 The dog task

Description
Hunter Bob often walks with his dog Ralph. Bob walks with a constant speed and his route is a polygonal line (possibly selfintersecting) whose vertices are specified by N pairs of integers (Xi, Yi) ? their Cartesian coordinates.
Ralph walks on his own way but always meets his master at the specified N points. The dog starts his journey simultaneously with Bob at the point (X1, Y1) and finishes it also simultaneously with Bob at the point (XN, YN).
Ralph can travel at a speed that is up to two times greater than his master's speed. While Bob travels in a straight line from one point to another the cheerful dog seeks trees, bushes, hummocks and all other kinds of interesting places of the local landscape which are specified by M pairs of integers (Xj',Yj'). However, after leaving his master at the point (Xi, Yi) (where 1 <= i < N) the dog visits at most one interesting place before meeting his master again at the point (Xi+1, Yi+1).
Your task is to find the dog's route, which meets the above requirements and allows him to visit the maximal possible number of interesting places. The answer should be presented as a polygonal line that represents Ralph's route. The vertices of this route should be all points (Xi, Yi) and the maximal number of interesting places (Xj',Yj'). The latter should be visited (i.e. listed in the route description) at most once.
An example of Bob's route (solid line), a set of interesting places (dots) and one of the best Ralph's routes (dotted line) are presented in the following picture:
InputThe first line of the input contains two integers N and M, separated by a space ( 2 <= N <= 100 ,0 <= M <=100 ). The second line contains N pairs of integers X1, Y1, ..., XN, YN, separated by spaces, that represent Bob's route. The third line contains M pairs of integers X1',Y1',...,XM',YM', separated by spaces, that represent interesting places.
All points in the input file are different and their coordinates are integers not greater than 1000 by the absolute value.
OutputThe first line of the output should contain the single integer K ? the number of vertices of the best dog's route. The second line should contain K pairs of coordinates X1'',Y1'' , ...,Xk'',Yk'', separated by spaces, that represent this route. If there are several such routes, then you may write any of them.
Sample Input4 5
1 4 5 7 5 2 2 4
4 2 3 9 1 2 1 3 8 3
Sample Output6
1 4 3 9 5 7 5 2 1 2 2 4
 其他相关推荐
 The Dog Task
 Hunter Bob often walks with his dog Ralph. Bob walks with a constant speed and his route is a polygonal line (possibly selfintersecting) whose vertices are specified by N pairs of integers (Xi, Yi)  their Cartesian coordinates.nnRalph walks on his own way but always meets his master at the specified N points. The dog starts his journey simultaneously with Bob at the point (X1, Y1) and finishes it also simultaneously with Bob at the point (XN, YN). nnRalph can travel at a speed that is up to two times greater than his master's speed. While Bob travels in a straight line from one point to another the cheerful dog seeks trees, bushes, hummocks and all other kinds of interesting places of the local landscape which are specified by M pairs of integers (Xj', Yj'). However, after leaving his master at the point (Xi, Yi) (where 1 <= i < N) the dog visits at most one interesting place before meeting his master again at the point (Xi+1, Yi+1). nnYour task is to find the dog's route, which meets the above requirements and allows him to visit the maximal possible number of interesting places. The answer should be presented as a polygonal line that represents Ralph's route. The vertices of this route should be all points (Xi, Yi) and the maximal number of interesting places (Xj', Yj'). The latter should be visited (i.e. listed in the route description) at most once.nnAn example of Bob's route (solid line), a set of interesting places (dots) and one of the best Ralph's routes (dotted line) are presented in the following picture:nnnnnInputnnThe first line