Golang FromXToGo micro - A microservice toolkit Other Easy parsing of Excel spreadsheet format with Swift's Codable protocols Front-End Performance Checklist 2019 [PDF, Apple Pages, MS Word] Questions for a new technology Mobile iOS Performance Trick…

梦想MxWeb3D协同设计平台 2019.02.28更新 SDK开发包下载地址: http://www.mxdraw.com/ndetail_10130.html 在线演示网址: http://www.mxdraw.com:3000/ 1.  编写API CHM帮助文档 2.  增加保存和回退功能 3.  增加图纸搜索位置设置函数,addFileSearchPath 4.  完善绘图函数 5.  优化后台缓存文件大小 6.  优化前后台代码 7.  增加清除函数…

上一篇:OI生涯回忆录 2017.9.10~2018.11.11 一次逆风而行的成功,是什么都无法代替的 ………… 历经艰难 我还在走着 一 NOIP之后,全机房开始了省选知识的自学. 动态DP,LCT,后缀数组,后缀自动机,多项式 NOIP获得全省第六名好成绩的我, 自以为省队纳入囊中只是时间问题 学习效率并不高效. 12月培训由于自学过 听课效果很好,实际上并没有收获和总结太多. 当时问郭神算法的问题, 他说: “算法不在多,而在于精” 事实证明, 此言得之. 太多的自满,太多的不深刻 太多…

Golang Roadomatic: Node vs. Go Quick Guide to Golang for Java Developers 3 Go Gotchas Web Choosing a Front End Framework: Angular vs. Ember vs. React Microservices Decoded: Best Practices and Stacks Mastering ASP.NET 5 without growing a beard 5 Tips…

Web Getting Started With Vapor: A Swift Web Framework Front-end vs Back-end vs Network Performance Spring Boot Application Connect to LDAP Userstore Other Keeping Your Code Clean While Logging Coding Best Practices and Solutions Writing Clean Code: A…

Mobile Android SDK: Working with Picasso View Debugging in Xcode 6 5 Common C# tasks in Apple Swift Programming Language Beyond JSON: Spearal Serialization Protocol for iOS .NET IDisposable: What Your Mother Never Told You About Resource Deallocation…

Other Back To Basics: Hashtables – Part2 How To Make A Game Part 1:Picking a Framework Modern Garbage Collection in Theory and Practice Operation Performance Evaluation Get rid of your StringUtils! Refactoring with Go Fmt .NET Building Performance Me…

传送门 题意:给一个串,每个位置有一个权值,当S[s...s+len−1]=S[t...t+len−1]&&S[s...s+len]̸=S[t..t+len]S[s...s+len-1]=S[t...t+len-1]\&\&S[s...s+len]\not=S[t..t+len]S[s...s+len−1]=S[t...t+len−1]&&S[s...s+len]̸​=S[t..t+len]时我们称两个字串是"lenlenlen"相似的,…

传送门 fftfftfft菜题. 题意简述:给一个数列aia_iai​,对于i=1→ni=1\rightarrow ni=1→n求出ansi=∑i<jai(i−j)2−∑i>jai(i−j)2ans_i=\sum_{i<j}\frac{a_i}{(i-j)^2}-\sum_{i>j}\frac{a_i}{(i-j)^2}ansi​=∑i<j​(i−j)2ai​​−∑i>j​(i−j)2ai​​ 思路: 考虑分开求减号前后的两组和. 前面的直接是一个卷积的形式,后面的可以…

传送门 跟这道题差不多. 只不过是让权值小的儿子做权值大的儿子的父亲而已. 代码…