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Description: crc任意位生成多项式
任意位运算
自适应算法
循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的
编码方式，这种方法把要发送的数据看成是一个多项式的系数
，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为：
bn-1Xn-1+bn-2Xn-2+…+b１X+b０
例如：数据“10010101”可以写为多项式
X７+X４+X２+1。
循环冗余校验CRC
循环冗余校验方法的原理如下：
(1) 设要发送的数据对应的多项式为P(x)。
(2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式
的最高次幂为r。
(3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x)
。(左移r位)
(4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x)
×G(x)+R(x)。
(5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数
据块的末尾加上余式所对应的数据得到的。
(6) 发送T(x)所对应的数据。
(7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x)
，若余式为0，则认为没有错误，否则认为有错。-crc-generating polynomial arbitrary Operators adaptive algorithm Cyclic Redundancy Check (CRC. Cyclic Redundancy Code) is the polynomial coder, This way the data to be sent as a polynomial coefficient data bn - 1bn-2 ... b1b0 (0 or 1), corresponding to the polynomial : bn - 1Xn-1 bn - 2Xn-2 ... b1X belts such as : data "10010101" polynomial can be written as a X7 X4 X2. Cyclic Redundancy Check Cyclic Redundancy Check method of principle as follows : (1) The data to be sent to the corresponding polynomial p (x). (2) the sender and the receiver agreed on a generator polynomial G (x), set up the generator polynomial of the highest power of r. (3) In the data block Add to the end of r-0, then the polynomial corresponding to M (x) = XrP (x). (R-bits) (4) M (x) divided by G (
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Size: 460182 |
Author: johu |
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Description: crc任意位生成多项式 任意位运算 自适应算法 循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的 编码方式，这种方法把要发送的数据看成是一个多项式的系数 ，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为： bn-1Xn-1+bn-2Xn-2+…+b１X+b０ 例如：数据“10010101”可以写为多项式 X７+X４+X２+1。 循环冗余校验CRC 循环冗余校验方法的原理如下： (1) 设要发送的数据对应的多项式为P(x)。 (2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式 的最高次幂为r。 (3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x) 。(左移r位) (4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x) ×G(x)+R(x)。 (5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数 据块的末尾加上余式所对应的数据得到的。 (6) 发送T(x)所对应的数据。 (7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x) ，若余式为0，则认为没有错误，否则认为有错-crc-generating polynomial arbitrary arbitrary Operators adaptive algorithm Cyclic Redundancy Check (CRC. Cyclic Redundancy Code) is the polynomial coder, This way the data to be sent as a polynomial coefficient data bn - 1bn-2 ... b1b0 (0 or 1), corresponding to the polynomial : bn - 1Xn-1 bn - 2Xn-2 ... b1X belts such as : data "10010101" polynomial can be written as a X7 X4 X2. Cyclic Redundancy Check Cyclic Redundancy Check method of principle as follows : (1) The data to be sent to the corresponding polynomial p (x). (2) the sender and the receiver agreed on a generator polynomial G (x), set up the generator polynomial of the highest power of r. (3) In the data block Add to the end of r-0, then the polynomial corresponding to M (x) = XrP (x). (R-bits) (4) M (x) divided by G (
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Author: yzs |
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Description: Uniform Generator
Computer simulations often require random numbers. One way to generate pseudo-random numbers is via a function of the form
where `` " is the modulus operator.
Such a function will generate pseudo-random numbers (seed) between 0 and MOD-1. One problem with functions of this form is that they will always generate the same pattern over and over. In order to minimize this effect, selecting the STEP and MOD values carefully can result in a uniform distribution of all values between (and including) 0 and MOD-1.
For example, if STEP = 3 and MOD = 5, the function will generate the series of pseudo-random numbers 0, 3, 1, 4, 2 in a repeating cycle. In this example, all of the numbers between and including 0 and MOD-1 will be generated every MOD iterations of the function. Note that by the nature of the function to generate the same seed(x+1) every time seed(x) occurs means that if a function will generate all the numbers between 0 and MOD-1, it will generate pseudo-random numbers uniformly with every MOD iterations.