of the input contains two integers N and M, separated by a space (2 <= N <= 100, 0 <= M <= 100). The second line contains N pairs of integers X1, Y1, ..., XN, YN, separated by spaces, that represent Bob's route. The third line contains M pairs of integers X1', Y1', ..., XM', YM', separated by spaces, that represent interesting places.nnAll points in the input are different and their coordinates are integers not greater than 1000 by the absolute value.nnnOutputnnThe first line of the output should contain the single integer K  the number of vertices of the best dog's route. The second line should contain K pairs of coordinates X1", Y1", ..., XK", YK", separated by spaces, that represent this route. If there are several such routes, then you may write any of them.nnnThis problem contains multiple test cases!nnThe first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.nnThe output format consists of N output blocks. There is a blank line between output blocks.nnSample Inputnn1nn4 5n1 4 5 7 5 2 2 4n4 2 3 9 1 2 1 3 8 3nnnSample Outputnn6n1 4 3 9 5 7 5 2 1 2 2 4n
 The dog task
 DescriptionnnHunter Bob often walks with his dog Ralph. Bob walks with a constant speed and his route is a polygonal line (possibly selfintersecting) whose vertices are specified by N pairs of integers (Xi, Yi) ? their Cartesian coordinates. nRalph walks on his own way but always meets his master at the specified N points. The dog starts his journey simultaneously with Bob at the point (X1, Y1) and finishes it also simultaneously with Bob at the point (XN, YN). nRalph can travel at a speed that is up to two times greater than his master's speed. While Bob travels in a straight line from one point to another the cheerful dog seeks trees, bushes, hummocks and all other kinds of interesting places of the local landscape which are specified by M pairs of integers (Xj',Yj'). However, after leaving his master at the point (Xi, Yi) (where 1 <= i < N) the dog visits at most one interesting place before meeting his master again at the point (Xi+1, Yi+1). nYour task is to find the dog's route, which meets the above requirements and allows him to visit the maximal possible number of interesting places. The answer should be presented as a polygonal line that represents Ralph's route. The vertices of this route should be all points (Xi, Yi) and the maximal number of interesting places (Xj',Yj'). The latter should be visited (i.e. listed in the route description) at most once. nAn example of Bob's route (solid line), a set of interesting places (dots) and one of the best Ralph's routes (dotted line) are presented in the following picture: n![](http://poj.org/images/1034/dog.gif)nInputnnThe first line of the input contains two integers N and M, separated by a space ( 2 <= N <= 100 ,0 <= M <=100 ). The second line contains N pairs of integers X1, Y1, ..., XN, YN, separated by spaces, that represent Bob's route. The third line contains M pairs of integers X1',Y1',...,XM',YM', separated by spaces, that represent interesting places. nAll points in the input file are different and their coordinates are integers not greater than 1000 by the absolute value. nOutputnnThe first line of the output should contain the single integer K ? the number of vertices of the best dog's route. The second line should contain K pairs of coordinates X1'',Y1'' , ...,Xk'',Yk'', separated by spaces, that represent this route. If there are several such routes, then you may write any of them.nSample Inputnn4 5n1 4 5 7 5 2 2 4n4 2 3 9 1 2 1 3 8 3nSample Outputnn6n1 4 3 9 5 7 5 2 1 2 2 4
 DOG=!DOG 这句有什么用?
 在导师的项目的程序里,经常看到:rnsbit DOG=P1^1;rnDOG=!DOGrnrn无论是主程序还是子程序,好象没几行就会出现,到底是什么意思呢?
 Top Dog
