视频圆桌骑士斧头9UP打宝通关视频
Travelport.Smartpoint.UserGuide
1.2
Draft Changes Final
June 2012
Sandra Brinkley Sandra Brinkley
1.3
July 2012
Inclusion of v2.2 updates Travelport Smartpoint app
June 2012
Version 1.3
பைடு நூலகம்
Page 2
Travelport Smartpoint App™
Global Product User Guide v2.2
June 2012
Version 1.3
Page 1
Revision History Version 1.0 1.1 Status Draft Final Date 21Sep11 10Nov 11 Update Summary Initial version Inclusion of TRAM and SW plus Updates By Julie Dennis Julie Dennis
Table of Contents
Overview ....................................................................................................................................... 7 Product Information ...................................................................................................................... 7 Product Instal
魔古山宝库完全攻略
BOSS前小怪：小怪分2种，成群的小狮子和独立的大狮子。
小狮子会随机标记，点名一个变大并减少附近小怪受到的AOE压力。
有2种处理方式：1是先单点标记变大的再A，2是把标记的拉远去掉AOE减伤效果。
大狮子技能是叠层石化不能动并清除仇恨，要避免全团同时石化脱离战斗怪重置。
部分大狮子会巡逻，拉小的时候不要误拉到。
还是2处理方式：1是集火同一个在石化前先杀再处理另外一个，2是安排部分人员后进本错开石化时间避免脱离。
团队配置：10人：2T 5DPS 3治疗坦克：由于BOSS对坦主要伤害为流血伤害。
所以骑士坦、武僧坦、死骑坦都有相对良好的表现。
治疗：奶骑补坦较好。
奶德、神牧补团较好。
DPS ：暗牧、鸟德、盗贼、火法、武器战为优势职业。
BOSS战综述：这场战斗的机制比较特殊。
场上有四只共享血量的狮子，有三只会参战。
开战后不久，其中一只狮子会被激活，激活的狮子会让全团人慢慢石化但是受到该狮子的魔法伤害降低99%，只有让其能量到100才能消除石化（完全石化等于灭团）。
那么狮子怎么涨能量呢？场上任意两只狮子距离12码以内就会一起涨能量，落单的那只不涨能量但99%免伤。
不过，一只狮子能量涨到100会进行AOE（即过载），激活的狮子AOE伤害极低，未激活的狮子AOE伤害可以导致灭团。
因此很显然，我们要做的就是，将激活的狮子与未激活的狮子拉在一起涨能量（同时对它们进行双目标输出），并通过合理的换坦嘲讽，保证激活的狮子能量先到100过载，造成极低AOE伤害的同时解除石化。
接下来下一个狮子激活后，继续同样的工作保证下一次超载的也是激活的狮子，如此循环直至战斗结束。
战术分配：a）坦克任务：坦克分别坦住守卫，其中一个坦克拉2只守卫。
坦克需要让其他未激活的守卫者与激活的守卫者在一起增长能量，并且使得激活的守卫者能量最先到达100而超载。
你可以点击这里查看换坦战术详细图解。
b）治疗任务：2治疗看坦克，1个治疗看中红线的人。
躲好紫水和蓝色冰晶，红狮子激活情况下中红线远离10码并保持15秒把红线拉断，非红狮子激活则中红线靠近；刷好中红线的人和踩紫水踩冰陷阱的人，刷好超载和青玉裂片的AOE。
密室逃脱23迷失俱乐部攻略
密室逃脱23迷失俱乐部攻略密室逃脱23迷失俱乐部一共有十一关，现将每个关卡的闯关攻略整理如下：第一关：根据游戏通关要求找到香烟。
首先根据提示找到放大镜碎片放大镜手柄+圆环+玻璃片，在物品栏里合成成放大镜，发现书桌上的密码数字35-91，按照提示输入，打开箱子，获得钥匙，可以打开左边书柜的抽屉，抽屉里有一盒香烟。
通关，门打开了，进入第二关。
第二关：这一关的通关要求是找到红宝石，下面就来说说找到红宝石的详细步骤：1、点开沙发上的靠枕，找到半片钥匙，拾取。
2、沙发旁边的盒子里，打开，另外半片钥匙。
3、将两个钥匙残片合成一把钥匙，点开靠窗户边的窗帘。
里面有一个柜子。
用钥匙把柜子打开。
得到一幅画。
跟墙上的蝴蝶画一样，但是颜色不同。
将两幅画重合到一起！得到一个全新颜色的蝴蝶画。
4、然后将桌子上的首饰，四个颜色调整成合成后的新蝴蝶图色板，机关打开，得到红宝石，过关！点开窗户，到下一关！第三关：1、要求：找到装着情报的信封。
2、拾取前台电脑旁边的花瓶。
3、用花瓶砸开这个红色的柜子，得到一张卡片。
拾取。
去前台刷卡那里，刷一下，得到数字字母线索：7G 。
4、然后分别点开这两个位置的书本，图案上面打X的记一下图形。
5、门边墙壁上挂的画，可以点击，不要上图的两个团。
打开了一个机关。
之后调整成上文得到的7G线索，得到信封。
通关。
打开门进入下一关！第四关：1、要求：找到皇室权杖。
2、首先我给大家圈一下本关的线索物品点。
根据我圈出的位置找到左下方抽屉上面残缺的两个金色金属片。
装上后调整位置，抽屉打开后得到一幅图。
3、拾取椅子旁边桌子上的木块，点击灯。
把木块放上去，调整成上图的位置，打开机关。
得到权杖，通关！第五关：1、要求：找到邮册。
2、首先我给大家圈一下本关的线索物品点。
根据我圈出的位置找出火柴和火油，放到壁炉内，点火，上方点开得到数字线索，打开右边的书柜，在书页上得到数字线索。
打开右边的柜子。
有一个宝箱。
输入密码7542，宝箱打开，得到邮册！第六关：1、要求：带走跑车。
国家标准 建筑设计防火规范(gb50016-2014)
3.
