发分裂,极端情况下可能就1个,这对业务来说并不是什么好事。如果设置较小则对小表友好,但一个大表就会在整个集群产生大量的region,这对于集群的管理、资源使用、failover来说都不是一件好事。 2、 IncreasingToUpperBoundRegionSplitPolicy 0.94版本~2.0版本默认切分策略 切分策略稍微有点复杂,总体看和ConstantSizeRegionSplitPolicy思路相同,一个region大小大于设置阈值就会触发切分

Is it possible to divide the value of a list in a certain amount of times

谁都会走 提交于 2020-02-25 22:37:50
问题 Hi I am trying to split the values in a (y2) list into (inv) amount of parts. x = [0,100,200,300,400,500,600,700,800,900,1000,1100] y2 = [58,55,49,12,6,5,4,4.5,35,48,56,58] interval = 5 inv = (x[0] + x[1])/interval I will like to split y2 into inv parts so my answer should be: [58 57.85 57.7 57.55 57.4 57.25 57.1 56.95 56.8 56.65 56.5 56.35 56.2 56.05 55.9 55.75 55.6 55.45 55.3 55.15 55...…….49...…………..12...….6 ………...5...……..] I will like to mention that inv will always be (x[0] + x[1])

Is it possible to divide the value of a list in a certain amount of times

六眼飞鱼酱① 提交于 2020-02-25 22:37:07
问题 Hi I am trying to split the values in a (y2) list into (inv) amount of parts. x = [0,100,200,300,400,500,600,700,800,900,1000,1100] y2 = [58,55,49,12,6,5,4,4.5,35,48,56,58] interval = 5 inv = (x[0] + x[1])/interval I will like to split y2 into inv parts so my answer should be: [58 57.85 57.7 57.55 57.4 57.25 57.1 56.95 56.8 56.65 56.5 56.35 56.2 56.05 55.9 55.75 55.6 55.45 55.3 55.15 55...…….49...…………..12...….6 ………...5...……..] I will like to mention that inv will always be (x[0] + x[1])

分布式计算框架

北战南征 提交于 2020-02-25 22:21:23
MapReduce 简介 概念 面向批处理的分布式计算框架 一种编程模型: MapReduce程序被分为Map(映射)和Reduce(化简)阶段 核心思想 分而治之, 并行计算 移动计算而非移动数据 特点 MapReduce有几个特点: 移动计算而不移动数据:分布式计算,计算跟着数据走,数据存放在哪就在哪里进行计算,极大的减少了IO的开销。 良好的扩展性:分布式计算框架拥有相当良好的扩展性,随着节点数量的增加,单个节点的计算量减小,整体的计算能力近乎线性的递增。 高容错性:计算任务失败,会自动恢复计算,拥有良好的容错性。 状态监控:提交任务之后,任务具体执行在哪个节点,具体执行到哪个阶段,在后台或者监控界面,我们都能实时的进行监控。 适合于海量数据的离线批处理:MapReduce在设计之初的目标就是面向离线批处理,特别是大吞吐量的离线处理场景,更适合于MapReduce。 降低了分布式编程的门槛:大部分操作MapReduce已经实现,我们仅仅需要在特定的部分编写我们自己的业务逻辑,极大的减少了工作量,同时也降低了编程的门槛。 MR原理 1. 作业(job):作业是客户端提交请求执行的一个单元,它包括数据、计算框架以及一些配置信息等。 2. 任务(task):是作业细分之后的细分工作单元,如MapReduce中的Map Task和Reduce Task。