狂野飙车8赛车数据 捷豹XC75数据介绍
狂野飙车8全车数据表
奔驰SLK55 道奇挑战者 讴歌 阿尔法罗密欧8C M1 612 奥迪RS4 小莲花 奔驰suv 路虎 英菲尼迪 280z 兰博SUV 福克斯 科迈罗GS 奥迪TTS 奔驰电动 奥迪RS3 D8 RX8 4C 斯巴鲁 蓝瑟 纳米流泉帝诺 S2000 马六 M3 吉利 308 小尼桑 野马 Z/28 370 奥迪S4 DS3/小屌丝 雷神 甲壳虫 小雷诺 阿尔法罗密欧GTA 迷你
序号
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
车外号
恶魔 复仇女神 屁王GTR 狼孙 雷诺 红彤彤 超级莫斯勒 恩佐 棺材 韦伯/龙虾 狗飞飞 迈凯伦F1 spano坑钱版 旺旺 自私 轩尼诗/毒液 SSC飓风 大标志 坦克 纳米流特铁 大蜥蜴 狼崽 阿格拉 飓风 银电 布加迪 风神 FXX 鲁夫3 斯帕诺 毒药 177 宝马3.0 捷豹XJ220S 科迈罗2016SS 宾利EXP10 小莫斯勒
Mclaren 675LT Sbarro Alcador Chevrolet Corvette C3 Lamborghini Aventador LP 700-4 Mercedes-AMG C 63 Coupe Touring Car 2014 Ferrari 330 P4 Tramontana XTR Mclaren 570S Volkswagen W12 Savage Rivale GTR Nissan GT-R NISMO Mazzanti Evantra Mclaren P1 Lamborghini Asterion Ferrari LaFerrari Jaguar C-X75 Ferrari FF Lucra L148 RUF RT 12 S Savage Rivale Roadyacht GTS Lamborghini Sesto Elemento Ferrari F12berlinetta Pagani Zonda R Mclaren 12C Spider Mercedes-Benz SLS AMG GT Coupe Final Edition Chevrolet Corvette C7 Marussia B2 Dodge Viper SRT10 ACR-X Ferrari F430 Mercedes-Benz SL 65 AMG Black Series Cadillac CTS-V Coupe Race Car Ford 2006 GT Falcon F7 Mercedes-AMG GT3 Lamborghini Estoque Holden Coupe 60 TVR Sagaris Kepler Motion BXR Bailey Blade GT1 Chevrolet SS Peugeot SR1
HSV-160U交流伺服驱动单元使用说明书-1.1
1．3．1 HSV-160U-020、030 伺服驱动单元外形尺寸 ........................................................... 11 1．3．2 HSV-160U-050、075 伺服驱动单元外形尺寸 ........................................................... 12 2．1 安装环境条件 ......................................................................................................................... 13 2．1．1 环境温度 ....................................................................................................................... 13 2．1．2 湿度 ............................................................................................................................... 13 2．1．3 海拔高度 ....................................................................................................................... 13 2．1．4 振动和冲击 ................................................................................................................... 13 2．1．5 水 ................................................................................................................................... 13 2．1．6 大气污染 ....................................................................................................................... 13 2．2 伺服驱动单元安装 ................................................................................................................. 13 2．2．1 安装方法 ....................................................................................................................... 13
极品飞车8-地下狂飙2攻略及秘籍
一、赛前部署1.好显卡相当于你的视力是 1.5～2.0——先谈谈玩家怎样具备一双“火眼金睛”。
笔者用华硕9570 TD显卡（GeForce FX 5700）测试该游戏，虽然CPU只是P4 2.0，但将全部视觉效果打开并使用800×600分辨率时，画面刷新率还算叫人满意。
游戏对老型号显卡采取的折衷效果是“慢帧”，也就是画面刷新速度变慢，这显然比那些“跳帧”或“停帧”的赛车游戏好多了。
即便你拥有一块上等显卡，你仍然可以考虑把特效关掉一些，因为这可以把弯道和建筑看得更清楚，切弯走线自然更准，当然这仅仅是建议，用不用你自个儿瞧着办。
2.糟糕的显示器相当于你得了夜盲症——虽然液晶显示器亮度较高，但由于像素略粗，加之高速行车时有拖影，因此一台高品质台式显示器才真正适合这款游戏。
当然你还需把显示器的亮度和对比度调到最高，如此才能获得更清晰的视觉效果。
如果你的显示器质量欠佳或年头较老，你可在Windows的控制面板→显示→设置→高级→显卡设置中将画面强行调亮。
3.方向盘和键盘“争风吃醋”——这款游戏可在驾驶时同时支持方向盘和键盘，但这两种操控工具会时不常地发生一些小冲突，比如键盘的转向键会突然失灵，此外拔掉提供方向盘力回馈的变压器会造成游戏死机。
如果你用方向盘，游戏画面刷新速度会稍慢，而使用键盘的话就不会有上述问题，前提是你必须把方向盘从机箱的USB口彻底拔下来。
其实键盘的操纵感比较直接，更何况这款游戏并不属于模拟类，因此键盘对大多数玩家来说已足够了。
注意选用那些按键清脆的键盘，不宜使用手感柔和的键盘，虽然感觉舒服，但不利于你作出准确的判断——键盘这东西，还得玩“硬”的。
4.上路前先把规矩听明白——本游戏的主体就是“生涯模式”（Career），在这个模式中你可以驾驶着你的跑车在夜幕中巨大的城市内自由奔驰。
其间要随时留意察看收到的SMS信息，同时根据大地图上显示的比赛位置（在地图上用不同颜色的圆圈表示）前往参加比赛（开进那个闪光的圆圈后按回车键）。
狂野飙车8车辆升级表Class D
品牌金钱效果评分金钱效果评分金钱效果0 5.00230.071.2000->1200 4.87 3.0400232.9317.8200 1.2051->22000 4.67 4.54500237.2126.72000 1.2132->33500 4.40 5.98000242.9335.63500 1.2233->46000 4.077.415000250.0744.56000 1.2354->512500 3.678.925000258.6453.412500 1.2505+ 3.30266.501.264总计2420029.752900178.224200279.35PRO升级5金钱卡片评分金钱卡片评分金钱卡片18.4 3.6210.4 4.5312.5 5.4414.6 6.2516.67.1总计0062.50026.800状况金币6000120002100027000430005600080000122500代币40608012018030038057510014020050060018000320006300014525014519532547070000495515极速分析零改加赞名字购买金钱性能升级加速(sec)极速(km/h)操控(Gs)极值(kph)零改4改加赞(最大零改满PRO加赞轮胎(平速+操控) - Tires 悬挂(氮增+操控) - Suspension 传动(平速+加速) - Dri257.22261.43291.99299.36761.80参赛级别1058.721237.27161400评分金钱效果评分金钱评分平速氮增27.150.1600.2293.030028.51 5.9110029.74.5350030.548.91200044.5加速操控5.9550033.2611.92050059.410.0010.007.41000036.6514.83700074.260.0020.008.92000040.7217.87000089.1平速氮增44.45价格倍率61.2128.5713.5729.73930059.4140600296.950.56评分金钱卡片评分卡片评分加速(sec)极速(kph)操控(Gs)8.4 3.623.90.065 1.390.002410.4 4.529.80.145 3.110.005512.5 5.4初级35.70.242 5.190.009114.6 6.2初级41.60.3557.600.013316.67.1初级47.50.48310.360.018162.50026.8178.5引擎种类168000275000294000420000550000860950125020003500175000980145048000075011001495225019500027500032450082595016752350225000980355000240092517502650性价比11总计MAX加赞MAX+满PRO加赞s)氮气(+km/h)配分比/增量单项属性增加值(相对PRO零改(最大化Kit)PRO改造缩水率 Drivetrain排气(氮增+加速) - Exhaust总计99.360.725总提升性能分0.829310.95341.50氮气(+kph)0.660.0491.480.1092.460.1823.610.2664.920.363强制系数1.000O零改时)0.725总提升。
狂野飙车8所有车型赏析47辆豪车全图鉴
狂野飙车8所有车型赏析47辆豪车全图鉴狂野飙车8所有车型赏析47辆豪车全图鉴狂野飙车8强势登陆ios/安卓系统，作为竞速类游戏中的标杆型游戏，狂野飙车8中你可以选择总共47辆豪华型的跑车中的一种进入赛道奔驰，下面就让我们一起来欣赏一下这些精美的豪华跑车们，赶快从中挑选一辆让他成为你独一无二座驾吧!D系车图鉴lC系车图鉴lB系车图鉴lA系车图鉴lS系车图鉴l 道奇 Dart GT前不久，道奇公布了2013款全新Dart GT的消息，在2103北美车展上，道奇终于带来了Dart GT的实车。
而道奇 Dart GT 也将在狂野飙车8的赛道上飞驰。
奥迪R8(Audi R8)奥迪R8(Audi R8)是一款中置引擎双座超级跑车，极速达316km/h，由德国汽车制造商奥迪于2006年推出。
奥迪R8是奥迪量产的首款中置引擎超级跑车，基于兰博基尼Gallardo的开发平台，融合了奥迪在多个运动赛事中取胜的经验，技术以及突破传统观念的完美设计。
在狂野飙车8里，它将伴随很多玩家的第一赛季。
特斯拉Model S特斯拉Model S是一款纯电动车型，外观造型方面，该车定位一款四门Coupe车型，动感的车身线条使人过目不忘。