If STEP = 15 and MOD = 20, the function generates the series 0, 15, 10, 5 (or any other repeating series if the initial seed is other than 0). This is a poor selection of STEP and MOD because no initial seed will generate all of the numbers from 0 and MOD-1.
Your program will determine if choices of STEP and MOD will generate a uniform distribution of pseudo-random numbers.
Input
Each line of input will contain a pair of integers for STEP and MOD in that order ( ).
Output
For each line of input, your program should print the STEP value right- justified in columns 1 through 10, the MOD value right-justified in columns 11 through 20 and either ``Good Choice" or ``Bad Choice" left-justified starting in column 25. The ``Good Choice" message should be printed when the selection of STEP and MOD will generate all the numbers between and including 0 and MOD-1 when MOD numbers are generated. Otherwise, your program should print the message ``Bad Choice". After each output test set, your program should print exactly one blank line.
Sample Input
3 5
15 20
63923 99999
Sample Output
3 5 Good Choice
15 20 Bad Choice
63923 99999 Good Choice
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Size: 360 |
Author: samurai1x4kyo@126.com |
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Description: 黑客的工具~ 好用。用做各种扫描！ 还有攻击目标 有知趣者下！-Hacker s tool, very useful,which can scan many ports and attack them.
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Size: 949248 |
Author: 秦书玉 |
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Description: 提到 RPG （角色扮演游戏，Role Play Game），在座各位恐怕没有不熟悉的。从古老经典的 DOS 版《仙剑奇侠传》到新潮花哨的《轩辕剑》系列，无不以曲折优美的故事情节，美丽可人的主角，悦耳动情的背景音乐，震撼了每一个玩家的心灵。而说到 RPG，就不能不提 DirectX，因为 PC 上大部分的 RPG 都是用这个冬冬开发的。早在《轩辕剑叁外传——天之痕》推出的时候，我就曾想过用 DirectX 写一个自己的 RPG，自己来安排故事情节的发展，却总是因为这样或那样的事情，一直没有能够实现这个心愿。在耗费了宝贵的几年青春，搞定了诸如考试、恋爱、出国等琐碎杂事之后，我终于可以在这个 SARS 肆虐的时代，坐在陪伴了我整个大学生涯的电脑前，听着颓废而又声嘶力竭的不知名歌曲，写一些一直想写却没有写的东西。-mentioned RPG (role-playing game, Role Play Game) No, I am afraid you are not familiar with. From the old classic DOS version of "goods" to the garish hip "X-Men" series, all twists and turns to a beautiful story, beautiful girl, the protagonist, the emotionally pleasant background music, each one shook the player's minds. Talking about the RPG, one has to mention DirectX. because most of the PC RPG has been used in the development of the boys. Back in the "X-Men San rumor-- Tianzhigen" launched, I once thought of using DirectX write their own RPG, to arrange their own story of development, because it is a way or another things have not been able to achieve this aspiration. Wasting precious years of their youth, to get as examinations, love and leave the
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Size: 1313792 |
Author: 李金鹏 |
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Description: POP3弱口令扫描
DELPHI开发制作-POP3 weak password production scanning DELPHI
Platform: |
Size: 911360 |
Author: jimsun |
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Description: crc任意位生成多项式
任意位运算
自适应算法
循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的
编码方式，这种方法把要发送的数据看成是一个多项式的系数
，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为：
bn-1Xn-1+bn-2Xn-2+…+b１X+b０
例如：数据“10010101”可以写为多项式
X７+X４+X２+1。
循环冗余校验CRC
循环冗余校验方法的原理如下：
(1) 设要发送的数据对应的多项式为P(x)。
(2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式
的最高次幂为r。
(3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x)
。(左移r位)
(4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x)
×G(x)+R(x)。
(5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数
据块的末尾加上余式所对应的数据得到的。
(6) 发送T(x)所对应的数据。
(7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x)