16 、
10. 1. 1 、 10. l. 2 、 10. 1. 5 、 10. 1. 6 、 10. 1. 8 、 10. 1. 10 (1 、 2 ）、 10. 2. 1 、 10. 2.4 、 10. 3. 1 、 10. 3.
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火规范）修订项目计划的函》（建标［2009] 94 号），由公安部天津消防研究所、四川消防研究所会同
有关单位，在《建筑设计防火规范》 GB 50016-2006 和《高层民用建筑设计防火规范》 GB 50045-95
(2005 年版〉的基础上，经整合修订而成。
旨”
本规范在修订过程中，遵循国家有关基本建设的方针政策，贯彻“预防为主，防消结合刀的捎
2结~唱9来自程实践和科学研究认真总结经验，注意积累资料，在执行过程中如有意见、建议和问题，请径寄公
』
.
安部捕防局（地址：北京市西城区广安门南街 70 号，邮政编码： 100054 ），以便今后修订时参考和
组织公安部天津、四川硝防研究所作出解释。
本规范主编单位、参编单位、主要起草人和审查人：
主编单位：公安部天津消防研究所
刘建华黄晓家李向东张凤新宋孝春寇九贵郑铁一 主要审查人：方汝清张耀泽赵惶刘跃红张树平张福麟何任飞金鸿祥王庆生 吴华潘一平苏丹夏卫平江刚党杰郭景范珑杨西伟 胡小援朱冬青龙卫国黄小坤
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200611-DIVING-PAM中文操作手册
DIVING-PAM 操作手册
泽泉科技有限公司，德国 WALZ 公司中国技术服务中心
目录
1 安全指导 ...................................................................................................................................- 4 1.1 一般安全指导 ................................................................................................................- 4 1.2 特殊安全指导 ................................................................................................................- 4 -
水下调制荧光仪——
DIVING-PAM
操作手册
2006 年 11 月版
泽泉国际集团（香港）有限公司 泽泉科技有限公司 德 国 WA L Z 公 司 中 国 技 术 服 务 中 心
中国（总部）：上海市中江路 879 号天地软件园 28 幢 402-403 座（200333）
电话：021-51556114/15/16/17/18
传真：021-51556111
E-Mail：sevice@
网址：
北京代表处：北京市海淀区花园北路 48 62304952/53/58，89110167
传真：010-62304952 转 20
9.1.1 荧光的激发与检测............................................................................................- 28 9.1.2 内置卤素灯 .......................................................................................................- 29 9.1.3 可充电电池 .......................................................................................................- 29 9.1.4 显示器 ...............................................................................................................- 30 9.1.5 电子元件 ...........................................................................................................