此外在前脸造型方面，该车也采用了自己的设计语言。
凯迪拉克AT S基于强劲的动力性能和出色的燃油经济性表现，凯迪拉克AT S运用全新方式诠释了凯迪拉克将艺术与科技完美融合的理念。
该车是凯迪拉克家族中最敏捷、最轻盈的车型，也是同级产品中最轻的车型之一，整备质量不到1542公斤。
凯迪拉克XT S作为凯迪拉克全新定位的大型豪华轿车，XT S首次以混合动力车的身份亮相于2010年的北京车展。
外观方面，流线型的车身长5131mm，轴距为2837mm，采用了20英寸11辐的轮毂及普利司通轮胎。
LED前大灯和安装在行李箱盖上的扰流板等部件都展现了其全新的设计风格。
赛恩FR-S全新赛恩FR-S是一款2+2座双门轿跑车，采用了传统的前置后驱布局。
CNG1000综合接入设备产品手册
本文件所有智能财产权归深圳市世纪网通通信技术开发有限公司所有，并受所有中华人民 共和国法律规定之相关条款及协议保护。深圳市世纪网通通信技术开发有限公司保有所有权 利。
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了解更多信息，请与深圳市世纪网通通信(H323/SIP)产品手册
目录
第一部分 产品概述 ...................................................................................................................... 7 1.1 CNG1000 综合接入设备(R1)产品特性 ..................................................................................... 8 1.2 CNG1000 综合接入设备(R1)技术规格 ..................................................................................... 9 1.3 产品包装................................................................................................................................... 10 1.4 产品外观................................................................................................................................... 10
海浦蒙特HD09变频器用户手册
版本修订记录
用户手册的版本位于封面的右下角。
改版时间：2014 年 3 月
改版版本：V1.1
修改章节 版本修订记录 第七章 功能参数速查表
9.6 应用举例
修改内容
增加： • 版本修订记录
增加：
• F00.14 增加：“千位：频率通道切换至模拟选择” • 0：不保存。 • 1：保存之前操作面板/端子设定的频率。 • 设定范围及出厂值改为：0000－1111【1001】
l1l2l3nbrpel1l2l3n制动电阻仅三相电源机型能选配制动电阻hd09变频器接触器电源接地mccb第四章电气安装深圳市海浦蒙特科技有公司12hd09系列用户手册v1143控制端子及连接图控制端子说明端子名称规格10外引电源最大输出电流100maai可编程模拟输入由拨码开关决定电压输入或电流输入电流020ma阻抗500ao可编程模拟输出电压010vgnd电源地模拟和数字地0vdi1di2di3可编程数字输入与gnd短接有效di4可编程数字输入与gnd短接有效高速脉冲输入f1503设为53号功能最高频率500khzf1617设定do可编程数字输出开路集电极输出高速脉冲输出f1519设为38号功能最高频率500khzf1626设定r1ar1br1c可编程继电器输出触点容量
修改： 2．写从机 2 的设定运行频率，掉电保存（例如：设定运行频率为 45.00Hz），寄 存器内容 0x11,0x94
地址 请求/应答帧 0x02
寄存器内容
0x11
0x94
校验和 0xDB
0x7E
9．读从机 2 的输出频率（例如：变频器输出频率为 50.00Hz），寄存器内容 0x13, 0x88
第五章 技术数据 .....................................................................................................................................................15 5.1 外围器件选型..............................................................................................................................................................15 5.2 技术规格 ....................................................................................................................................................................... 16
狂野飙车8赛道大全 巴塞罗那赛道全景图
狂野飙车8赛道大全巴塞罗那赛道全景图狂野飙车8中每个赛道路径、难度区别都非常的大。
如果想取得更好的成绩，除了不停的练习操作，对赛道的认知与理解也是必不可少的。
下面小编就为大家带来狂野飙车8赛道全景图。
在西班牙有一个艺术殿堂的城市——巴塞罗那，那里有美轮美奂的博物馆与大剧院，更有世
界知名的高迪建筑群，当然，他们的足球队也是不可忽略的存在。
在狂野飙车8中收录了一条来自巴塞罗那的赛道，玩家们在这条赛带中将领略巴塞罗那的街头风光
好了，所有内容都介绍完毕了，希望大家能玩的开心。
更多游戏资讯与攻略请关注百度攻略&口袋巴士《狂野飙车8》专区
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1。
北京斯达峰控制技术有限公司 SF-2100S-AH 切割机数控系统说明书
*使用系统前请认真阅读本手册使用注意事项:1. 包装箱打开后，请检查系统在运输过程中有无破损，装箱单上所列明细内容与箱内物品是否一致。
2. 本说明书适用于北京斯达峰控制技术有限公司生产的SF-2100S-AH切割机数控系统。
3. 请检查电网电压是否正确。
在电网与系统之间要使用AC220V的隔离变压器,以确保系统可靠工作和人员的安全。
4. 数控系统要求工作环境温度为 0℃～ +40℃,相对湿度为0～85% 。
如在高温、高湿和有腐蚀性气体的环境下工作,需要采取特殊的防护。
5. 数控系统各部分接线要正确,地线接触良好。
6. 数控系统不允许带电插拔机箱后部的所有电缆插头，由此产生的后果，本公司拒绝保修。
7. 数控系统后部输出端口的线缆，不允许和其他电源线短路；否则将烧毁数控。
8. 在高粉尘环境下,整机需做粉尘防护,并且定期清理灰尘,尽量保证数控系统的清洁。
9. 数控系统应由专人管理,对操作人员应进行岗前培训。
10. 不允许将数控系统内部使用的交流/直流电源连接到其它外部电器上。
11. 如遇问题,请与本公司联系。
切勿在不熟悉的情况下自行拆装、改造系统。
12. 维护系统和机床，每班执行一次日常维护和检查；每月执行一次二级维护；每六个月执行一次一级维护。
13.数控系统设置的各项参数，要严格按照本说明书或订货时的补充说明进行设置；如设置的参数超出规定范围，会使数控系统工作不正常，甚至损坏。
14.系统的液晶屏为易碎物品，使用过程中注意对液晶进行防护。
15.本系统技术指标如有变更，恕不另行通知。
16.注意：系统的USB口输出功率很小，只能供U盘使用，不能接其他的USB设备，以防损坏。
17.特殊声明：本产品的保修期为自出厂之日起十二个月内。
保修期外和保修范围以外发生故障的处理为收费服务。
以下情况不在保修范围内：A：违反使用要求的人为损坏；B：不可抗力导致的损坏；不可抗力通常包括两种情况：一种：是自然原因引起的，如雷击、水灾、旱灾、暴风雪、地震等；另一种：是社会原因引起的，如战争、罢工、政府禁令等；C：未经许可，擅自拆卸、改装、修理等行为导致的损坏。
侠盗猎车手圣安地列斯全车辆特点性能详解
侠盗猎车⼿圣安地列斯-全车辆特点性能详解 侠盗猎车⼿圣安地列斯⾥⾯的各种载具想必是玩家体验游戏的⼀⼤乐趣，不同车辆的体验也会不⼀样，到底每种车辆有些什么样的区别呢，看看下⾯全车辆特性详细解析吧。
全车辆特点性能详解 400:L a n d s t a l 改装：可以 推荐改装： 因为是越野车所以⽤了越野轮胎。
速度：中 控制：良 越野车但是越野性能不太好，适合跑路，L Z亲测路⾯状况，车⼦还可以，过弯可以控制，但是不能漂移。
401:B r a v u r a 改装：可以 推荐改装： 改装前看着是⼀种烧饼的样⼦没想到改装后有了⼀种街头赛车的赶脚。
速度：中 控制：优 因为是街头赛车所以加速过弯什么的完全没有问题 402：B u f f a l o 改装：可以 推荐改装 因为这车是在改装什么的没有⼏个，所以... 这车速度性能也不错，控制也可以，尤其是引擎让我这个作为各种车迷的L Z⼗分喜欢，怎么看也有种肌⾁车的感觉吧 这车漂移什么的实在让⼈⼤跌眼镜，打滑什么的最讨厌了 403：L i n e r u n 这是经典的美式车头啊--改装什么的当然不可以啦 漂移加速直接跳过，⽤它来撞车不错。
404：P e r e n 又是⼀款⼩⼩的旅游车~~ 然后可以改装。
今天我才发现原来改装后这么风骚。
注意还有尾翼(阻流器) 它的速度不怎么样，开在乡村⾥感觉还不错，飘逸的话呢。
适应的话就可以啦~~这个车平稳性还是不错滴 405：S e n t i n e l 又是⼀款商务型办公车辆 可以改装 推荐改装效果。
怎么让我改完的车都有⼀种街头赛车的感觉 楼主上路测的时候感觉这部车⼦还不错，速度什么的都可以，转弯也可以，适应就好，总体来说可以6分~就是不太喜欢这样⼦有点丑 406：D u m p e r 拉⼟车~⼀看名字就蛮霸⽓的 ⽤来碾压车辆不错。
可以这样玩。
407：F i r e t r u k 消防车，样⼦蛮霸⽓的。
SD Host Controller 2.0设计说明书
SD Host Controller 2.0设计说明书（仅供内部使用）拟制： 张 杰 日期： 2008-8-13审核： 日期：审核： 日期；批准： 日期；创辉电脑深圳代表处香港创辉电脑有限公司深圳代表处研究管理部文档中心 产品版本 密级 V1.0绝密 产品名称: SD Host Controller 2.0修订记录日期修订版本描述作者2008-8-13 1.0 初稿完成张杰目录目录 .................................................................................................................................................. ３第1章总体设计. (5)§1.1 SD卡控制器综述 (5)§1.2 SD卡控制器应用结构图 (5)§1.3 顶层结构图 (6)§1.4 SD卡控制器主要功能点 (6)§1.5 SD卡控制器接口描述 (7)§1.6 SD卡控制器寄存器定义 (8)第2章模块设计与验证 (29)§2.1 命令控制模块（CCL）设计与验证 (29)§2.1.1 模块说明 (29)§2.1.2 模块接口描述 (30)§2.1.3 状态机设计 (32)§2.1.4 功能仿真 (43)§2.2 中断控制/时钟管理模块（Clockgen/INT）设计与验证 (44)§2.2.1 模块说明 (44)§2.2.2 模块接口描述 (44)§2.2.3 模块结构图 (47)§2.2.4 功能仿真 (47)§2.3 数据控制模块（DCL）设计与验证 (50)§2.3.1 模块说明 (50)§2.3.2 详细功能描述 (50)§2.3.3 模块接口描述 (52)§2.3.4 状态机设计 (56)§2.3.5 功能仿真 (56)第3章系统级功能仿真和FPGA验证 (57)§3.1 功能仿真结果表格 (57)§3.2 FPGA验证 (57)第4章LINUX下SD卡驱动程序 (59)§4.1 SDIO驱动程序流程图 (59)§4.2 SD/MMC/SDHC Host 2.0驱动程序流程图 (61)§4.3 相关Card驱动流程图（2.0标准） (67)第5章附录（SD卡控制器验证计划书） (70)§5.1 SD控制器功能验证方案 (70)§5.1.1 验证环境 (70)§5.1.2 验证平台 (70)§5.1.3 验证流程 (71)§5.2 功能点统计 (72)§5.3 详细测试步骤和寄存器设置 (73)§5.4 FPGA验证 (77)§5.4.1 验证目标 (77)§5.4.2 验证方法 (78)§5.4.3 详细验证计划 (78)第6章参与设计人员 (81)第1章总体设计§1.1SD卡控制器综述本设计遵循SD Host Controller Simplified Specification Version2.00版本。