，若余式为0，则认为没有错误，否则认为有错。-crc-generating polynomial arbitrary Operators adaptive algorithm Cyclic Redundancy Check (CRC. Cyclic Redundancy Code) is the polynomial coder, This way the data to be sent as a polynomial coefficient data bn- 1bn-2 ... b1b0 (0 or 1), corresponding to the polynomial : bn- 1Xn-1 bn- 2Xn-2 ... b1X belts such as : data "10010101" polynomial can be written as a X7 X4 X2. Cyclic Redundancy Check Cyclic Redundancy Check method of principle as follows : (1) The data to be sent to the corresponding polynomial p (x). (2) the sender and the receiver agreed on a generator polynomial G (x), set up the generator polynomial of the highest power of r. (3) In the data block Add to the end of r-0, then the polynomial corresponding to M (x) = XrP (x). (R-bits) (4) M (x) divided by G (
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Size: 459776 |
Author: johu |
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Description: delphi中读写XML文件,例子中说明如何更加msxml类实现读xml文件，与写xml文件的方法
-delphi were read and write XML documents, examples on how to achieve more mjs15451 category Reading xml document, and write the way xml
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Size: 416768 |
Author: 周林 |
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Description: crc任意位生成多项式 任意位运算 自适应算法 循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的 编码方式，这种方法把要发送的数据看成是一个多项式的系数 ，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为： bn-1Xn-1+bn-2Xn-2+…+b１X+b０ 例如：数据“10010101”可以写为多项式 X７+X４+X２+1。 循环冗余校验CRC 循环冗余校验方法的原理如下： (1) 设要发送的数据对应的多项式为P(x)。 (2) 发送方和接收方约定一个生成多项式G(x)，设该生成多项式 的最高次幂为r。 (3) 在数据块的末尾添加r个0，则其相对应的多项式为M(x)=XrP(x) 。(左移r位) (4) 用M(x)除以G(x)，获得商Q(x)和余式R(x)，则 M(x)=Q(x) ×G(x)+R(x)。 (5) 令T(x)=M(x)+R(x)，采用模2运算，T(x)所对应的数据是在原数 据块的末尾加上余式所对应的数据得到的。 (6) 发送T(x)所对应的数据。 (7) 设接收端接收到的数据对应的多项式为T’(x)，将T’(x)除以G(x) ，若余式为0，则认为没有错误，否则认为有错-crc-generating polynomial arbitrary arbitrary Operators adaptive algorithm Cyclic Redundancy Check (CRC. Cyclic Redundancy Code) is the polynomial coder, This way the data to be sent as a polynomial coefficient data bn- 1bn-2 ... b1b0 (0 or 1), corresponding to the polynomial : bn- 1Xn-1 bn- 2Xn-2 ... b1X belts such as : data "10010101" polynomial can be written as a X7 X4 X2. Cyclic Redundancy Check Cyclic Redundancy Check method of principle as follows : (1) The data to be sent to the corresponding polynomial p (x). (2) the sender and the receiver agreed on a generator polynomial G (x), set up the generator polynomial of the highest power of r. (3) In the data block Add to the end of r-0, then the polynomial corresponding to M (x) = XrP (x). (R-bits) (4) M (x) divided by G (
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Size: 1024 |
Author: yzs |
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Description: 五笔加加伴侣发布说明：
《五笔加加伴侣》是在针对“五笔加加1.0”和“五笔加加PLUS2.x”的工具软件，主要考虑以下原因而制作：
一、如果说极点码表对用户是透明的，微软输入法码表是半透明的，那么五笔加加的码表是不透明的。本工具使五笔加加的码表成为是半透明的。
二、五笔加加的词库由于存在导入反序问题，自已的专业词库导入加加不方便。本工具提供一个自行解决的途径。。
三、五笔加加的皮肤只有一个，本工具提供了一个小图标的例子。
安装特别提示：
本软件是绿色软件，不要注册表和系统中留下任何信息。使用本软件应先备份您正在使用的五笔加加。-five strokes increasing or partner Release : "Wubi do is partner" in the "five-stroke increase and 1.0" and "Wubi do is PLUS2.x" tool software , to consider the following main reasons for the production : one, if pole-table will be transparent to users. Microsoft input code table is translucent, then five strokes added on the stopwatch is opaque. The tools do is make five strokes of the code table is to become translucent. Two, five strokes increasing or thesaurus because of the introduction reverse order, their own professional thesaurus into increasing or inconvenient. The tool provides a way to resolve. . 3, five strokes do is the only one skin, the tool provides a small icon example. Installing special Tip : The software is green software, and system regis
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Size: 466944 |