- 30 9.1.6 接口介绍 ...........................................................................................................- 30 9.2 标准光纤 DIVING-PAM/F 和微光纤 DIVING-PAM/F1 ...........................................- 32 9.3 光量子传感器 ..............................................................................................................- 32 9.4 深度传感器 ..................................................................................................................- 33 9.5 水温传感器 ..................................................................................................................- 33 9.6 水下通用样品架 DIVING-USH ..................................................................................- 33 9.6.1 介绍 ...................................................................................................................- 33 -
伏魔战记3.9详细攻略
圆桌骑士攻略砍出魔杖
善用连击技巧：通过连续攻击，快速消灭敌人，获得更高的战斗收益。
利用环境因素：利用场景中的障碍物、高低差等环境因素，制造出有利的战斗条件。
灵活运用技能：根据敌人的类型和战斗环境，灵活运用各种技能，提高战斗胜率。
敌人弱点与应对策略
敌人弱点：不同的敌人有不同的弱点，例如部分敌人对物理攻击的防御力较低，部分敌人对魔法攻击的防御力较低。
应对策略：根据敌人的弱点，选择合适的攻击方式，例如使用魔法攻击来击败对物理防御力较低的敌人。
战斗技巧：在战斗中，利用技巧可以增加攻击力和防御力，例如使用连击和躲避技能。
观察与试探：通过观察敌人的攻击模式和试探敌人的弱点，可以更好地制定战斗策略。
战斗辅助技能与道具使用
地图探索与任务完成
PART 06
添加标题
支线任务：与主线任务相关的分支任务，可以获得额外奖励和经验值。
主线任务：游戏的主要剧情任务，需要按照顺序完成以推动剧情发展。
日常任务：每天都可以接取的任务，可以获得大量经验和奖励。
活动任务：特定时间段内开放的任务，可以获得丰厚奖励和经验值。
任务奖励积累经验值可以解锁更高级的副本和地图
圆桌骑士攻略砍出魔杖
汇报人：云
目录
01
添加目录项标题
02
游戏背景与规则
03
角色选择与升级
04
装备获取与强化
05
战斗策略与技巧
06
地图探索与任务完成
添加章节标题
PART 01
游戏背景与规则
PART 02
游戏背景介绍
游戏背景：中世纪欧洲，亚瑟王与圆桌骑士的传奇故事
游戏设定：玩家扮演圆桌骑士，通过砍杀敌人获取魔杖，挑战最终BOSS
装备强化：通过强化装备，提升角色属性和战斗力，同时解锁装备附加效果。
大山孕育天生神力 山岭巨人(小小)Tiny大型攻略
大山孕育天生神力山岭巨人（小小）Tiny大型攻略一、【英雄简介及评价】【山岭巨人】-【Stone Giant】【简称】-【Tiny】【昵称】-【小小】、【山岭】在艾森娜的岩石峭壁中的隐约可见的山岭巨人由大山孕育而生。
山岭巨人小小是一个新生的巨人。
身形虽小，却有着山一般的力量。
拥有超强力量的他可以把任何东西都扔到空中，比如投掷起大量的泥石以造成山崩来包围他的敌人。
他岩石般坚硬的外表，让攻击他的敌人反受其害。
随着时间的推移，来自他花岗岩心脏深处的吸引力能够吸引四周的石头土块，让山岭巨人“发育长大”，成为名副其实的山岭巨人。
【初始数据】：HP：606/606 MP：182/182 攻击距离：128弹道速度：近战攻击：61-67 攻击前/后摇：0.49/1施法前/后摇：0.001/0力量：24(+3) 敏捷：9(+0.9) 智力：14(+1.6)护甲：0（0.26）移动速度：285 基础攻击间隔: 1.7攻击间隔：1.56s(+9%IAS) 白天/夜晚视野：1800/800【25级时数据】：HP：1973/1974 MP：676/676 攻击: 133-139力量: 96 敏捷：30 智力: 52 护甲值：31级与25级裸装无黄点及25级有黄点对比【英雄评价】【总成长】：成长值之和5.5。