TW2操作手册
TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室- 1 -TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室目录1前言 (7)2操作办法综述 (9)2.1调车计划的储存作业 (9)2.1.1调车计划的储存 (9)2.1.2调车计划的检查和修改 (10)2.2溜放作业的办理 (10)2.2.1“溜放开始”操作 (10)2.2.2双推单溜的自动切换 (11)2.2.3溜放计划的修改 (11)2.2.4溜放中的“钓鱼”处理 (12)2.2.5溜放中的“峰下摘勾”处理 (12)2.2.6溜放中的“追勾”处理 (12)2.2.7溜放中的“途停”处理 (12)2.2.8溜放中的“满线”处理 (12)2.2.9溜放中的“堵门”处理 (13)2.2.10溜放中的“道岔恢复”处理 (13)2.2.11溜放中的设备故障报警处理 (13)2.2.12股道封锁与解锁的应用 (13)2.2.13机车上下峰进路的办理 (14)2.2.14去禁溜线或迂回线的控制 (14)2.2.15分路道岔手动 (15)2.3调车作业办理 (15)2.3.1与溜放作业的冲突 (15)2.3.2以调车线为始终端的情况 (15)2.3.3进路上的减速器 (15)2.3.4仅可办理长调车进路的情况 (15)2.3.5分路道岔区调车进路的特点 (16)2.3.6调车线出岔 (16)2.4调速作业 (16)2.4.1自动作业方式 (17)2.4.2半自动作业方式 (17)2.4.3手动作业方式 (17)2.4.4放头拦尾 (18)2.4.5油轮车、大轮车和薄轮车 (18)- 2 -TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室2.4.6调速特殊情况的观察与防范 (19)2.4.7调速修正量的应用 (19)2.5推峰速度控制 (20)3主窗体使用说明 (21)3.1概述 (21)3.2主窗体菜单的使用说明 (21)3.2.1开始菜单 (21)3.2.2溜放1菜单 (23)3.2.3溜放2菜单 (23)3.2.4调车单菜单 (23)3.2.5图形1菜单 (23)3.2.6图形2菜单 (23)3.2.7帮助菜单 (23)3.3主窗体按钮的操作与显示 (23)3.3.1 A、B机工作状态 (23)3.3.2板故障 (24)3.3.3修正 (24)4调车单窗使用说明 (25)4.1概述 (25)4.1.1自动存储方式和人工存储方式 (25)4.1.2车次列表窗和勾车编辑窗 (25)4.1.3调车作业计划单 (26)4.1.4调车单的显示方式 (26)4.2对调车单窗的操作 (27)4.2.1如何新建一个车次计划 (27)4.2.2如何取消一个车次计划 (27)4.2.3如何保存一个车次计划 (27)4.2.4如何修改一个车次计划 (27)4.3菜单的操作说明 (27)4.3.1计划菜单 (27)4.3.2勾编辑菜单 (28)4.3.3窗口菜单 (30)4.3.4选项菜单 (31)4.3.5帮助菜单 (33)- 3 -TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室5溜放窗使用说明 (34)5.1概述 (34)5.2溜放窗的操作 (35)5.2.1如何办理溜放开始 (35)5.2.2如何办理溜放暂停 (35)5.2.3如何办理溜放取消 (35)5.2.4如何修改溜放窗 (36)5.3溜放窗菜单的操作 (36)5.3.1操作菜单 (36)5.3.2勾编辑菜单 (37)5.3.3窗口菜单 (38)5.3.4选项菜单 (39)5.3.5帮助菜单 (39)6图形窗使用说明 (42)6.1概述 (42)6.2图形窗的显示 (42)6.2.1信号机 (42)6.2.2道岔 (43)6.2.3区段的显示 (44)6.2.4测长 (45)6.2.5减速器 (45)6.3图形窗的操作 (47)6.3.1办理进路的操作 (47)6.3.2办理去往禁溜线、迂回线 (49)6.3.3办理信号重复开放 (49)6.3.4办理进路总取消或总人解 (49)6.3.5办理故障解锁 (49)6.3.6办理道岔单独操纵 (49)6.3.7办理道岔单锁和单解 (50)6.3.8场间联系按钮的显示 (50)6.3.9驼峰信号机的操纵 (50)6.3.10办理人工定速和定速取消 (51)6.3.11办理减速器的单独操纵 (51)6.3.12办理减速器的单锁或单解 (51)6.3.13场间调车联系作业 (51)- 4 -TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室6.4图形窗菜单 (51)6.4.1系统管理菜单 (52)6.4.2命令操作菜单 (52)6.4.3用户设置菜单 (52)6.4.4帮助菜单 (54)6.5设备弹出菜单 (54)6.5.1场间联系按钮 (54)6.5.2重力式减速器 (54)6.5.3非重力式减速器 (55)6.5.4一般无岔区段 (55)6.5.5警冲标区段 (56)6.5.6驼峰主体信号机 (56)6.5.7调车信号机 (57)6.5.8自动集中道岔(分路道岔) (57)6.5.9电气集中道岔(非分路道岔) (57)7测长窗 (59)8报警记录查询窗使用说明 (60)8.1概述 (60)8.2操作说明 (60)8.2.1树型窗 (60)8.2.2选择窗： (61)8.2.3参数设置窗： (61)8.2.4报警信息窗 (62)8.3报警信息窗中的菜单操作 (62)8.3.1新的查询 (62)8.3.2退出查询 (62)8.3.3打印结果 (62)8.3.4信息保存 (63)8.3.5查询暂停与查询继续 (63)8.3.6剔除该项与恢复剔除 (63)9统计报告窗操作说明 (64)9.1调车作业计划表： (64)9.2计划执行概要表： (65)9.3计划速度控制表： (65)- 5 -TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室9.4车站运营状态表： (66)10手动盘 (69)11主体窗显示的提示与报警内容 (70)12操作注意事项 (75)12.1应急台手柄位置 (75)12.2计划中的辆数 (75)12.3计划中的重量 (75)12.4溜放开始时机 (75)12.5清除残留 (75)12.6关于途停 (75)12.7关于防侧冲 (75)12.8手动界定 (76)12.9注意溜放车 (76)12.10溜放中的超速现象 (76)12.11减速器的预先制动 (76)12.12关于中途改变定速 (76)12.13雷达信号受阻 (76)12.14空线打靶 (76)12.15减速器单锁与解锁 (76)12.16减速器单操 (77)12.17测长显示 (77)- 6 -TW-2型驼峰自动化系统操作手册北京全路通信信号研究设计院驼峰二室1前言TW-2型组态式驼峰自动控制系统是用于驼峰进路及调速自动控制的控制装置，它由控制微机、雷达、测长、减速器、转辙机、信号机、轨道电路、操作工作站及报警打印机等环节组成，它实现以自动、半自动、手动相结合的控制模式，其主要功能如下：办理与集中场的场间联系；办理峰上及峰下机车走行进路的联锁；调车作业计划的自动接收或人工输入及计划的临时改变，按作业需要顺序调用计划；调用计划同时建立推送进路，自动或人工控制驼峰信号机；自动按计划执行下列性质的勾作业：●溜放勾●上下峰勾●禁溜及迂回进路取送勾自动控制间隔制动位减速器（一二部位），调整勾车间隔及保障三部位减速器人口速度；自动控制目的制动位减速器（三部位），调整勾车速度与股道内的停留车安全联挂；办理股道满线封锁及编发线发车锁闭。
狂野飙车8新版本世界顶级跑车性能解析
狂野飙车8新版本世界顶级跑车性能解析狂野飙车8新版本世界顶级跑车性能解析《狂野飙车8》是⼀款充满速度与激情的赛车游戏，最新版本中为玩家们带来了Devel sixteen prototype、T RION NEMESIS 等众多神车，当然售价也不菲。
对于⼤多数玩家来说，狂野飙车8世界顶级跑车哪辆性能好成了⽐较困惑的问题，⼀时间被众多神车挑花了眼睛。
今天，⼩编为⼤家带来的狂野飙车8新版本世界顶级跑车性能解析来对新版本的神车做全⾯的分析，希望可以帮助到⼤家。
车名：Devel sixteen prototype性能分：1810～1840类型：精英赛车售价：14500代币引擎：V16*24同级别竞争对⼿：1805复仇者、1773宝马优点解析：极速快，⾃重⼤，氮效供应较为充⾜，可主动攻击对⼿赛车，⾏驶中不会轻易被撞翻擦翻，擅长冰岛、东京、摩纳哥、巴塞罗那、内华达等⼤多数A8中长距离赛道，且可以根据不同玩家的需求⾃⾏调节车辆性能。
是⼀辆具备了攻击性，持久性，及稳定性的⾼分段顶尖赛车。
缺点解析：⾃重过⼤导致了加速较慢，超⾼的极速也导致了操控的不灵敏，此车漂移半径⼤，硬拐反应迟钝，所以对玩家操控技术有着较⾼的要求，要想将⽐赛耗时缩减到最短，那么弯道漂移时多要搭配点飘来弥补，也只有⾼技术的玩家才能将此车的性能发挥到极致。
性价⽐分析：14500代币可算是A8商店中售价最⾼的赛车了。
就性能来说这辆挂着“精英赛车”招牌的恶魔也还算是对得起这个价格。
不过这辆恶魔⽆论单机还是联⽹，每驾驶三次就要进⾏调节，否则性能会⼀再下降。
每次调节要两个半⼩时，使⽤代币跳过的话是250枚代币，换句话说就是250代币开三次。
所以楼主认为，这辆车性能虽好，但使⽤代价较⼤，如果你只攒够了买车的代币⽽没有为它的将来做准备，那还是再考虑考虑吧。
世界赛胜率测试：10胜7世界赛综合评分：85分总凭：世界赛⾼分段第⼀集团车辆，推荐代币充裕的玩家⼊⼿。
车名：T RION NEMESIS性能分：1805类型：普通赛车售价：400万⾦币引擎：增压式V8*60同级别竞争对⼿：Devel恶魔、1773宝马优点解析：⼀辆打破A8⾼分段平衡的赛车，加速极快，稳定性极⾼，氮效充⾜。
神钢sk75超8详细资料
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捷豹XF SPORTBRAKE新款车型说明书
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目录第一章安全知识 (1)1.驾驶员须知 (1)2.安全和车辆损坏警告 (1)3.锂电池组的回收 (2)4.高压系统 (2)5.发生事故时的注意事项 (3)6.灭火器 (3)第二章车辆驾驶信息 (5)1.仪器仪表说明 (5)2.车载影音娱乐系统 (12)3.左组合开关 (13)4.右组合开关 (14)5.外后视镜 (15)6.内部照明 (16)7.内后视镜 (16)8.档位开关装置 (17)9.空调装置 (18)10.警示灯开关 (25)11.玻璃升降器开关 (25)12.应急开关 (26)13.转向盘锁 (27)14.机舱盖 (27)15.充电接口 (28)16.后背门 (29)17.左、右侧车门 (31)18.座椅 (32)19.安全带 (34)20.转向盘调节 (36)第三章车辆使用 (36)1.实用驾驶技巧 (36)2.动力电池组充电 (41)3.车辆启动 (46)4.车辆牵引 (51)5.后台数据监控 (52)6真空助力失效及真空泵震动噪声说明 (52)第四章车辆维护 (53)1.定期保养 (53)2.液面检查 (54)3.电池保养 (55)4.玻璃升降器 (58)5.车轮和轮胎 (59)6.车身 (62)7.保险丝盒 (64)第五章技术参数 (70)1.车辆信息 (71)2.结构概述 (72)3.整车基本技术参数 (73)3.1整车主要尺寸参数 (73)3.2整车质量参数 (74)3.3整车要求性能参数 (75)3.4其他要求 (75)3.5整车配置 (76)第一章安全知识1.驾驶员须知您在中华人民共和国境内驾驶本车辆时，请您必须先考取公安交通管理部门颁发的驾驶执照。
A Practical Guide to Quantitative Finance Interview (Part)