Author: david |
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Description: 1、将makefile中的
CROSS =/usr/local/arm/2.95.3/bin/arm-linux-
CFLAGS+=-I/usr/local/src/2.4.18-rmk7/include
CFLAGS+=-I/usr/local/src/2.4.18-rmk7/include/linux
CFLAGS+=-I/usr/local/arm/2.95.3/arm-linux/include
路径改为你宿主机上的。
2、#make，来编译驱动模块。
3、#export PATH=$PATH:/usr/local/arm/2.95.3/bin //取决于你宿主机上的交叉编译工具路径
#arm-linux-gcc -o gpio_test gpio_test.c //编译应用程序
最终在你的当前文件夹下会生成gpiodrv.o与gpio_test两个文件。
4、#mkdir /tmp/1
5、通过ftp的方法，把gpio_test与gpiodrv.o文件下载到你的开发板上的/tmp/1目录中。
6、#mknod /dev/gpio c 220 0 //创建文件节点
#insmod gpiodrv.o
#chmod +x gpio_test
#./gpio_test
然后，就可以看到四个LED灯在闪烁了。:)-1, the makefile in the CROSS =/usr/local/arm/2.95.3/bin/arm-linux- CFLAGS+ =- I/usr/local/src/2.4.18-rmk7/include CFLAGS+ =- I/usr/local/src/2.4.18-rmk7/include/linux CFLAGS+ =- I/usr/local/arm/2.95.3/arm-linux/include changed the path on your host. 2,# make, to compile driver modules. 3,# export PATH = $ PATH:/usr/local/arm/2.95.3/bin// depends on your host on the path of the cross compiler tool# arm-linux-gcc-o gpio_test gpio_test.c// compile the application procedures for the ultimate in your current folder will be generated under the two documents gpiodrv.o and gpio_test. 4,# mkdir/tmp/1 5, through the ftp way to gpiodrv.o File Download gpio_test with the development of your board/tmp/1 directory. 6,# mknod/dev/gpio c 220 0// create a file node# insmod gpiodrv.o# chmod+ x gpio_test#./Gpio_test then four LED lights can be seen in the flicker of. :)
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Size: 9216 |
Author: 马工 |
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Description: 1.求双向循环链表的长度
2.在双向循环链表指定位置插入一个元素
3.给定双向循环链表中一个结点元素值e，要求将数据y插入到元素e的前面
4.给定双向循环链表中一个结点元素值e，要求将数据y插入到元素e的后面
5.给定双向循环链表中一个结点元素值e，要求将数据y插入到元素e的后面
6.删除双向循环链表中指定位置的元素
7.给定双向循环链表中一个结点元素值e，要求将其对应结点删除
8.给定双向循环链表中一个结点元素值e，要求删除对应结点的直接前驱
9.给定双向循环链表中一个结点元素值e，要求删除对应结点的直接后继
10.求双向循环链表中第i个元素
11.判断双向循环链表H中是否存在元素x，若存在返回1，否则返回0
12.将双向循环链表L就地逆置
13.将双向循环链表逆序输出
14.求双向循环链表H中结点p的直接前驱
15.求双向循环链表H中结点p的直接后继-1. Seek a two-way circular list the length of the
2. In a two-way circular linked list insert an element in the specified location
3. For a given two-way loop in a linked list node element of the value of e, requires that the data y is inserted into the front element e
4. For a given two-way loop in a linked list node element of the value of e, requires that the data y is inserted into the back of element e
5. For a given two-way loop in a linked list node element of the value of e, requires that the data y is inserted into the back of element e
6. Remove the two-way circular linked list of elements in the specified location
7. For a given two-way loop in a linked list node element of the value of e, request to delete the corresponding nodes
8. For a given two-way loop in a linked list node element of the value of e, to delete the corresponding node immediate precursor
9. For a given two-way loop in a linked list node element of the value of e, to delete the correspondi
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Size: 2048 |
Author: 张敏 |
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Description: Navicat For MySQL Enterprise Edition V8.2.19简体中文绿色版
Navicat是一个强大的MySQL数据库管理和开发工具。Navicat为专业开发者提供了一套强大的足够尖端的工具，但它对于新用户仍然是易于学习。Navicat,使用了极好的图形用户界面（GUI），可以让你用一种安全和更为容易的方式快速和容易地创建、组织、存取和共享信息。 用户可完全控制 MySQL 数据库和显示不同的管理资料，包括一个多功能的图形化管理用户和访问权限的管理工具，方便将数据从一个数据库移转到另一个数据库中（Local to Remote、Remote to Remote、Remote to Local），进行档案备份。 Navicat 支援 Unicode，以及本地或遥距 MySQL 伺服器多连线，用户可浏览数据库、建立和删除数据库、编辑数据、建立或执行 SQL queries、管理用户权限（安全设定）、将数据库备份/复原、汇入/汇出数据（支援 CSV, TXT, DBF 和 XML 档案种类）等。新版与任何 MySQL 5.0.x 伺服器版本兼容，支援 Triggers，以及 BINARY VARBINARY/BIT 数据种类 等的风范。
绿化说明:
1.双击绿化+注册.exe程序.就可以使用了.