——评分：2.5星【各项成长】：【力量】：力量成长3。
山岭巨人力量成长不错，排名DOTA全英雄前几名，初始也达到了24点，所以血量上来说，山岭巨人血量很高。
——评分：5星【敏捷】：敏捷成长0.9。
山岭巨人的敏捷成长就是个悲剧，这导致了他不仅攻速慢，护甲也极其低。
——评分：1星【智力】：智力成长1.6。
山岭巨人的智力成长也偏低，因为作为一个前期靠法系输出吃饭的英雄，不论初始智力还是智力成长相对于山岭巨人的技能魔耗来说都太低了，这导致了山岭巨人及其缺魔。
——评分：1.5星【总体评分】：2星【初始属性分析】：HP：606，MP：182，攻击：61–67，攻击间隔：1.56s(+9%IAS)，护甲：0，移速：285，施法前/后摇：0.001/0，攻击前/后摇0.49/1。
圆桌骑士
第一关（进攻战火之村）1—1山上，向前，先劈开下面的桶，等红色小兵SOLDIER在地上滚的时候劈开400钱箱（可能出红震），劈出红震不要吃，先杀小兵再吃红震，2个带头盔的绿色小兵SOLDIER，2轻刀+1重刀。
开栅栏，600分不劈直接吃（劈开90%只有540）。
2个剑士SWORD MAN，各2个轻刀+1重刀，后出现的2个小兵只砍死1个，然后放保险（A+B）旋死1小兵SOLDIER 2剑士SWORD MAN（320+1600+1600）。
1—2过桥，有2个带头盔的绿色小兵，别砍他们，吃血吃分（不考虑劈红绿2震，耽误时间）。
后出现的2个不带头盔的红色小兵要赶快做掉1个，退后放保险旋倒绿色小兵（注意不要旋死刚才前面未杀完的红色小兵），之后顺序杀死3小兵（先杀红后杀绿多40分）。
BOSS战（SCORN）砍BOSS 5重刀左侧出现2个带头盔的绿色小兵，赶快做掉1个，另外1个赏1重刀不管他继续砍BOSS 2重刀，又出现1小兵(左侧)和1剑士（右侧），小兵依旧1重刀，剑士2轻刀+2重刀，之后保险旋死2小兵1剑士（3200+400+400），12000升级，之后让BOSS下班（整关速度要快，否则影响时间分）过关分数：25000+。
第二关（大乱战）2—1干掉迎面出现的2小兵，向前，2个剑士各2轻刀+1重刀，然后就坚决不碰他们(当然，让他们再挨1-2个轻刀也无所谓)。
2小兵就赐死吧，2粉色飞人MASK MAN各2重刀，放保险1600+1600+1600+1600入帐。
继续向前，吃分，杀胖子（4轻刀连杀，可能留180分珠宝），再干掉前面2个绿色小兵（1轻刀+1重刀）向前，出现2小兵2剑士，迅速解决前面的小兵，然后3轻刀+1重刀杀死后面的小兵形成4连（显示800分，有可能留600钱箱），现在该大干一场了，让敌人砍你几刀（只留五分之一以下的血，坚决不能放保险）将红牛BUSTER.F从马上砍下来，最好只留刚好踩死的血（马下砍1轻刀+2重刀），上马踩死2400（有可能留600箱子），剩下的2剑士、1胖子FAT MAN、1鸟人BIRD MAN、1灰牛BUSTER.S、1小兵、通通踩死而且不能把马弄丢了(1600+1600+1200+1200+3200+400入账可能再加800珠宝箱1个)。
亚瑟传说攻略
Schelfaniel 露西莉斯
主人公狄洛，住在叫做莫古里的小村里，父亲仲马是个猎人。但是父亲最后一次进山到现在还没有回来，听说最近野外 出现怪兽，担心父亲安危的狄洛决定去找父亲。但是村长让人看守西村口，想出去只好求助于好朋友乔亚（村长的儿 子）。
---去村长家找乔亚同行。 乔亚以村长让他和狄洛给猎人们送治疗药剂精灵之泪为借口，骗过守村口的人，和狄洛出村。 ---在村口打怪练级。 5级以后：
---到山洞1遇到 女 。 ---到山洞2遇到两个人封 入洞的入口。 ---到黑森林，蕾比决定加入队 共同
file:///F:/亚瑟传说攻略%20-%20世嘉街机专区%20-%20虎纹猫家园☆2011%20[我们4岁了]%20梦战·游戏·攻略·音乐·原创·专题·... 2011/12/22
亚瑟传说攻略 - 世嘉街机专区 - 虎纹猫家园☆2011 [我们4岁了] 梦战·游戏·攻略·音乐·原创·专题·文化·闲侃·竞技 - Powered by D... 页码，3/30
亚瑟传说攻略 - 世嘉街机专区 - 虎纹猫家园☆2011 [我们4岁了] 梦战·游戏·攻略·音乐·原创·专题·文化·闲侃·竞技 - Powered by D... 页码，4/30
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出水之殿进入地图有些时候我想抓住那一闪而逝的流星去照亮我渴望共鸣沉默的心心湖上结了冰人墙也没感应茫茫人海谁愿意与我结伴而行举杯同饮把所有美梦化整为零回复引用top页码1430亚瑟传说攻略虎纹猫家园2011我们4岁了梦战游戏攻略音乐原创专题文化闲侃竞技20111222file