 You are a topsecret, governmentemployed software engineer assigned to TOPDOG, the latest military intelligence program. TOPDOG handles everything from mapping out enemy territory and position to parsing highly encrypted messages from the Commander in Chief.nThe operator of TOPDOG needs to be able to transfer information from remote computer workstations onto diskette in case of hardware problems or wellplaced enemy fire. All TOPDOG information is stored in the Oracattle database, but the only way to access the database is through the Dispatcher, an infamous and powerful software layer that only allows access to the database on a "needtoknow" basis. You are in charge of writing a piece of software which will get the desired information from the Oracattle database and store it on a "flat" file which will later be sent out to disk.nnnInput and OutputnnThe first line of each case contains three integers  the number of lines for the three parts.nnThe first part of each case contains the desired tables to be exported. This part will simply contain the names of each table to be exported on separate lines. There will be no blank lines, and each table name will be unique.nnFor each table, you must search the second part which will contain all column names, types, and sizes (if VARCHAR) for each table in the Oracattle database. This is the only information which the Dispatcher allows you access to without begging.nnNote that, for security reasons, not every table in the first part will always be available in the second part. Under each table name is the column name (one unique word), data type, and size (for VARCHAR), each separated by one space. The # sign indicates there are no more columns for the table. There will never be consecutive # signs immediately following each other, and each table will contain at least one column name with the size. The only four data types are VARCHAR, INT, DATE, and LONGINT.nnAn Oracattle SQL statement must be built in order to query the Dispatcher for the desired table data. The Oracattle SQL statement must be precisely built in order to keep the Dispatcher happy (we wouldn't want the Dispatcher to be confused). The statement begins with "SELECT", followed by each column name and generic data type in parenthesis, separated with commas, terminated with "FROM", the table name, and a semicolon. The generic data type is CHAR for VARCHAR and DATE, and NUM for INT and LONGINT. The generic name must be used because the Dispatcher only understands data as CHAR or NUM (it may be powerful, but it's not extremely intelligent). These SELECT statements must be put into the first part of the output.nnIf a table name cannot be found in the second part, a "" is returned for rows with empty fields.nnNote that each line in the third part may contain any number of spaces between words unless it was declared as a NUM. Also, all data is returned by row and table name in the same order it was presented to the Dispatcher. If no data exists in the table, a # sign immediately follows the table name (as in GROUPSPI in the example data).nnYou must now finally integrate all the information you have received from the Dispatcher into the second part of the output. This part will contain all data needed to describe the database tables. This file will later be imported using the Oracattle SqlImporter (OSI), a text to database utility. Lucky for you, all you need to do is get the second part of the output into the OSI format. This can be a little tricky. The first argument to be supplied is the table name, followed by the number of columns in parentheses, followed by the number of records (rows) in parentheses (no spaces on this line).nnNext comes the column name and then the data in quotation marks (a single space should separate the column name and it's data). When the maximum length of the column data is greater than 64, the size must also be supplied in parenthesis immediately following the column name (no space inbetween). This is so OSI can allocate more memory for large data.nnNo blank lines are to exist in the second part of the output, and all data must remain on the same line as the column name (no endofline characters inbetween quotation marks). Once this file has been created, you are all done! There will be a maximum of 100 columns in a single table, but there may exist any number of rows in a single