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Join the GitHub Game Off!
Game Bytes

This year’s Game Off theme is LEAPS AND BOUNDS.

You’re welcome to interpret the theme however you wish, but if you need some ideas to kickstart your creativity, maybe your game could:

  • Make your character evolve to overcome challenges
  • Add obstacles to test your hero’s ability to leap into action
  • Deduct points for anything going out of bounds
  • Duck, dip, dive, and dodge an enemy that’s changing or moving rapidly
  • Take place in the year 2020, the next leap year

Let your imagination run wild and create a game loosely-based or inspired by that theme using whatever game engines, libraries or languages you prefer before December 1st, 13:37 PT.

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AWS re:Invent is the largest global cloud computing conference bringing together 50,000+ technologists.
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gizmodus commented May 8, 2019

Hey Cube.Js team!
I get the following yarn warnings when installing dependencies. Can these be fixed? Thank you.

warning @cubejs-backend/server-core > @cubejs-backend/schema-compiler > joi@14.3.1: This module has moved and is now available at @hapi/joi. Please update your dependencies as this version is no longer maintained an may contain bugs and security issues.

warning @cubejs-backe
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【互联网一线大厂Java 工程师面试+学习指南】进阶知识完全扫盲:涵盖高并发、分布式、高可用、微服务等领域知识,作者风格幽默,看起来津津有味,把学习当做一种乐趣,何乐而不为,后端同学必看,前端同学我保证你也看得懂,看不懂你加我微信骂我渣男就好了。

  • Updated Nov 22, 2019
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一款基于代码生成器的JAVA快速开发平台,开源界“小普元”超越传统商业企业级开发平台!采用前后端分离架构:SpringBoot 2.x,Ant Design&Vue,Mybatis-plus,Shiro,JWT。强大的代码生成器让前后端代码一键生成,无需写任何代码! 引领新的开发模式(OnlineCoding模式-> 代码生成器模式-> 手工MERGE智能开发),帮助Java项目解决70%的重复工作,让开发更多关注业务逻辑。既能快速提高开发效率,帮助公司节省成本,同时又不失灵活性。JeecgBoot还独创在线开发模式(No代码):在线表单配置(表单设计器)、移动配置能力、在线工作流配置(流程设计器)、在线报表配置、在线图表配置、插件能力(可插拔)等等

  • Updated Nov 22, 2019
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Night-Reading-Go《Go 夜读》 > Share the related technical topics of Go every week through zoom online live broadcast, every day on the WeChat/Slack to communicate programming technology topics. 每周通过 zoom 在线直播的方式分享 Go 相关的技术话题,每天大家在微信/Slack 上及时沟通交流编程技术话题。

  • Updated Nov 22, 2019
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azl397985856
azl397985856 commented Oct 28, 2019

数轴上放置了一些筹码,每个筹码的位置存在数组 chips 当中。

你可以对 任何筹码 执行下面两种操作之一(不限操作次数,0 次也可以):

将第 i 个筹码向左或者右移动 2 个单位,代价为 0。
将第 i 个筹码向左或者右移动 1 个单位,代价为 1。
最开始的时候,同一位置上也可能放着两个或者更多的筹码。

返回将所有筹码移动到同一位置(任意位置)上所需要的最小代价。

 

示例 1:

输入:chips = [1,2,3]
输出:1
解释:第二个筹码移动到位置三的代价是 1,第一个筹码移动到位置三的代价是 0,总代价为 1。
示例 2:

输入:chips = [2,2,2,3,3]
输出:2
解释:第四和第五个筹码移动到位置二的代价都是 1,所以最小总代价为 2。
 

提示:

1 <= chips.length <= 1

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