亚瑟传说攻略 - 世嘉街机专区 - 虎纹猫家园☆2011 [我们4岁了] 梦战·游戏·攻略·音乐·原创·专题·文化·闲侃·竞技 - Powered by D... 页码，1/30
《欢乐数学之游戏大闯关：一本充满》记录
《欢乐数学之游戏大闯关：一本充满》读书札记目录一、内容综述 (2)二、书籍概述 (3)三、正文内容解析 (4)3.1 作者与出版社信息 (5)3.2 书籍背景及内容简介 (6)3.3 书籍结构安排 (7)四、深入解读游戏大闯关中的数学元素 (8)4.1 第一关 (9)4.2 第二关 (10)4.3 第三关 (12)4.4 第四关 (13)五、数学游戏在提升数学能力方面的作用分析 (14)5.1 激发数学学习兴趣 (15)5.2 锻炼逻辑思维与问题解决能力 (16)5.3 提升数学实践与创新能力 (17)5.4 培养数学思维的乐趣与自信 (18)六、《欢乐数学之游戏大闯关》的优缺点分析 (20)6.1 优点分析 (21)6.2 缺点探讨 (22)6.3 对未来类似书籍的展望与建议 (23)七、结语 (24)一、内容综述《欢乐数学之游戏大闯关》是一本以数学为主题的儿童益智游戏书籍。
本书旨在通过有趣的游戏和谜题，激发孩子们对数学的兴趣和热爱，让他们在轻松愉快的氛围中学习和掌握数学知识。
本书的内容丰富多样，涵盖了从基础数学到高级数学的多个领域。
通过精心设计的游戏和关卡，引导孩子们逐步深入数学的奇妙世界。
在基础数学知识部分，本书通过简单的游戏和例题，介绍了孩子们日常生活中所能接触到的数学概念，如数的基本概念、加减法、比较大小等。
这些游戏形式多样，让孩子们在玩耍中自然而然地掌握这些基础知识。
在中级数学挑战部分，本书引入了一些更具挑战性的游戏和谜题，如逻辑推理、图形空间等。
这些游戏旨在培养孩子们的逻辑思维能力和空间想象力，让他们在面对复杂问题时能够灵活运用数学知识。
在高级数学探险部分，本书通过一系列复杂的游戏和关卡，引导孩子们深入探索数学的奥秘。
这些游戏涉及函数、代数、几何等高级数学知识，让孩子们在解决问题的过程中拓宽视野，提升数学素养。
本书的特色亮点在于其独特的设计理念和丰富的游戏形式，作者巧妙地将数学知识融入游戏中，使孩子们在玩耍的过程中自然而然地掌握数学知识。
功夫熊猫游戏流程攻略
第4章 Protect the Palace
这关开始是用鹤飞行（飞行游戏），左摇杆是控制上下左右（上下是反的），X 是攻击，躲过黑色云彩，知道看见太阳。这里 有个技巧，有的玩家看不清前方的路，就一直跟着金币走就行。 GOINGAMER 极击网社区论坛9 Q' N: x2 j; t d 看到太阳，剧情过后，来到一个大厅，下来熊猫要根据提示做好多小任务。
任务：有的野猪是负责打架的，有的野猪是负责偷花瓶的，在这里暂称偷花瓶的野猪为老鼠 1：打老鼠防止偷花瓶，被偷后右下角会显示剩余个数，老鼠总共有三个地方可逃，两个是开始地点背后墙上的小洞（左右 各一），另一个是在水池右边院子的墙上有个小洞。 2：打野猪 8 y& R8 e1 M' s- B 3：打老鼠 4：打野猪战神3,如龙4,合金装备,符文工房3,最终幻想,魔女贝尼特,神秘海域,牧场物语,宵星传说,忍者龙剑传,生化危机5,如龙3,雷顿教授, 游戏,剧情,攻略,高清视频$ N9 r: L+ e3 e 5：关红房间天花板，在任务开始地点左边有个长廊，那里通向天花板打开的红房间（金币会指引你进去），进去看到两扇天 花板开了，先上右边武器架，上到顶回身一踢关住一扇（如图），另一扇上箱子跳起一脚就可关住。 \1 M' }: o3 O+ n. H1 f GOINGAMER 极击网社区论坛/ Y& k% v1 r# J" \; r1 g0 C' P/ d
BOSS 战：BOSS 是个大野猪挥舞着大刀，大家只要用 LT+A（翻滚躲避），然后用 XXY 招，3遍即可搞定它
第二章 Tournament of the Dragon Warrior 战神3|如龙4|合金装备|符文工房3|最终幻想|神秘海域|牧场物语|宵星传说|忍者龙剑传|生化危机5|如龙3|雷顿教授|游戏|剧情|攻略|高 清视频|下载1 }, ?' |+ n$ `, `- j# ]9 ~$ H 游戏攻略高清剧情视频9 k9 x9 j7 @* m- e. C5 k' ?3 ]' M# @ 被挡在门外，咱另想办法。右边尽头有辆马车，马车后面有个印章可以拾取，左边尽头有个蹦床，我们可以利用蹦床进去， 两边马车上有金币，我们可以把旁边的箱子用胖胖的身体挤过来跳上去得到（如图）
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?。
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圆桌骑士 魔杖 1UP 2UP 攻略