table. The maximum column name and table name length is 25, and the maximum data length is 100. All input will always contain data in an the expected format (as described in these specifications), so there is no need for error checking. Remember, case is iMpOrTaNt  is not the same as .nnProcess to the end of file.nnnSample Inputnn4 25 28nINTELSYSnGROUPSPInSYSINTELnDEPLOYREGnGROUPSPInGRCODE INTnGRSUBNET VARCHAR 20nGRREGION VARCHAR 25nGRACTION VARCHAR 100nGRREF VARCHAR 100n#nINTELSYSnISDATE DATEnISNUM INTnISGEN LONGINTnISSUBGEN VARCHAR 25n#nSYSINTELnSITRANS INTnSISUBLET LONGINTnSINUM INTnSIGEN VARCHAR 10nSIACTION VARCHAR 50nSINOTES VARCHAR 100n#nQUICKFInQFDATE DATEnQFDATA VARCHAR 100n#nINTELSYSn122922T DEC 94n1n2nn111111Z DEC 01n3n4nCONFIRMEDn010101Z DEC 02n5n6nn020202Z DEC 03n7n8nCAN'T SAYn#nGROUPSPIn#nSYSINTELn342n3498938n000nSCOUTAnPURGE DATABASEnUNABLE TO COMPLY WITH A2DD UNDER GENERAL BURK'S COMMANDn#nnnSample OutputnnSELECT (CHAR) ISDATE, (NUM) ISNUM, (NUM) ISGEN, (CHAR) ISSUBGEN FROM INTELSYS;nSELECT (NUM) GRCODE, (CHAR) GRSUBNET, (CHAR) GRREGION, (CHAR) GRACTION, (CHAR) GRREF FROM GROUPSPI;nSELECT (NUM) SITRANS, (NUM) SISUBLET, (NUM) SINUM, (CHAR) SIGEN, (CHAR) SIACTION, (CHAR) SINOTES FROM SYSINTEL;n " statement must be substituted for the SELECT statement in the first part of the output. No blank lines are to be in this file, and only single spaces are to separate SELECT, column names, FROM, and the table name in the SELECT statements. The entire SELECT statement must be on one line.nnSince you currently do not have access to the Dispatcher, we will assume that you have correctly built the first part of the output and that the Dispatcher has processed it and created the table information in the third part. This file contains the table name followed by the data from each row in the Oracattle Database table. ""nISDATE "020202Z DEC 03"nISNUM "7"nISGEN "8"nISSUBGEN "CAN'T SAY"nGROUPSPI(5)(0)nSYSINTEL(6)(1)nSITRANS "342"nSISUBLET "3498938"nSINUM "000"nSIGEN "SCOUTA"nSIACTION "PURGE DATABASE"nSINOTES(100) "UNABLE TO COMPLY WITH A2DD UNDER GENERAL BURK'S COMMAND" nINTELSYS(4)(4)nISDATE "122922T DEC 94"nISNUM "1"nISGEN "2"nISSUBGEN ""nISDATE "111111Z DEC 01"nISNUM "3"nISGEN "4"nISSUBGEN "CONFIRMED"nISDATE "010101Z DEC 02"nISNUM "5"nISGEN "6"nISSUBGEN "
 Dog & Gopher
 Description n一只地鼠想沿直线跑到位于点(x2,y2)的一个坑中，但是在(x1,y1)有一条狗想吃它，狗的速度是地鼠速度的两倍，如果狗比地鼠提前或者同时到达(x2,y2)，地鼠就会被狗吃了，地鼠可以选一些坑作为起点，问地鼠能否成功逃脱，如果可以，输出离(x2,y2)最近的坑，如果最近距离有多个则输出首先输入的坑 nInput n第一行为四个浮点数分别表示x1,y1,x2,y2，即狗的初始位置和地鼠的目标位置，之后输入多行，每行两个浮点数表示一个地鼠可以选择的初始坑，以文件尾结束输入 nOutput n如果地鼠从任何一个坑出发都会被狗吃掉则输出The gopher cannot escape.如果地鼠能够成功逃脱，则输出The gopher can escape through the hole at (x,y).（其中(x,y)为离(x2,y2)最近的坑，如果最近距离有多个则输出先输入的坑） nSample Input n1.000 1.000 2.000 2.000 n1.500 1.500 nSample Output nThe gopher cannot escape.
 DOG简析
 Difference of Gaussian(DOG)是高斯函数的差分。我们已经知道可以通过将图像与高斯函数进行卷积得到一幅图像的低通滤波结果，即去噪过程，这里的Gaussian和高斯低通滤波器的高斯一样，是一个函数，即为正态分布函数。那么difference of Gaussian 即高斯函数差分是两幅高斯图像的差，一维表示：二维表示：opencv实现：using namespace std; u
 mapgis DOG
 中地公司MAPGIS的软件狗 从控制面板中,添加新硬件
 task task 实体
 jbpm 流的控制 过程 task 实体流的定义
 task与自定义task
 gradle管理本身自带了一系列的task，要了解有哪些task，可以用命令行–gradle tasks查看 自定义task–我们可以自定义一些task例如：task copyReleaseAPKToDistributeFolder(type: Copy, dependsOn:'assembleRelease') { def manifestFile = file("src/main/An
 Task中启动task的风险分析
 这两天遇到一个task中参数传递错误的问题，折腾了一天多，总算找到原因了，记录之。
 lucky dog 恶作剧
 一个很简单的恶作剧程序。原理很简单，但挺有意思
 看门狗模式 watch dog
 linux看门狗 c c++ arm linux drivers开源代码 看门狗其实就是一个可以在一定时间内被复位的计数器。当看门狗启动后，计数器开始自动计数，经过一定时间，如果没有被复位，计数器溢出就会对CPU产生一个复位信号使系统重启（俗称“被狗咬”）。系统正常运行时，需要在看门狗允许的时间间隔内对看门狗计数器清零（俗称“喂狗”），不让复位信号产生。如果系统不出问题，程序保证按时“喂狗”，一旦程序跑飞，没有“喂狗”，系统“被咬”复位。
 teach a dog to rest
 该文档可以指导开发者构建restful api ,文档内容轻松，简短，与大家分享
 Difference of Gaussian DoG
 DoG (Difference of Gaussian)实现角点检测。 效果见http://blog.csdn.net/abcjennifer/article/details/7639488#comments
 Cat VS Dog(二分匹配)
 The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a likeanimal and a dislikeanimal, if the child's likeanimal is a cat, then his/hers dislikeanimal must b...