Table of ContentsChapter 1 General Principles (1)1. Build a broad knowledge base (1)2. Practice your interview skills (1)3. Listen carefully (2)4. Speak your mind (2)5. Make reasonable assumptions (2)Chapter 2 Brain Teasers (3)2.1 Problem Simplification (3)Screwy pirates (3)Tiger and sheep (4)2.2 Logic Reasoning (5)River crossing (5)Birthday problem (5)Card game (6)Burning ropes (7)Defective ball (7)Trailing zeros (9)Horse race (9)Infinite sequence (10)2.3 Thinking Out of the Box (10)Box packing (10)Calendar cubes (11)Door to offer (12)Message delivery (13)Last ball (13)Light switches (14)Quant salary (15)2.4 Application of Symmetry (15)Coin piles (15)Mislabeled bags (16)Wise men (17)2.5 Series Summation (17)Clock pieces (18)Missing integers (18)Counterfeit coins I (19)2.6 The Pigeon Hole Principle (20)Matching socks (21)Handshakes (21)Have we met before? (21)Ants on a square (22)Counterfeit coins II (22)Contentsii 2.7 Modular Arithmetic (23)Prisoner problem (24)Division by 9 (25)Chameleon colors (26)2.8 Math Induction (27)Coin split problem (27)Chocolate bar problem (28)Race track (29)2.9 Proof by Contradiction (31)Irrational number (31)Rainbow hats (31)Chapter 3 Calculus and Linear Algebra (33)3.1 Limits and Derivatives (33)Basics of derivatives (33)Maximum and minimum (34)L’Hospital’s rule (35)3.2 Integration (36)Basics of integration (36)Applications of integration (38)Expected value using integration (40)3.3 Partial Derivatives and Multiple Integrals (40)3.4 Important Calculus Methods (41)Taylor’s series (41)Newton’s method (44)Lagrange multipliers (45)3.5 Ordinary Differential Equations (46)Separable differential equations (47)First-order linear differential equations (47)Homogeneous linear equations (48)Nonhomogeneous linear equations (49)3.6 Linear Algebra (50)Vectors (50)QR decomposition (52)Determinant, eigenvalue and eigenvector (53)Positive semidefinite/definite matrix (56)LU decomposition and Cholesky decomposition (57)Chapter 4 Probability Theory (59)4.1 Basic Probability Definitions and Set Operations (59)Coin toss game (61)Card game (61)Drunk passenger (62)A Practical Guide To Quantitative Finance InterviewsN points on a circle (63)4.2 Combinatorial Analysis (64)Poker hands (65)Hopping rabbit (66)Screwy pirates 2 (67)Chess tournament (68)Application letters (69)Birthday problem (71)100th digit (71)Cubic of integer (72)4.3 Conditional Probability and Bayes’ formula (72)Boys and girls (73)All-girl world? (74)Unfair coin (74)Fair probability from an unfair coin (75)Dart game (75)Birthday line (76)Dice order (78)Monty Hall problem (78)Amoeba population (79)Candies in a jar (79)Coin toss game (80)Russian roulette series (81)Aces (82)Gambler’s ruin problem (83)Basketball scores (84)Cars on road (85)4.4 Discrete and Continuous Distributions (86)Meeting probability (88)Probability of triangle (89)Property of Poisson process (90)Moments of normal distribution (91)4.5 Expected Value, Variance & Covariance (92)Connecting noodles (93)Optimal hedge ratio (94)Dice game (94)Card game (95)Sum of random variables (95)Coupon collection (97)Joint default probability (98)4.6 Order Statistics (99)Expected value of max and min (99)Correlation of max and min (100)Random ants (102)Chapter 5 Stochastic Process and Stochastic Calculus (105)iiiContentsiv 5.1 Markov Chain (105)Gambler’s ruin problem (107)Dice question (108)Coin triplets (109)Color balls (113)5.2 Martingale and Random walk (115)Drunk man (116)Dice game (117)Ticket line (117)Coin sequence (119)5.3 Dynamic Programming (121)Dynamic programming (DP) algorithm (122)Dice game (123)World series (123)Dynamic dice game (126)Dynamic card game (127)5.4 Brownian Motion and Stochastic Calculus (129)Brownian motion (129)Stopping time/ first passage time (131)Ito’s lemma (135)Chapter 6 Finance (137)6.1. Option Pricing (137)Price direction of options (137)Put-call parity (138)American v.s. European options (139)Black-Scholes-Merton differential equation (142)Black-Scholes formula (143)6.2. The Greeks (149)Delta (149)Gamma (152)Theta (154)Vega (156)6.3. Option Portfolios and Exotic Options (158)Bull spread (159)Straddle (159)Binary options (160)Exchange options (161)6.4. Other Finance Questions (163)Portfolio optimization (163)Value at risk (164)Duration and convexity (165)Forward and futures (167)Interest rate models (168)A Practical Guide To Quantitative Finance Interviews Chapter 7 Algorithms and Numerical Methods (171)7.1. Algorithms (171)Number swap (172)Unique elements (173)Horner's algorithm (174)Moving average (174)Sorting algorithm (174)Random permutation (176)Search algorithm (177)Fibonacci numbers (179)Maximum contiguous subarray (180)7.2. The Power of Two (182)Power of 2? (182)Multiplication by 7 (182)Probability simulation (182)Poisonous wine (183)7.3 Numerical Methods (184)Monte Carlo simulation (184)Finite difference method (189)vChapter 2 Brain TeasersIn this chapter, we cover problems that only require common sense, logic, reasoning, and basic—no more than high school level—math knowledge to solve. In a sense, they are real brain teasers as opposed to mathematical problems in disguise. Although these brain teasers do not require specific math knowledge, they are no less difficult than other quantitative interview problems. Some of these problems test your analytical and general problem-solving skills; some require you to think out of the box; while others ask you to solve the problems using fundamental math techniques in a creative way. In this chapter, we review some interview problems to explain the general themes