【所有宝物，必须大砍和必杀杀死关键人物~】1-1.下桶400|红震出来不打兵,先打开下面的桶,保持斜角距离，等红兵走几步后滚地的过程开400,就出红震#刷分经验，场景最后一幕，把巨刀手砍大半血，走到屏幕最右端，开AB，一次过清兵过场景，可叠加获分1-2.上栅栏水果|绿震杀掉上面个绿兵,把320蔬菜栅栏打开,自己站在旁边,等下面的绿兵站在你的左下往上斜跑到320蔬菜附近出刀攻击时候马上开320蔬菜,就出800绿震1-2.下栅栏400|红震打掉左上角的小兵，然后调戏右下角的绿兵，等他右跑+冲刺，AB旋倒，打开栅栏，开400出红震#刷分经验，第一幕，四个小兵，杀一个，用必杀释放倍数，然后杀三个，加BOSS的一个小兵，然后等巨刀手削血，必杀，可以升1级2-1.#刷分经验：这里一开始处在两剑士那里有一个400分箱，虽然可以打出红震，但不如杀四小兵旋杀两剑士和两忍者拿四倍分，因此这里的红震我就不写了.初段：杀四小兵旋杀两剑士和两忍者拿四倍分中段：杀四小兵，骑马，踏人加栅栏处的三剑士2-1初段下桶|红震（C步，即上下移动一小步，人物从C字母的上端看似往前绕了一个曲线才到C字母的下端）先消灭右上角的小胖，调戏右下角的小胖，等其冲刺后，走下移步，然后回旋步，最后就是C步时，AB旋倒，胖子可以不死，即可开桶2-1末段3桶处方法1：左桶|100%红震+上桶|100%魔杖三桶保留，首先处理右上角的小兵，砍4轻刀0，然后保证一刀（不管大小刀）可以死，想办法让小兵位于你的水平线向右跑步然后冲刺（我的方法是一直站在最上面那条水平线，先向前引诱出来，往回走一小段，再往前砍几刀，然后往回走一段，让左边留有一段空间，来回移动，然后通过跳动，大部分时间处于他的左方并且站立，迫使他向右跑，再水平向我冲刺），在冲刺过程KO他，这个小兵是为了400出红震。
打开400箱子一、出红震，震死右下角的小兵，直接开上面的800，出魔杖。
二、不出红震，往下引出下面的绿兵，同样4轻刀，等其在地上打滚的后期用AB旋死（一定要掌握好节奏，滚地后期，这个兵是为了开魔杖）开中间600，出6份，开800箱子，出魔杖。
《圆桌武士》65万分攻略
《圆桌武⼠》65万分攻略转载于N e w w i s e任天堂世界 功略危险系数较⼤，⽬标是分，不是保命(注：本功略不使⽤天兵加分法) 使⽤⾓⾊:l a n c e l o t 第⼀关 场景1：⼭上，向前，⼀个重⼑做掉不带头盔的战⼠，把分吃了。
两个带头盔的战⼠，2轻⼑+1重⼑。
开栅栏，600分不劈直接吃。
两个剑⼠，各2个轻⼑+1重⼑；后出现的两个⼩兵只砍死⼀个，然后放保险。
场景2：⼭下，有两个带头盔的⼩兵，千万别碰他们，吃⾎吃分。
后出现的两个不带头盔的⼩兵要赶快 做掉，并且不能伤害前⾯被⽆视的两个。
放保险(别问我为什么)，⼲掉那俩⼈。
B O S S战：5个重⼑出现2个带头盔⼩兵，快速⼲掉。
继续砍B O S S，来了1剑⼠+1⼩兵，砍剑⼠3个回⾝斩+1重⼑，⼩兵1重⼑，放保险(如果你R P好，这时候已经升级了)最后⼲什么呢？废话，当然是让B O S S下班啦 第⼆关 场景1：⼲掉迎⾯出现的⼩兵，向前，两个剑⼠各2轻⼑+1重⼑，然后就坚决不碰他们(当然，让他们再挨1-2个轻⼑也⽆所谓)。
2⼩兵就赐死吧，2飞⼈各2重⼑，放保险。
继续向前，吃分，杀胖⼦，再⼲掉那3个⼩兵(引诱第三个⼩兵的时候要慢点⾛，最好别把其他的也引出来)。
现在该⼤⼲⼀场了，不理前⾯出现的2剑⼠，最快速度⼲掉⼩兵形成4连，将红⾊B U S T E R从马上砍下来，剩下的任务就是踩死这些⼈⽽且不能把马弄丢了(什么？太危险？没看见顶楼头⼀句话吗？)。
3个剑⼠，在马上⼀⼈给3个轻⼑，然后⼀定要⼀起踩死(什么？还嫌危险？去去去，那就别玩了)兄弟，60000分了吧。
什么？没到？那你⼀定没按我功略⾛。
g e t d o w n t o b u s i n e s s。
开栅栏，震死两只可怜的鸟⼉，救兄弟，但先不杀死飞⼈，先吃分，再杀⼈(原因：怕劈出魔杖影响连分)做掉2⼩兵再⽤马踩死飞⼈(如果马在前⾯丢了就先杀飞⼈再⼩兵)。
开栅栏，杀死后⾯的两个⼩兵2胖⼦+1魔术师全踩死。
圆桌武士高概率实用固定打宝集锦
