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This is a sample agenda for a sprint planning meeting. Depending on your context you will have to change the details, just make sure the outcomes stay the same. Meeting purpose: Plan and prepare for the upcoming sprintMeeting duration: ca. 1 hour for…
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原文链接:http://domaintree.me/?p=1037 By Robert Thibodeau –  Starting a business can be a very daunting adventure if a proper plan is not put in place. Most entrepreneurs start up their businesses without putting adequate plans in place to succeed. No wo…
Employment Planning Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5292 Accepted Submission(s): 2262 Problem DescriptionA project manager wants to determine the number of the workers needed in eve…
recourse: "Software Engineering", Ian Sommerville Keywords for this chapter: planning scheduling cost estimation Project planning takes place at three stages in a project life cycle: At the proposal stage (When you are bidding for a contract to…
原创博文:转载请标明出处:http://www.cnblogs.com/zxouxuewei 最近有不少人询问有关MoveIt!与OMPL相关的话题,但是大部分问题都集中于XXX功能怎么实现,XXX错误怎么解决.表面上看,解决这些问题的方法就是提供正确的代码,正确的编译方法,正确的运行步骤. 然而,这种解决方法只能解决这个特定的问题,而且解决之后我们也无法学到一些实际的东西.要想彻底明白,需要从源头入手,也就是说,不要问"MoveIt! 怎么把机械手从空间一个点移到另一个点?",而是要…
1. General review. Professor Webster published this article in Urban Planning Forum, one of the top Chinese urban planning academic journals, before he went to take the office as Dean of the Faculty of Architecture in the University of Hong Kong. Obv…
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本文将介绍FPGA中和时钟有关的相关概念,阅读本文前需要对时序收敛的基本概念和建立.保持关系有一定了解,这些内容可以在时序收敛:基本概念,建立时间和保持时间(setup time 和 hold time)中找到. 系列目录      时序收敛:基本概念     建立时间和保持时间(setup time 和 hold time)     OFFSET约束(OFFSET IN 和OFFSET OUT)     Clock Skew , Clock uncertainly 和 Period     特…
传送门 Description Let S be a number string, and occ(S,x) means the times that number x occurs in S. i.e. S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1. String u,w are matched if for each number i, occ(u,i)=occ(w,i) always holds. i.e. (1,2,2,1,3)≍(1,3,…
Period 题目大意:给定一个字符串,要你找到前缀重复了多少次 思路,就是kmp的next数组的简单应用,不要修正next的距离就好了,直接就可以跳转了 PS:喝了点酒用递归实现除法和取余了...结果tle不知道怎么回事... #include <iostream> #include <functional> #include <algorithm> #include <string> using namespace std; typedef int Po…
cargo插件,报错:failed to finish deploying within the timeout period [120000] 解决方法:配置timeout为0 <plugin> <groupId>org.codehaus.cargo</groupId> <artifactId>cargo-maven2-plugin</artifactId> <version></version> <configu…
/** * get period for last year * @param time * @return */ public static DatePeriodDTO getLastYear(long time) { Calendar calendar = Calendar.getInstance(); //get last year calendar.add(Calendar.YEAR, -1); int lastYear = calendar.get(Calendar.YEAR); ca…
今天碰到了一个查询异常问题,上网查了一下,感谢原创和译者 如果你使用的数据库连接类是 the Data Access Application Blocks "SqlHelper" 或者 SqlClient Class , 你在执行一个很费时的SQL 操作时候,可能就会碰到下面的超时异常. --------------------------- ---------------------------Timeout expired.  The timeout period elapsed…
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Tourism Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1051    Accepted Submission(s): 460 Problem Description Several friends are planning to take tourism during the next holiday. Th…
Period Time Limit: 3000MSMemory Limit: 30000K Total Submissions: 12089Accepted: 5656 Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefi…
[LA3026]Period 试题描述 For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 ≤ i ≤ N) we want to know the l…
Period Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2120    Accepted Submission(s): 1039 Problem Description For each prefix of a given string S with N characters (each character has an ASCI…
背景: 在最近开发中遇到一个问题,对一个数据库进行操作时,我采用64个并行的任务每个任务保证一个数据库连接对象:但是每个任务内部均包含有24个文件需要读取,在读取文件之后,我们需要快速将这24个文件批量入库到数据库中. 于是我这样开发我的程序: 主任务处理方式:最多允许64并行主任务: 主任务内部子任务采用串行方式:24个文件依次读取,和当前主任务均使用同一个数据库连接字符串. 每个主任务都需要24个文件入库到各自的物理分表中,采用的是串行读取文件资源,串行入库,没有能并行插入24个批处理文件到…
最近用Swift写了个小工具Scrum Planning Card,如果你也用scrum管理项目, 或许用这个工具可以提高你的工作效率. 另外欢迎提建议和反馈,谢谢. 欢迎关注我的微信公众号 Hope this helps, Michael…
Period Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2866    Accepted Submission(s): 1433 Problem Description For each prefix of a given string S with N characters (each character has an ASCI…
题目链接 找循环位数是奇数的数有多少个 这个自己很难写出来,完全不能暴力 维基百科链接 维基百科上面说的很好,上面的算法实现就好了. 就是上面的 Java程序: package project61; public class P64{ void run(){ int count = 0; int m = 0; int d = 1; int a0 = 0; int a = 0; int period = 0; for(int S = 2;S<10000;S++){ period = 0; m =…
如何设置适当的值并不轻易. 首先,假如要使用大量线程的话,ramp-up period 一般不要设置成零. 因为假如设置成零,Jmeter将会在测试的开始就建立全部线程并立即发送访问请求, 这样一来就很轻易使服务器饱和,更重要的是会隐性地增加了负载,这就意味着服务器将可能过载,不是因为平均访问率高而是因为所有线程的第一次并发访问而引起的不正常的初始访问峰值,可以通过Jmeter的聚合报告监听器看到这种现象.这种异常不是我们需要的,因此,确定一个合理的ramp-up period 的规则就是让初始…
UVAlive 3026 Period 题目: Period   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive…
Period 题意:一个长为N (2 <= N <= 1 000 000) 的字符串,问前缀串长度为k(k > 1)是否是一个周期串,即k = A...A;若是则按k从小到大的顺序输出k即周期数: Sample Input 3 aaa 12 aabaabaabaab 0   Sample Output Test case #1 2 2 3 3   Test case #2 2   2 6   2 9   3 12  4  题目其实是来自于LA的..挺好的一道题,用的是原版的kmp.. 写…
PTTFM-X467G-P7RH2-3Q6CG-4DMYB 数据中心版:PTTFM-X467G-P7RH2-3Q6CG-4DMYB   测试可用 开 发者 版:MC46H-JQR3C-2JRHY-XYRKY-QWPVM 企    业 版:R88PF-GMCFT-KM2KR-4R7GB-43K4B 标    准 版:B68Q6-KK2R7-89WGB-6Q9KR-QHFDW 工 作 组版:XQ4CB-VK9P3-4WYYH-4HQX3-K2R6Q WEB    版:FP4P7-YKG22-WGRV…