of brain teasers that you are likely to encounter in quantitative interviews.2.1 Problem SimplificationIf the original problem is so complex that you cannot come up with an immediate solution, try to identify a simplified version of the problem and start with it. Usually you can start with the simplest sub-problem and gradually increase the complexity. You do not need to have a defined plan at the beginning. Just try to solve the simplest cases and analyze your reasoning. More often than not, you will find a pattern that will guide you through the whole problem.Screwy piratesFive pirates looted a chest full of 100 gold coins. Being a bunch of democratic pirates, they agree on the following method to divide the loot:The most senior pirate will propose a distribution of the coins. All pirates, including the most senior pirate, will then vote. If at least 50% of the pirates (3 pirates in this case) accept the proposal, the gold is divided as proposed. If not, the most senior pirate will be fed to shark and the process starts over with the next most senior pirate… The process is repeated until a plan is approved. You can assume that all pirates are perfectly rational: they want to stay alive first and to get as much gold as possible second. Finally, being blood-thirsty pirates, they want to have fewer pirates on the boat if given a choice between otherwise equal outcomes.How will the gold coins be divided in the end?Solution: If you have not studied game theory or dynamic programming, this strategy problem may appear to be daunting. If the problem with 5 pirates seems complex, we can always start with a simplified version of the problem by reducing the number of pirates. Since the solution to 1-pirate case is trivial, let’s start with 2 pirates. The seniorBrain Teasers4pirate (labeled as 2) can claim all the gold since he will always get 50% of the votes from himself and pirate 1 is left with nothing.Let’s add a more senior pirate, 3. He knows that if his plan is voted down, pirate 1 will get nothing. But if he offers private 1 nothing, pirate 1 will be happy to kill him. So pirate 3 will offer private 1 one coin and keep the remaining 99 coins, in which strategy the plan will have 2 votes from pirate 1 and 3.If pirate 4 is added, he knows that if his plan is voted down, pirate 2 will get nothing. So pirate 2 will settle for one coin if pirate 4 offers one. So pirate 4 should offer pirate 2 one coin and keep the remaining 99 coins and his plan will be approved with 50% of the votes from pirate 2 and 4.Now we finally come to the 5-pirate case. He knows that if his plan is voted down, both pirate 3 and pirate 1 will get nothing. So he only needs to offer pirate 1 and pirate 3 one coin each to get their votes and keep the remaining 98 coins. If he divides the coins this way, he will have three out of the five votes: from pirates 1 and 3 as well as himself.Once we start with a simplified version and add complexity to it, the answer becomes obvious. Actually after the case 5,n a clear pattern has emerged and we do not need to stop at 5 pirates. For any 21n pirate case (n should be less than 99 though), the most senior pirate will offer pirates 1,3,, and 21n each one coin and keep the rest for himself.Tiger and sheepOne hundred tigers and one sheep are put on a magic island that only has grass. Tigers can eat grass, but they would rather eat sheep. Assume: A . Each time only one tiger can eat one sheep, and that tiger itself will become a sheep after it eats the sheep. B . All tigers are smart and perfectly rational and they want to survive. So will the sheep be eaten?Solution: 100 is a large number, so again let’s start with a simplified version of the problem . If there is only 1 tiger (1n ), surely it will eat the sheep since it does not need to worry about being eaten. How about 2 tigers? Since both tigers are perfectly rational, either tiger probably would do some thinking as to what will happen if it eats the sheep. Either tiger is probably thinking: if I eat the sheep, I will become a sheep; and then I will be eaten by the other tiger. So to guarantee the highest likelihood of survival, neither tiger will eat the sheep.If there are 3 tigers, the sheep will be eaten since each tiger will realize that once it changes to a sheep, there will be 2 tigers left and it will not be eaten. So the first tiger that thinks this through will eat the sheep. If there are 4 tigers, each tiger will understandA Practical Guide To Quantitative Finance Interviews5that if it eats the sheep, it will turn to a sheep. Since there are 3 other tigers, it will be eaten. So to guarantee the highest likelihood of survival, no tiger will eat the sheep.Following the same logic, we can naturally show that if the number of tigers is even, the sheep will not be eaten. If the number is odd, the sheep will be eaten. For the case 100,n the sheep will not be eaten.2.2 Logic ReasoningRiver crossingFour people, A ,B ,C and D need to get across a river. The only way to cross the river is by an old bridge, which holds at most 2 people at a time. Being dark, they can't cross the bridge without a torch, of which they only have one. So each pair can only walk at the speed of the slower person. They need to get all of them across to the other side as quickly as possible. A is the slowest and takes 10 minutes to cross; B takes 5 minutes; C takes 2 minutes; and D takes 1 minute.What is the minimum time to get all of them across to the other side?1Solution: The key point is to realize that the 10-minute person should go with the 5-minute person and this should not happen in the first crossing, otherwise one of them have to go back. So C and D should go across first (2 min); then send D back (1min); A and B go across (10 min); send C back (2min); C and D go across again (2 min). It takes 17 minutes in total. Alternatively, we can send C back first and then D back in the second round, which takes 17 minutes as well.Birthday problemYou and your colleagues know that your boss A ’s birthday is one of the following 10 dates:Mar 4, Mar 5, Mar 8Jun 4, Jun 7Sep 1, Sep 5Dec 1, Dec 2, Dec 8A told you only the month of his birthday, and told your colleague C only the day. After that, you first said: “I don’t know A ’s birthday; C doesn’t know it either.” After hearing 1 Hint: The key is to realize that A andB should get across the bridge together.Brain Teasers6what you said, C replied: “I didn’t know A ’s birthday, but now I know it.” You smiled and said: “Now I know it, too.” After looking at the 10 dates and hearing your comments, your administrative assistant wrote down A ’s birthday without asking any questions. So what did the assistant write?Solution: Don’t let the “he said, she said” part confuses you. Just interpret the logic behind each individual’s comments and try your best to derive useful information from these comments.Let D be the day of the month of A ’s birthday, we have {1,2,4,5,7,8}.D If the birthday is on a unique day, C will know the A ’s birthday immediately. Among possible D s, 2 and 7 are unique days. Considering that you are sure that C does not know A ’s birthday, you must infer that the day the C was told of is not 2 or 7. C onclusion: the month is not June or December. (If the month had been June, the day C was told of may have been 2; if the month had been December, the day C was told of may have been 7.) Now C knows that the month must be either March or September. He immediately figures out A ’s birthday, which means the day must be unique in the March and September list. It means A ’s birthday cannot be Mar 5, or Sep 5. C onclusion: the birthday must be Mar 4, Mar 8 or Sep 1.Among these three possibilities left, Mar 4 and Mar 8 have the same month. So if the month you have is March, you still cannot figure out A ’s birthday. Since you can figure out A ’s birthday, A ’s birthday must be Sep 1. Hence, the assistant must have written Sep 1.Card gameA casino offers a card game using a normal deck of 52 cards. The rule is that you turn over two cards each time. For each pair, if both are black, they go to the dealer’s pile; if both are red, they go to your pile; if one black and one red, they are discarded. The process is repeated until you two go through all 52 cards. If you have more cards in your pile, you win $100; otherwise (including ties) you get nothing. The casino allows you to negotiate the price you want to pay for the game. How much would you be willing to pay to play this game?2Solution: This surely is an insidious casino. No matter how the cards are arranged, you and the dealer will always have the same number of cards in your piles. Why? Because each pair of discarded cards have one black card and one red card, so equal number of2Hint: Try to approach the problem using symmetry. Each discarded pair has one black and one red card. What does that tell you as to the number of black and red cards in the rest two piles?A Practical Guide To Quantitative Finance Interviews7red and black cards are discarded. As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same. The dealer always wins! So we should not pay anything to play the game.Burning ropesYou have two ropes, each of which takes 1 hour to burn. But either rope has different densities at different points, so there's no guarantee of consistency in the time it takes different sections within the rope to burn. How do you use these two ropes to measure 45 minutes?Solution: This is a classic brain teaser question. For a rope that takes x minutes to burn, if you light both ends of the rope simultaneously, it takes /2x minutes to burn. So we should light both ends of the first rope and light one end of the second rope. 30 minutes later, the first rope will get completely burned, while that second rope now becomes a 30-min rope. At that moment, we can light the second rope at the other end (with the first end still burning), and when it is burned out, the total time is exactly 45 minutes. Defective ballYou have 12 identical balls. One of the balls is heavier OR lighter than the rest (you don't know which). Using just a balance that can only show you which side of the tray is heavier, how can you determine which ball is the defective one with 3 measurements?3Solution: This weighing problem is another classic brain teaser and is still being asked by many interviewers. The total number of balls often ranges from 8 to more than 100.Here we use 12nto show the fundamental approach. The key is to separate the original group (as well as any intermediate subgroups) into three sets instead of two. The reason is that the comparison of the first two groups always gives information about the third group.Considering that the solution is wordy to explain, I draw a tree diagram in Figure 2.1 to show the approach in detail. Label the balls 1 through 12 and separate them to three groups with 4 balls each. Weigh balls 1, 2, 3, 4 against balls 5, 6, 7, 8. Then we go on to explore two possible scenarios: two