 基于DoG的边缘检测
 边缘检测 difference of gaussion
 计算机视觉——DoG和LoG算子
 计算机视觉—DoG和LoG算子 brycezou@163.com 阅读本文，需要有一定的数字图像处理基础，否则不太容易明白数学公式想要传达的物理意义。希望通过仅此一篇文章就能让你理解图像处理中的高斯滤波（也叫高斯平滑、高斯模糊、高斯卷积）、DoG算子、LoG算子，以及它们之间的关系。下面先讲理论，再讲实际应用。在理论部分，一切语言都显得过于苍白，因此我只给出了最核心的、最简
 DOG算子的特征提取（二）
 DOG（Difference of Guassian）：简称 高斯函数的差分，是灰度图像增强和角点检测的一种方法。 （一）理论基础： 下面详细介绍DOG的角点检测（也称特征点提取）的理论过程： 首先，通过将目标图像与高斯函数进行卷积运算得到一幅目标图像的低通滤波结果，此过程称为去燥。（注：这里的Gaussian和高斯低通滤波器的高斯是一
 Cat和Dog的问题。
 using System;rnusing System.Collections; rnnamespace ConsoleApplication3rnrn abstract class Animalrn rn public abstract void MakeNoise();rn rnrn class Cat:Animalrn rn public override void MakeNoise()rn rn Console.WriteLine("Meow!");rn rn rnrn class Dog:Animalrn rn public override void MakeNoise()rn rn Console.WriteLine("Woof");rn rn rnrn class Class1rn rn static int Main(string[] args)rn rn ArrayList zoo;rn zoo = new ArrayList(3); rnrn Cat Sasha,Koshka;rn Sasha = new Cat();rn Koshka = new Cat();rnrn Dog Milou;rn Milou = new Dog();rnrn zoo.Add(Milou);rn zoo.Add(Sasha);rn zoo.Add(Koshka);rnrn foreach (Animal a in zoo) //这里将Animal改为Cat或Dog就出错，为什么？ rn rn a.MakeNoise();rn rnrn Console.WriteLine("Capacity = 0.",zoo.Capacity);rn Console.WriteLine("Hit Enter to terminate...");rn Console.Read();rn return 0;rn rn rnrn
 DoG算子和LoG算子
 DoG(Difference of Gaussian)算子和LoG(Laplacian of Gaussian)算子是常用的极值点检测(Blob Detection)两种方法，高斯卷积是为了进行尺度变换，那么LapLacian呢。 因此这里首先引入LapLacian算子。 图像边缘检测 因此进行边缘检测有两种方法。 一阶导数的极值 梯度算子定义为：G(x,y)=∇xf(x,y)2+∇yf...
 Dog类c++（练习题）
 #include &lt;iostream&gt;#include&lt;cmath&gt;using namespace std;class Dog{ public: Dog(); Dog(int a,double w):age(a),weight(w) {cout&lt;&lt;"Consyructor called"&lt;&lt;endl;}...
 面向对象Dog（一）
 package Demo1;public class Demo { private String name = "大白"; private int health = 100;//健康 private int love = 0;//亲密度 private String strain = "拉布拉多犬"; //狗方法 public void Dog(){ System.out.println("我的...
 cat vs dog
 cat vs dog 经典的二分类深度学习
 DoG角点检测（C++&openCV2.4.6版本）
 为C++结合opencv2.4.6版本，配置好VS后，改变路径，直接运行即可
 DOG与寻找极值点
 代码较全 提供给刚接触Matlab的同学 效果明显
 Big Dog 的介绍
 四足仿生机器人BIGDOG的简介。学习一下国外的机器人技术。BIGDOG是美国波士顿公司研发的。
 LoG and DoG Filters
 LoG and DoG Filters. Good introduction for CV beginners to understand how some block filter is generated ! And sth else
 高斯差分（DoG）MATLAB代码
 一个简单的高斯差分算法实现，运行时设置输入图像大小即可
 标定差值图像 DoG
 该代码用matlab编写，主要是对图像用不同方差的高斯过滤器进行处理，然后相减获取DoG图像。由于差值图像会出现负值的情况，所有需要进行标定处理，代码出列出了两种标定算法。
 Arnie's Dog Biscuits
 Our discerning gourmet puppy Arnie is turning to you for a program to help him split his dog biscuits. Each biscuit is shaped like a rectangle and perforated into equal sized squares:nnnnUnfortunately, Arnie will only eat squareshaped biscuits; therefore, he must break the biscuit into squares. Each break, termed a split, is applied to one rectangle, runs along one straight perforated line, and separates the rectangle into two pieces:nnnnInputnnThe first line of the input contains one positive integer n, the number of biscuits to split. Each of the next n lines contains two positive integers r and c, the number of rows and columns of one biscuit, separated by white space.nnOutputnnThe output contains one line for each biscuit specifying the minimal number of splits required to break the biscuit into squares.nnSample Inputnn2n6 7n5 5nThis defines two biscuits: the one shown above which requires four splits, and a square biscuit which requires no splits.nnSample Outputnn4n0n
