描述 Given an N x N square grid (2 <= N <= 15) and each grid has some beans in it. You want to write at most K (1 <= K <= 2N - 2) horizontal or vertical lines going across the entire grid, which to partition the grid into some piles. But you wan…

The link to problem:Problem - D - Codeforces   D. Range and Partition  time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input output: standard output Given an array a of n integers, find a range of values [x,y] (x≤y…

描述 FJ now is a postman of a small town in the hills. The town can be represented by a N×N matrix. Each field is represented with:-'K' character: a house;-'P' character: the post office;-'.' character: a pasture. Moreover, each field is assigned with…

一个二分partition算法,将整个数组分解为小于某个数和大于某个数的两个部分,然后递归进行排序算法. 法一: int partition(vector<int>&arr, int begin, int end){ int pivot = arr[begin]; // Last position where puts the no_larger element. int pos = begin; ; i!=end; i++){ if(arr[i] <= pivot){ pos+…

DZY Loves Partition 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5646 Description DZY loves partitioning numbers. He wants to know whether it is possible to partition n into the sum of exactly k distinct positive integers. After some thinking he f…

题目链接: Heap Partition Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge A sequence S = {s1, s2, ..., sn} is called heapable if there exists a binary tree T with n nodes such that every node is labelled with exactly one element from…

Heap Partition Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge A sequence S = {s1, s2, ..., sn} is called heapable if there exists a binary tree T with n nodes such that every node is labelled with exactly one element from the se…

E. Prairie Partition It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let's call that representation prairie partition of x. For example, the p…

二分查找时间复杂度 partition时间复杂度 O(n) = O(n) + O(n/2) + O(n/4)+.... 然后用等比求和公式得出是O(2n),即O(n)…

题意:计算机中有一些固定大小的内存,内存越大,处理速度越快.对于一个程序,加入不同的内存空间,处理所需时间不同.现给出m个内存空间,n个程序,对于每个程序程序,有k组数据(s,t),分别表示当程序 i 在某个内存环境s下对应的运行时间t.当有一个内存空间为si,一个程序的2组数据a,b,满足as<=si<bs,那么这个程序放到该内存中的运行时间是at(取下界).已知不同的内存空间可以同时运行,同一个内存空间的程序在某一时刻只能有一个在运行.又知道一个程序从提交给计算机到运行结束这段时间为“运行…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5646 题意:将n分成k个正整数之和,要求k个数全部相同:并且这k个数的乘积最大为多少?结果mod 1e^9+7; 思路:由于是mod,不能通过模拟进行比较来判断是否为最优解:那么我们就必须直接计算出这个最优解的序列: 由于当a <= b-2时(相等时表示中间空一位),a*b < (a+1)*(b-1);所以最优解序列要不就是一串连续的序列,要不就是中间空一位,分成两段连续的序列: 因为如果存在空格超过…

描述 有个脑筋急转弯是这样的:有距离很近的一高一低两座桥,两次洪水之后高桥被淹了两次,低桥却只被淹了一次,为什么?答案是:因为低桥太低了,第一次洪水退去之后水位依然在低桥之上,所以不算“淹了两次”.举例说明:假定高桥和低桥的高度分别是5和2,初始水位为1第一次洪水:水位提高到6(两个桥都被淹),退到2(高桥不再被淹,但低桥仍然被淹)第二次洪水:水位提高到8(高桥又被淹了),退到3.没错,文字游戏.关键在于“又”的含义.如果某次洪水退去之后一座桥仍然被淹(即水位不小于桥的高度),那么下次洪水来临水…

题意:给你n个数字s1~sn,要你把它们组成一棵棵二叉树,对这棵二叉树来说,所有节点来自S,并且父节点si<=子节点sj,并且i<j,问你树最少几棵二叉数.树 思路:贪心.我们往multiset加还能加子节点的节点,二分查找一个个大于等于当前插入节点的节点,然后插入,若找不到则重新建一棵树. 没想到set自带lower_bound(),第一次迭代器遍历TLE就想着手动写二分...然后发现自带二分... 代码: #include<iostream> #include<stdio…