2.该版本已去除注册机程序。（貌似杀软误报这个文件，所以去除了！）
3.该版中加入绿化卸载程序！ -Navicat For MySQL Enterprise Edition V8.2.19 green simplified Chinese version
Navicat is a powerful MySQL database management and development tools. Navicat for professional developers with a powerful set of tools for sophisticated enough, but it is still easy for new users to learn. Navicat, the use of an excellent graphical user interface (GUI), allows you to use a safe and easier way to quickly and easily create, organize, access and share information. Users can completely control the MySQL database and display different management information, including a multi-purpose graphical management of users and access rights management tools to facilitate the transfer of data from one database to another database (Local to Remote, Remote to Remote , Remote to Local), for file backup. Navicat supports Unicode, as well as local or remote MySQL server, multi-connection, the user can browse the database, creating and deleting a database, edit data, creating or perform SQL queries, manage us
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Size: 6144000 |
Author: 陈海龙 |
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Description: 汽车加油行驶问题（习题3-16）
问题描述：
给定一个N*N 的方形网格，设其左上角为起点，坐标为（1，1），X 轴向右为正，Y 轴向下为正，每个方格边长为1。一辆汽车从起点出发驶向右下角终点，其坐标为（N，N）。
在若干个网格交叉点处，设置了油库，可供汽车在行驶途中加油。汽车在行驶过程中应遵守如下规则：
（1）汽车只能沿网格边行驶，装满油后能行驶K 条网格边。出发时汽车已装满油，在起点与终点处不设油库。
（2）当汽车行驶经过一条网格边时，若其X 坐标或Y 坐标减小，则应付费用B，否则免付费用。
（3）汽车在行驶过程中遇油库则应加满油并付加油费用A。
（4）在需要时可在网格点处增设油库，并付增设油库费用C（不含加油费用A）。
（5）(1)～(4)中的各数N、K、A、B、C均为正整数。
算法设计：
求汽车从起点出发到达终点的一条所付费用最少的行驶路线。
-Driving automobile fuel problem (Exercise 3-16)
Problem description:
Given an N* N square grid, set the upper left corner as a starting point, coordinates (1,1), X-axis to the right is positive, Y axis under
Positive side of each square is 1. A car starting from the starting point towards the lower right corner of the end, the coordinates (N, N).
In a number of grid intersection points, set the oil tanks, fuel for the car is moving his way. Car in motion the process to be observed, such as
The following rules:
(1) The car can only travel along the grid edge, full of oil can travel the grid section K edge. Starting when the car was filled with oil, in the beginning and the end
No point depot.
(2) When the vehicle driving through a mesh edge, if the X coordinate or Y coordinate decreases, the fees payable B, or toll-free
Use.
(3) vehicles in the process of moving the event should fill up and pay for fuel oil storage costs A.