圆桌武⼠⾼概率+实⽤+固定打宝集锦为了满⾜⼤家的打宝⼼愿,我将我⾃⼰,黄Z Z兄,D B兄还有群内业界研究⼈⼠的多年研究成果和⼼⾎总结出来,⼤家可以试着在街机上⼀个个实现,我只说最好⽤和常见的,以免有⼈嫌⿇烦,记不住(前⾯带*的是只要操作正确就绝对100%) . 以下是⾼概率(⼀般都有50%以上,圆桌⾥概率50%就已经很好了,2打1中)+实⽤(⽐较简明扼要,不是很⿇烦)+固定(都是栅栏和桶之类留的东西). 第⼀关: 1-1.800红震 出来不打兵,先打开下⾯的桶,等⼩红兵⾛⼏步后前滚时候开400,就出红震 1-2.800绿震 杀掉上⾯个绿兵,把320蔬菜栅栏打开,⾃⼰站在旁边,等下⾯的绿兵站在你的左下往上斜跑到320蔬菜附近出⼑攻击时候马上开320蔬菜,就出800绿震 1-2.800红震 原理和[1-2.800绿震]类似 第⼆关: 2-1.6006开+800魔杖 地点在2个鸟之后的红忍与⼈质那⾥,你可以之前把忍杀了,400分的桶也要开了吃掉,之后剩下2个桶,2个站哨的绿兵,先引出1个随便杀了(底下⼀个不要惊动,之后再引出另外⼀个,轻砍2下或者3下到⼀保险就死的程度,引他在地上滚的时候保险死,先砍下⾯600桶,开600成6开(没有6开仍然有魔杖),再砍上⾯800桶,开800出魔杖 2-2.2U P (以后凡是涉及到2U P的⽣命都必须00,否则绝对不出2U P) 慢慢往⾥⾛,知道⽔果盘的栅栏能看见⼀个横⽊截⾯的程度,去砍开,320⽔果盘就被吸进屏幕了,之后慢慢引出B O S S,但凉着的320⽔果盘还能看见⼀点,等B O S S⾛V步拉马时候砍开,出2U P,⽽且刚刚吃到 2-2.800红震 原理和[2-2.2U P]⼀样,也是B O S S⾛V步拉马时候开 第三关: 3-1.2U P 说明:⽅法2我们俗称[斧男5跳出宝法],对于其他宝物仍然适⽤ *⽅法1(只限斧男):前⼀关B O S S处马不要放掉,引诱⼀⼜⽓的B O S S上马,等他拉马(你挡他马上拉马不算)时候保险死,吃600分,上马,背景⾳乐快结束时候输⼊1次前前踏马指令,发现马竟然踏了2次,第三关出来直接跳25次,之后不惊动守哨的胖⼦去开320⽔果盘栅栏,开⽔果盘,出2U P *⽅法2(只限斧男):类似于⽅法3,鸭⼦兵冲过来砍死之前去砍开320⽔果盘栅栏,之后等鸭⼦兵冲过来砍死,不管出什么分,⼤分⼤⾎不拆整吃,跳5下开320⽔果盘,出2U P ⽅法3:⽼⽅法,等鸭⼦兵冲过来砍死,出的如果是180,就去砍320⽔果盘栅栏,砍开320⽔果盘,出2U P 3-1.魔杖 *⽅法1:[斧男5跳出宝法] ⽅法2:同[3-1.2U P ⽅法3] 3-2.⼈质留的东西 只要留的东西能出宝,都可以打出宝,等红⽜蓄⼒准备跃起的时候砍,就出对应的该宝物,红⽜起跳位置最好隔留的东西近点,时机也要掌握好,理解红⽜的准备跃起状态 3-2.下⽅栅栏1U P *⽅法1:[斧男5跳出宝法] ⽅法2:同[3-1.2U P ⽅法3] ⽅法3:慢慢⾛,不引出鸭⼦,打开下⽅栅栏,留红⽜不要杀,之后处理⽅法同[3-2.⼈质留的东西] 3-2.(斧男)第⼀卷闸门前的魔杖 杀2⼤剑,别打破任何桶,绿⼩兵滚的时候保险死(先⼩砍2下,不然他不滚),跳4下,开800桶,开800分,出魔杖 *3-4.斧男2只⼤鸡(斧男⽆限⼤鸡前奏) 地点在B O S S之前4个桶那⾥,杀掉⽼虎后,引出1个绿兵,也是轻砍⼏下,然后等他在地上滚的时候保险死,然后跳4下,打开400的桶,开400 情况1:出800红震,就前前跑步把天井上的红兵和另外1个守哨的绿兵震死,之后跳6下,开2个桶(不开⾥⾯的分),最后开装⼤鸡的桶,砍之必定得到2个⼤鸡 情况2:不出800红震,那就⽐较⿇烦(之前⾃⼰的⾎量⾄少要⼀半才能保证这4个保险),慢慢⾛,背对另外1个守哨的绿兵,在不惊动他的情况下把他保险出来,之后他⼀站起来就保险死,不能让他哈⽓是关键(若让他哈了⽓,之后少开1个装分的桶就可以了),之后慢慢⼩步前⾛,有个红兵下来,保险死,⼀定要把他保险死在空中,之后跳17下,开2个桶(不开⾥⾯的分),最后开装⼤鸡的桶,砍之必定也得到2个⼤鸡 第五关: 5-1.开头魔杖 开始2个⼩红兵杀⼀个,把另外⼀个带到800分栅栏处,让此⼩红兵接近栅栏,之后砍掉栅栏,⼩兵⾛⼏步滚的时候开800分,出魔杖 5-1.第⼀灰⽜处1U P +魔杖(3⾓⾊通⽤) 这个恐怕是⼤家最关⼼的了,其实很简单,使此灰⽜的⾎在⼀个保险就死的量,⾃⼰⾎也要20%以下,最好只有保险⼀次的⾎,把马带过来让灰⽜坐上去,在灰⽜⾃发性拉马时候保险死他(你挡他拉不算),之后就会掉800分与不掉2种情况 100%1U P *情况1:掉800的话,整个吃了,开400桶,开400分,如果是4个140分,那么再开上⾯800桶和分,不管出不出魔杖,骑马踏12次,开底下500桶,开500,出1U P *情况2:不掉800,开400桶,开400分如果是4个140分,那么再开上⾯800桶和分,出魔杖的话,骑马踏28次,开底下500桶,开500,出1U P *情况3:也是不掉800,开400桶,开400分如果是3个140分,那么再开上⾯800桶和分,不管出不出魔杖,骑马踏28次,开底下500桶,开500,出1U P 100%魔杖 *情况1:掉800分的话,打开400桶+分,如果是3个,再砍下⾯500桶+盘⼦,出1U P的话,吃了1U P后反回去砍开先前灰⽜掉的800分(别让它消失),再骑马踏12次,砍上⾯800桶+800,出魔杖 *情况2:掉800分的话,打开400桶+分,如果是3个,再砍下⾯500桶+盘⼦,不出1U P的话,直接骑马踏12次,砍上⾯800桶+800,出魔杖 *情况3:掉800分的话,打开400桶的分,如果是3个,再砍下⾯500桶+盘⼦,不出1U P的话,反回去砍开先前灰⽜掉的800分,再骑马踏9次,砍上⾯800桶+800,出魔杖 由此看出:这⾥灰⽜必须都要掉800分,并且打开400分都是3个140分,才能100%打出魔杖,所以这⾥的魔杖⽐这⾥的1U P容易多了,不过在100%魔杖⾥⾯,500盘⼦出1U P的概率也是很⼤的,经过乱数原理测试,概率为6/8 *5-1.