 Cat vs. Dog
 nn```n DescriptionnThe latest reality show has hit the TV: ``Cat vs. Dog''. In this show, a bunch of cats and dogs compete for the very prestigious Best Pet Ever title. In each episode, the cats and dogs get to show themselves off, after which the viewers vote on which pets should stay and which should be forced to leave the show.nnEach viewer gets to cast a vote on two things: one pet which should be kept on the show, and one pet which should be thrown out. Also, based on the universal fact that everyone is either a cat lover (i.e. a dog hater) or a dog lover (i.e. a cat hater), it has been decided that each vote must name exactly one cat and exactly one dog.nnIngenious as they are, the producers have decided to use an advancement procedure which guarantees that as many viewers as possible will continue watching the show: the pets that get to stay will be chosen so as to maximize the number of viewers who get both their opinions satisfied. Write a program to calculate this maximum number of viewers.nInputnOn the first line one positive number: the number of testcases, at most 100. After that per testcase:nn* One line with three integers c, d, v (1 ≤ c, d ≤ 100 and 0 ≤ v ≤ 500): the number of cats, dogs, and voters.n* v lines with two pet identifiers each. The first is the pet that this voter wants to keep, the second is the pet that this voter wants to throw out. A pet identifier starts with one of the characters `C' or `D', indicating whether the pet is a cat or dog, respectively. The remaining part of the identifier is an integer giving the number of the pet (between 1 and c for cats, and between 1 and d for dogs). So for instance, ``D42'' indicates dog number 42.nOutputnPer testcase:nn* One line with the maximum possible number of satisfied voters for the show.nSample Inputn2n1 1 2nC1 D1nD1 C1n1 2 4nC1 D1nC1 D1nC1 D2nD2 C1nSample Outputn1n3n```n
 分布式系统sheepdog之dog执行流程
 dog部分主要是执行客户端的命令行请求，然后对命令进行解析，通过指定socket发送请求到sheep端，将请求交sheep端处理。具体流程请参考下图。 init_commands(&commands)函数将dog支持的命令都初始化在commands中进行调用，包括对vdi、cluster、node的命令操作， setup_commands()函数先比较主命令，然后比较subvommma
 log算子 和dog 算子
 log算子 和dog 算子
 001IT dog英语词汇
 style样式 solid实线 float浮动 position位置 margin外边距 margin 10px 5px 4px 15px（上右下左） border边界（边界粗细） padding内间距 absolute决对 clip剪刀 overflow溢出 visibility可见性，占空间 display显示，不占空间 space空间 variant不同的 italic斜体 repeat重复...
 hdu2768Cat VS Dog（最大独立集）
 这个题目就是把喜欢猫的和喜欢狗的分开，然后用结构体表示一下喜欢和讨厌，这个地方用的string，方便以后判断相等的地方，把喜欢猫里面不喜欢狗和喜欢狗的相比找一样的，或者喜欢猫里和讨厌猫相比找一样的，然后建边，求最大独立集就是总的点数最大匹配数#include <iostream> #include <string.h> #include <stdio.h> using namespace std;
 DOG高斯差分
 matlab实现的高斯差分滤波，DoG滤波
 Soft Dog edit32
 Soft Dog Edit32,Safe Net 专用写狗工具
 Cat VS Dog 最大独立集
 The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a likeanimal and a dislikeanimal, if the child’s likeanimal is a cat, then his/hers dislikeanimal must b...
 厚积薄发与Lucky Dog
 晚上在宿舍看TOEFL单词，“千万不要说我蛋疼”，由于下周三要考英语了。感受是：很多单词面熟，是经常遇到的常用词，但就是不能像熟人那样和它们打招呼，只能做我熟悉的陌生人。 突然来的感受，想写篇文章。至于目的，起警示、激励作用。 我似有先天多动症，是介于单线程人类和多线程人类之间的一种生物：做一件事很难完全集中精力，绝大多数情况下会有第二件事或者更多事会让我分心。 我很少会把一件事进行到底，总
 程序Dog的大梦想
 一、我是程序狗 “怎么又是任宏啊，每天都起这么早，要命啊……” “人家任宏可是要成为学霸的男人，咱们这些凡夫俗子啊，理解不了这么伟大的理想……”————微电影《初心》 上面这个场景，来自于本狗写的一个微电影剧本。人家都说，艺术源于生活，却又高于生活。没错的，虽然我的梦想并不是成为学霸。 IT行业，码农们被戏称为程序猿，更有“码农有三好，钱多话少死得早”这样的言论。虽然不知道码农的确切定义，...