(4) when required grid points in the addi
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Description: 给定一个N*N 的方形网格，设其左上角为起点◎，坐标为（1，1），X轴向右为正，Y轴向下为正，每个方格边长为1。一辆汽车从起点◎出发驶向右下角终点▲，其坐标为（N，N）。在若干个网格交叉点处，设置了油库，可供汽车在行驶途中加油。汽车在行驶过程中应遵守如下规则：
（1）汽车只能沿网格边行驶，装满油后能行驶K 条网格边。出发时汽车已装满油，在起点与终点处不设油库。
（2）当汽车行驶经过一条网格边时，若其X 坐标或Y 坐标减小，则应付费用B，否则免付费用。
（3）汽车在行驶过程中遇油库则应加满油并付加油费用A。
（4）在需要时可在网格点处增设油库，并付增设油库费用C（不含加油费用A）。
（5）(1)～(4)中的各数N、K、A、B、C均为正整数。
编程任务：
求汽车从起点出发到达终点的一条所付费用最少的行驶路线。
数据输入：
第一行是N，K，A，B，C的值，2 <=N<=100，2<=K<=10。
第二行起是一个N*N 的0-1方阵，每行N 个值，至N+1行结束。方阵的第i
行第j 列处的值为1 表示在网格交叉点（i，j）处设置了一个油库，为0 时表示未设油库。各行相邻的2 个数以空格分隔。
结果输出:
第1 行中的数是最小费用值。
-Given an N* N square grid, set the upper left corner as a starting point ◎, coordinates (1,1), X-axis to the right is positive, Y axis, is positive, each square side length of 1. Starting a car approached from the lower right corner of the end point ◎ ▲, its coordinates (N, N). In a number of grid intersection points, set the oil tanks, fuel for vehicles in a moving way. Vehicle in motion the process should follow the following rules:
(1) car can only travel along the grid edges, filled with oil can travel after the K edge of the grid. Starting when the car has been filled with oil, not in the starting and ending at the depot is located.
(2) when the vehicle is traveling through a mesh edge, if the X coordinate or Y coordinate decreases, then the fees payable B, or toll-free use.
(3) the process of moving car in case of fuel oil tanks should fill up and pay the cost of A.
(4) when needed at additional grid point depot, depot and pay the additional cost of C (excluding fuel costs A)
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Author: 张开 |
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Description: USACooverfencing 题解，描述 农夫John在外面的田野上搭建了一个巨大的用栅栏围成的迷宫。幸运的是，他在迷宫的边界上留出了两段栅栏作为迷宫的出口。更幸运的是，他所建造的迷宫是一个“完美的”迷宫：即你能从迷宫中的任意一点找到一条走出迷宫的路。给定迷宫的宽度W(1<=W<=38)及高度H(1<=H<=100)。 2*H+1行，每行2*W+1的字符以下面给出的格式表示一个迷宫。然后计算从迷宫中最“糟糕”的那一个点走出迷宫所需的步数(就是从最“糟糕”的一点，走出迷宫的最少步数）。（即使从这一点以最优的方式走向最靠近的出口，它仍然需要最多的步数）当然了，牛们只会水平或垂直地在X或Y轴上移动，他们从来不走对角线。每移动到一个新的方格算作一步（包括移出迷宫的那一步）这是一个W=5,H=3的迷宫：-USACooverfencing Solution description in the fields outside, Farmer John, built a huge fenced maze. Fortunately, he set aside in the maze of border export of two fence maze. Fortunately, that he had built the labyrinth is a " perfect" maze: from any point in the maze to find a way out of the maze. To the maze-set the width of the W (1 < = W < = 38) and height H (1 < = H < = 100).* H+1 line, the character of each line* W+1 a maze in the format given below. And then calculate the most from the maze of " bad" a point out of the maze required number of steps (from " bad" , the minimum number of steps out of the labyrinth). (From this point the best way to the nearest export, it still needs the most number of steps) Of course, cattle are only level or vertically in the X or Y axis move, they do not have to go diagonal . Squares counted every move to a new step (including that step out of the maze) This is a W = 5, H = 3 of the maze:
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Description: sgu265：Wizards(计算几何)
题目大意：
一个空间直角坐标系内，对一个点有三种操作：
1.平移(x,y,z)⇒ (x+a,y+b,z+c)；
2.缩放(x,y,z)⇒ (ax,by,cz)；
3.绕一个向量(x,y,z)顺时针旋转α角(角度制)。
对空间内的m个点给出n个操作，求每个点在操作后的坐标。-265. Wizards
time limit per test: 0.5 sec.
memory limit per test: 65536 KB
input: standard
output: standard
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