groups balance, as expressed using an “=” sign, or 1, 3Hint: First do it for 9 identical balls and use only 2 measurements, knowing that one is heavier than the rest.Brain Teasers82, 3, 4 are lighter than 5, 6, 7, 8, as expressed using an “<” sign. There is no need to explain the scenario that 1, 2, 3, 4 are heavier than 5, 6, 7, 8. (Why?4)If the two groups balance, this immediately tells us that the defective ball is in 9, 10, 11 and 12, and it is either lighter (L ) or heavier (H ) than other balls. Then we take 9, 10 and 11 from group 3 and compare balls 9, 10 with 8, 11. Here we have already figured out that 8 is a normal ball. If 9, 10 are lighter, it must mean either 9 or 10 is L or 11 is H . In which case, we just compare 9 with 10. If 9 is lighter, 9 is the defective one and it is L ; if 9 and 10 balance, then 11 must be defective and H ; If 9 is heavier, 10 is the defective one and it is L . If 9, 10 and 8, 11 balance, 12 is the defective one. If 9, 10 is heavier, than either 9 or 10 is H, or 11 is L.You can easily follow the tree in Figure 2.1 for further analysis and it is clear from the tree that all possible scenarios can be resolved in 3 measurements. In general if you have the information as to whether the defective ball is heavier or 4 Here is where the symmetry idea comes in. Nothing makes the 1, 2, 3, 4 or 5, 6, 7, 8 labels special. If 1, 2, 3, 4 are heavier than 5, 6, 7, 8, let’s just exchange the labels of these two groups. Again we have the case of 1, 2, 3, 4 being lighter than 5, 6, 7, 8.1/2L or 6H5H or 3L4L or 7/8H 12L or 12H 9/10H or 11L 9/10L or 11H 5,6,7,8+?1,2,53,6,9_1/2/3/4 L or 5/6/7/8 H1L 6H 2L 8H 4L 7H 3L 5H 2_18_79_39,108,11_9/10/11/12 L or H 9L 11H 10L 12H 11L 12L 10H 9H_912_8_9Figure 2.1 Tree diagram to identify the defective ball in 12 ballsA Practical Guide To Quantitative Finance Interviews9lighter, you can identify the defective ball among up to 3n balls using no more than n measurements since each weighing reduces the problem size by 2/3. If you have no information as to whether the defective ball is heavier or lighter, you can identify the defective ball among up to (33)/2n balls using no more than n measurements. Trailing zerosHow many trailing zeros are there in 100! (factorial of 100)?Solution: This is an easy problem. We know that each pair of 2 and 5 will give a trailing zero. If we perform prime number decomposition on all the numbers in 100!, it is obvious that the frequency of 2 will far outnumber of the frequency of 5. So the frequency of 5 determines the number of trailing zeros. Among numbers 1,2,,99, and 100, 20 numbers are divisible by 5 (5,10,,100 ). Among these 20 numbers, 4 are divisible by 52 (25,50,75,100). So the total frequency of 5 is 24 and there are 24 trailing zeros.Horse raceThere are 25 horses, each of which runs at a constant speed that is different from the other horses’. Since the track only has 5 lanes, each race can have at most 5 horses. If you need to find the 3 fastest horses, what is the minimum number of races needed to identify them?Solution: This problem tests your basic analytical skills. To find the 3 fastest horses, surely all horses need to be tested. So a natural first step is to divide the horses to 5 groups (with horses 1-5, 6-10, 11-15, 16-20, 21-25 in each group). After 5 races, we will have the order within each group, let’s assume the order follows the order of numbers (e.g., 6 is the fastest and 10 is the slowest in the 6-10 group)5. That means 1, 6, 11, 16 and 21 are the fastest within each group.Surely the last two horses within each group are eliminated. What else can we infer? We know that within each group, if the fastest horse ranks 5th or 4th among 25 horses, then all horses in that group cannot be in top 3; if it ranks the 3rd, no other horse in that group can be in the top 3; if it ranks the 2nd, then one other horse in that group may be in top 3; if it ranks the first, then two other horses in that group may be in top 3. 5 Such an assumption does not affect the generality of the solution. If the order is not as described, just change the labels of the horses.Chapter 4 Probability TheoryC hances are that you will face at least a couple of probability problems in most quantitative interviews. Probability theory is the foundation of every aspect of quantitative finance. As a result, it has become a popular topic in quantitative interviews. Although good intuition and logic can help you solve many of the probability problems, having a thorough understanding of basic probability theory will provide you with clear and concise solutions to most of the problems you are likely to encounter. Furthermore, probability theory is extremely valuable in explaining some of the seemingly-counterintuitive results. Armed with a little knowledge, you will find that many of the interview problems are no more than disguised textbook problems.So we dedicate this chapter to reviewing basic probability theory that is not only broadly tested in interviews but also likely to be helpful for your future career. 