斧男2只⼤鸡(斧男⽆限⼤鸡前奏) 地点在B O S S之前2个600桶那⾥,留者不要砍,过了TA L L M A N后,杀⼀⼩红兵,引出下⾯蹲着的1只鸭⼦杀了,不要惊动上⾯的,之后才引出上⾯的,接着等鸭⼦冲过来砍死,之后注意站位,如图的位置最佳(图会后⾯放上来),然后前+跳+砍,砍死滚来的⼤红兵,反回来跳2次,开上⾯⼀个600桶不要开600分,再开⼤鸡的栅栏,开⼤鸡,出2个 第七关: *7-1.斧男2只⼤鸡(斧男⽆限⼤鸡前奏) 刚开始出来下⾯600栅栏不要砍,2个桶不要砍,砍了2个胖⼦后,同时引出第3个胖⼦和1个鸭⼦,先杀死胖⼦后等鸭⼦冲过来杀死,保持现场还有个鸭⼦不动的情况下,跳6次,砍开600栅栏和600,砍开下⾯320蔬菜桶,砍开蔬菜,绝对出绿震,之后慢慢⾥⾛,别惊动鸭⼦,背对者把鸭⼦反旋出来,等他起来哈⽓时候背对背旋第⼆次,当他再次起来哈⽓时候吃掉绿震,鸭⼦变⼤鸡了,马上跳7次,开上⾯的320⽔果盘,别砍开,再去砍⼤鸡,出2个 7-2.TA L L M A N震成的1U P 地点在7-2进城之后,我来说明 2个⼤红兵+3个⼤黄兵,留1个⼤黄兵,其余的杀!桶⼀个也别打破,往前⾛有3⼤剑,也杀!总之留1个⼤黄兵就可以了 步骤如下: 0.确保不要引出TA L L M A N的笑声,桶⼀个也别打破 1.黄⾊⼤兵打到⼀点⾎时候等他直跑出⼑攻击或者斜跑出⼑攻击的时候保险死 2.把屏幕拖到刚刚能打到左上⾓的红震⽊筒为⽌,打开此⽊桶,看到红镇落到屏幕⾥了,注意不要引出TA L L M A N的笑声 3.打开右上⾓的红震⽊桶,注意也不要引出笑声 4.打开中间绿震的⽊桶,再次注意不要引出笑声,然后吃掉绿震 5.打开左下⾓400的⽊桶后,可以引出笑声了,之后打开400分,之后就视出不出红震⽽分2种情况 情况1:出震: 1.直接吃掉这个红震,TA L L M A N变500盘⼦ 2.打开右下800的桶,不打开800分 3.打开TA L L M A N变的500盘⼦,出1U P *情况2:不出震: 1.吃掉左上⾓的红镇,TA L L M A N变500盘⼦ 2.吃掉右上⾓的红镇 3.打开右下800的桶,并且打开800分 4.打开TA L L M A N变的500盘⼦,出1U P 注意:整个过程⼀定要快,否则左上⾓的红震会消失 逗游⽹——中国2亿游戏⽤户⼀致选择的”⼀站式“游戏服务平台。
圆桌骑士打宝原理
圆桌骑士打宝原理圆桌骑士打宝是一款风靡全球的冒险解谜类游戏，玩家需要在游戏中探索各种地图，解开谜题，找到宝藏。
在这个过程中，玩家需要运用一定的原理和技巧来提高自己的游戏水平，下面我们就来详细探讨一下圆桌骑士打宝的原理。
首先，要想在圆桌骑士打宝中取得成功，最基本的原理就是要善于观察和思考。
游戏中的谜题和宝藏往往隐藏在各种看似普通的场景中，只有通过细致的观察和深入的思考才能找到线索，解开谜题。
因此，玩家在游戏中要保持警觉，仔细观察每一个细节，思考每一个可能的线索，这样才能更快地找到宝藏。
其次，在圆桌骑士打宝中，合理运用道具也是非常重要的原理之一。
游戏中会有各种各样的道具，有些道具可以帮助玩家解开谜题，有些道具可以帮助玩家克服障碍，因此玩家需要根据情况合理地运用这些道具。
有时候，一个看似普通的道具可能会成为解决难题的关键，因此玩家需要善于发现道具的潜在作用，并且合理地运用它们。
另外，团队合作也是圆桌骑士打宝中的重要原理之一。
在游戏中，有些谜题需要多人合作才能解开，因此玩家需要善于与队友沟通，协作解决问题。
团队合作不仅可以提高解谜的效率，还可以增强游戏的乐趣，因此玩家需要注重团队合作的重要性，积极与队友合作，共同攻克难关。
最后，耐心和毅力也是圆桌骑士打宝中不可或缺的原理。
游戏中的谜题往往需要玩家花费大量的时间和精力来解决，有时候甚至需要反复尝试才能找到正确的解法。
因此，玩家需要保持耐心，不断尝试，不断思考，直到找到正确的解决方案。
总的来说，圆桌骑士打宝是一款需要玩家不断观察、思考、合理运用道具、团队合作以及保持耐心和毅力的游戏。
只有掌握了这些原理，玩家才能在游戏中取得更好的成绩，找到更多的宝藏。
希望以上内容能够帮助玩家更好地理解圆桌骑士打宝的原理，并在游戏中取得更好的表现。