描述 FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID num…

描述 Farmer John is well known for his great cows. Recently, the cows have decided to participate in the Incredible Cows Puzzle Contest (ICPC). Farmer John wants to divide the cows into two teams, and he wants to minimize the difference of Puzzle Solvi…

很多笔试面试都喜欢考察快排,叫你手写一个也不是啥事.我很早之前就学了这个,对快速排序的过程是很清楚的.但是最近自己尝试手写,发现之前对算法的细节把握不够精准,很多地方甚至只是大脑中的一个映像,而没有理解其真正的本质意图.于是今天结合了<数据结构>(严蔚敏),和<算法导论>进行一番探究. 首先先给出快速排序的严蔚敏版的实现(实际上这部分的partition也是算法导论里面思考题的实现方式,细节可能不一样): public class QuickSort implements Sort…

转自:http://blog.csdn.net/jewes/article/details/42970799 引言 Kafka中的Message是以topic为基本单位组织的,不同的topic之间是相互独立的.每个topic又可以分成几个不同的partition(每个topic有几个partition是在创建topic时指定的),每个partition存储一部分Message.借用官方的一张图,可以直观地看到topic和partition的关系. partition是以文件的形式存储在文件系统中…

http://poj.org/problem?id=2318 TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10178   Accepted: 4880 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child…

博文链接:http://haoyuanliu.github.io/2016/12/18/Partition%E7%AE%97%E6%B3%95%E5%89%96%E6%9E%90/ 对,我是来骗访问量的!O(∩_∩)O~~ partition算法从字面上就非常好理解,就是分割算法嘛!简单讲就是可以把数组按照一定的分成几个部分,其中最常见的就是快速排序中使用的partition算法,这是一个二分partition算法,将整个数组分解为小于某个数和大于某个数的两个部分,然后递归进行排序算法. 上述只…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10281   Accepted: 4924 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

topic中partition存储分布 Topic在逻辑上可以被认为是一个queue.每条消费都必须指定它的topic,可以简单理解为必须指明把这条消息放进哪个queue里.为了使得 Kafka的吞吐率可以水平扩展,物理上把topic分成一个或多个partition,每个partition在物理上对应一个文件夹,该文件夹下存储 这个partition的所有消息和索引文件.partiton命名规则为topic名称+有序序号,第一个partiton序号从0开始,序号最大值为partitions数量减…

<Algorithms Unlocked>是 <算法导论>的合著者之一 Thomas H. Cormen 写的一本算法基础,算是啃CLRS前的开胃菜和辅助教材.如果CLRS的厚度让人望而生畏,这本200多页的小读本刚好合适带你入门. 书中没有涉及编程语言,直接用文字描述算法,我用 JavaScript 对书中的算法进行描述. 二分查找 在排好序的数组中查找目标值x.在p到r区间中,总是取索引为q的中间值与x进行比较,如果array[q]大于x,则比较p到q-1区间,否则比较q+1到…

Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John i…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14433   Accepted: 6998 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13262   Accepted: 6412 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

POJ_2318_TOYS&&POJ_2398_Toy Storage_二分+判断直线和点的位置 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them.…

转载自:  https://yq.aliyun.com/ziliao/65771 参考:  Kafka集群partition replication默认自动分配分析    如何为kafka选择合适的partitions 1.前言 一个商业化消息队列的性能好坏,其文件存储机制设计是衡量一个消息队列服务技术水平和最关键指标之一. 下面将从Kafka文件存储机制和物理结构角度,分析Kafka是如何实现高效文件存储,及实际应用效果. 2.Kafka文件存储机制 Kafka部分名词解释如下: Broker…

题目: Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toy…

Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing…

消息的存储原理: 消息的文件存储机制: 前面我们知道了一个 topic 的多个 partition 在物理磁盘上的保存路径,那么我们再来分析日志的存储方式.通过 [root@localhost ~]# ls /tmp/kafka-logs/firstTopic-1/命令找到对应 partition 下的日志内容:       00000000000000000000.index 00000000000000000000.log        00000000000000000000.timein…