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one good turn deserves another 礼尚往来 gets a good salary 有一份很好的薪水 never pays it back 从不归还 deserve  应得的 lawyer 律师 immediately 立刻马上 There aren’t enough chairs for us all.Please bring another one. 这边的椅子对于我们来说不够,请给我们再拿一把.…
Text I was having dinner at a restaurant when Tony Steele came in. Tony worked in a layer's office years ago, but he is now working at a bank. He gets a good salary, but he always borrows money from his friends and never pays it back. Tony saw me and…
Lesson 9 A cold welcome 冷遇 What does 'a cold welcome' refer to?On Wednesday evening, we went to the Town Hall. It was the last day of the year and a large crowd of people had gathered under the Town Hall clock. It would strike twelve in twenty minute…
Proverbs(谚语) 作者:凯鲁嘎吉 - 博客园 http://www.cnblogs.com/kailugaji/ 更多请查看:English 1. Every man is the master of his own fortune. 每个人都是自己命运的主宰. 2. It's good to learn another man's cost. 前车之鉴. 3. The onlooker sees the game best. 当局者迷,旁观者清. 4. To err is human,…
大 多数Android设备有内置的传感器,来测量运动,方向和各种环境条件.这些传感器能提供高精度和准确度的原始数据,如果你想监控设备三维运动或者位 置,或者你想监控设备周围的环境变化,是非常有用的.例如,游戏可能跟踪设备重力传感器的数据,来推断复杂的用户首饰和动作,例如倾斜,震动,旋转,或者 振幅.同样的,天气应用可能使用设备的温度传感器和湿度传感器的数据来计算和报告结露点,或者旅行应用可能使用磁场传感器和加速度传感器来报告一个指南针 方位. Android平台支持三大类的传感器: 位移传感器…
1.A private conversation Last week I went to the theatre. I had a very good seat. The play was very interesting. I did not enjoy it. A young man and a young woman were sitting behind me. They were talking loudly. I got very angry. I could not hear th…
keep half an eye on something分神留意splash out随意花钱 大肆挥霍half a mind有想做某事go Dutch v. 各自付帐,打平伙chance in a million n. 百万分之一的机遇half the battle n. 成功了一半(有助于成功的条件)in dribs and drabs adj. 一小部分,一点点of a kind同类的, 徒有其名的take the bull by the horns采取果敢行动应付艰险局面, 明知山有虎,…
webRTC支持点对点通讯,但是webRTC仍然需要服务端:  . 协调通讯过程中客户端之间需要交换元数据,    如一个客户端找到另一个客户端以及通知另一个客户端开始通讯.  . 需要处理NAT(网络地址转换)或防火墙,这是公网上通讯首要处理的问题.    所以我们需要了解服务端相关的知识:信令.Stun.trun.ice. 一.什么是信令 信令就是协调通讯的过程,为了建立一个webRTC的通讯过程,客户端需要交换如下信息:  . 会话控制信息,用来开始和结束通话,即开始视频.结束视频这些操作…
原题:  Turn the pokers       思路:假设正面为0,反面为1.牌就像这样 000000....... .考虑到假如可以实现最终反面个数为m, 牌共n张, 则这n张排任取m个为反面其余都为正面的状况都能实现.于是转化为考虑最终可能出现1的个数的集合有哪些.       因为可能的个数集合是连续的(在最大最小值之内相差2的都可能), 所以每一次翻转之后的上下限都可以根据上一次所得的上下限推出.       最后算排列组合的适合需要用到组合数递推公式和费马小定理推论\( a^{p…
很好用的一款插件jQuery+turn.js翻书.文档和杂志3种特效演示 在线预览 下载地址 实例代码 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"…