学习当前较主流的开源云基础设施管理软件by Ruiy summarize publish; 我擦,有不时候Ruiy干事也就那吊风格,啥事也就那么随口一说,你们太sensitivity,同网络延迟对存储的性能依赖一样,貌似还有过之而无不及; open source cloud computing(openStack,cloudStack,openNebula,Eucalyptus)及paas/saas/iass于2014/05/24 running mechanism; 熟悉了当前打多少主流的云…

Open Source Project for iOS 所有和iOS相关的Open Source Project的汇总. 功能点 开源项目   iOS Gallery RMGallery https://github.com/robotmedia/RMGallery FGallery https://github.com/gdavis/FGallery-iPhone KYElegantPhotoGallery https://github.com/KittenYang/KYElegantPhot…

http://www.ibm.com/developerworks/cn/opensource/os-cn-cas/…

Kong for Enterprise | Kong - Open-Source API and Microservice Management Layer undefined…

Docker 应用容器平台 “电源管理和虚拟化以相同的方式允许我们从服务器利用率中获取最大的利益.如何真正的解决虚拟化,这世界第一难题仍然是普遍存在的.Docker 自从 2013 年开源以来,刚好在这开源改革的关键时刻适应了时势的发展,提供扩展构建块,允许类似我们这样的工程师和架构师能够以亘古未有的方式进行分布式平台的扩 展.”—Richard Morrell, Senior software engineer Petazzoni on the breathtaking growth of D…

1.Eucalyptus: Eucalyptus is a Linux-based software architecture that implements scalable private and hybrid clouds within your existingIT infrastructure. Eucalyptus allows you to use your own collections of resources (hardware, storage, and network)…

E. Caramel Clouds time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output It is well-known that the best decoration for a flower bed in Sweetland are vanilla muffins. Seedlings of this plant need…

Service discovery is a key component of most distributed systems and service oriented architectures. The problem seems simple at first: How do clients determine the IP and port for a service that exist on multiple hosts? Usually, you start off with s…

https://www.linkedin.com/pulse/open-source-cybersecurity-infrastructure-adrian/ The increased maturity and level of support of open source solutions make the deployment of an open-source based security architecture a potentially viable solution for m…

[CF833E]Caramel Clouds(线段树) 题面 CF 洛谷 题解 首先把区间一段一段分出来,那么只有四种情况. 要么没有被任何一朵云被覆盖,那么直接就会产生这一段的贡献. 要么被一朵云覆盖,要么被两朵云覆盖. 要么被三朵及以上的云所覆盖,那么这段的贡献永远取不到. 对于每朵云预处理出只被其覆盖的区间长度\(len[i]\),这样子就能处理只选择一朵云的贡献了. 现在考虑如何处理选择两朵云. 这里有两种情况. 第一种是两朵云无交,那么贡献就是\(len[i]+len[j]\). 否则…