1 The knowledge is applied to real interview problems to demonstrate the power of probability theory. Nevertheless, the necessity of knowledge in no way downplays the role of intuition and logic. Quite the contrary, common sense and sound judgment are always crucial for analyzing and solving either interview or real-life problems. As you will see in the following sections, all the techniques we discussed in Chapter 2 still play a vital role in solving many of the probability problems.Let’s have some fun playing the odds.4.1 Basic Probability Definitions and Set OperationsFirst let’s begin with some basic definitions and notations used in probability. These definitions and notations may seem dry without examples—which we will present momentarily—yet they are crucial to our understanding of probability theory. In addition, it will lay a solid ground for us to systematically approach probability problems.Outcome (Ȧ):the outcome of an experiment or trial.Sample space/Probability space ( ):the set of all possible outcomes of an experiment.1 As I have emphasized in Chapter 3, this book does not teach probability or any other math topics due to the space limit—it is not my goal to do so, either. The book gives a summary of the frequently-tested knowledge and shows how it can be applied to a wide range of real interview problems. The knowledge used in this chapter is covered by most introductory probability books. It is always helpful to pick up one or two classic probability books in case you want to refresh your memory on some of the topics. My personal favorites are First Course in Probability by Sheldon Ross and Introduction to Probability by Dimitri P. Bertsekas and John N. Tsitsiklis.Probability Theory60()P Z : Probability of an outcome (()0,,()1P P Z Z Z Z :t : ¦).Event:A set of outcomes and a subset of the sample space.()P A : Probability of an event A , ()()AP A P Z Z ¦.A B : Union A B is the set of outcomes in event A or in event B (or both).A B or AB : Intersection A B (or AB ) is the set of outcomes in both A and B .c A : The complement of A , which is the event “not A ”.Mutually Exclusive :A B ) where ) is an empty set.For any mutually exclusive events 12,,N E E E ,11().N N i i i i P E P E §· ¨¸©¹¦ Random variable: A function that maps each outcome (Ȧ) in the sample space ( ) into the set of real numbers.Let’s use the rolling of a six-sided dice to explain these definitions and notations. A roll of a dice has 6 possible outcomes (mapped to a random variable): 1, 2, 3, 4, 5, or 6. Sothe sample space :is {1,2,3,4,5,6} and the probability of each outcome is 1/6 (assuming a fair dice). We can define an event A representing the event that the outcomeis an odd number {1,3,5},Athen the complement of A is {2,4,6}.c A Clearly ()P A (1)(3)(5)1/2.P P P Let B be the event that the outcome is larger than 3: {4,5,6}.B Then the union is {1,3,4,5,6}A B and the intersection is {5}.A B One popular random variable called indicator variable (a binary dummy variable) for event A is defined as the following:1,{1,3,5}.0,{1,3,5}A if x I if x ® ¯ Basically 1A I when A occurs and 0A I if c A occurs. The expected value of A I is []()A E I P A .Now, time for some examples.A Practical Guide To Quantitative Finance Interviews61Coin toss gameTwo gamblers are playing a coin toss game. Gambler A has (1)n fair coins; B has n fair coins. What is the probability that A will have more heads than B if both flip all their coins?2Solution: We have yet to cover all the powerful tools probability theory offers. What do we have now? Outcomes, events, event probabilities, and surely our reasoning capabilities! The one extra coin makes A different from B . If we remove a coin from A ,A and B will become symmetric. Not surprisingly, the symmetry will give us a lot of nice properties. So let’s remove the last coin of A and compare the number of heads in A ’s first n coins with B ’s n coins. There are three possible outcomes:1E :A ’s n coins have more heads than B ’s n coins;2E :A ’s n coins have equal number of heads as B ’s n coins;3E :A ’s n coins have fewer heads than B ’s n coins.By symmetry, the probability that A has more heads is equal to the probability that B has more heads. So we have 13()().P E P E Let’s denote 13()()P E P E x and 2().P E y Since ()1,P Z Z :¦ we have 2 1.x y For event 1,E A will always have more headsthan B no matter what A ’s (1)n th coin’s side is; for event 3E ,A will have no moreheads than B no matter what A ’s (1)n th coin’s side is. For event 2E ,A ’s (1)n thcoin does make a difference. If it’s a head, which happens with probability 0.5, it will make A have more heads than B . So the (1)n th coin increases the probability that A has more heads than B by 0.5y and the total probability that A has more heads is 0.50.5(12)0.5x y x x when A has (1)n coins.Card gameA casino offers a simple card game. There are 52 cards in a deck with 4 cards for each value jack queen king ace 2,3,4,5,6,7,8,9,10,,,,J Q K A . Each time the cards are thoroughly shuffled (so each card has equal probability of being selected). You pick up a card from the deck and the dealer picks another one without replacement. If you have a larger number, you win; if the numbers are equal or yours is smaller, the house wins—as in all other casinos, the house always has better odds of winning. What is your probability of winning? 2 Hint: What are the possible results (events) if we compare the number of heads in A ’s first n coins withB ’s n coins? By making the number of coins equal, we can take advantage of symmetry. For each event, what will happen if A ’s last coin is a head? Or a tail?。
地平线4捷豹cx75调校代码
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推荐使用。
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适合直线较少的拉力赛。
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狂野飙车8赛车数据捷豹XC75数据介绍狂野飙车8是一款非常火热的赛车游戏，以精致画面、真实的操作和丰富的游戏内容而受广
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