Introduction 041602204 : 我是喜欢狗狗(particularly Corgi & Shiba Inu.)的丁水源 : 我的爱好是音乐.电影.英语(100%!!!!).吉他(but just a tiro!!!) :我最喜欢蔬菜吧!(各大餐厅都不错~) : 我最想说的一句话是: 抓住每一个maktub!好好学习!好好生活!冲鸭!(争取早日赚到自己的F*** money!!)希望在今后的生活中,在自己生活的还算decent 的前提之下,能投身公益~~虽然计算机是我十分喜欢的一…

1.Team Leader Link:        柯奇豪:点我 2.NABCD Model: Need(需求)——客户需求是什么? *. 希望能够有一款集成日常办公所需功能的软件(如:想法搜集.投票.签到.文章策划书论文第三方修改,提意见等的日常办公功能)                  *.希望使用方便,最好是能依托与一些我们日常使用的平台,允许跨设备的即时即用,尽量减少新软件的载入.                  *. 希望办公方便,精简高效.(如:使用微信办公.)   Appro…

1. a.This is the chart b. initial numPrimes = 4, t1 would over the loop. c. t = ( n = 1) d. node coverage: TR = {1,2,3,4,5,6,7,8,9,10,11,12} Test Paths :[1,2,3,4,5,6,3,7,8,9,2,10,11,10,12] edge coverage:TR = {(1,2),(2,10),(10,12),(10,11),(11,10),(2,3…

Problem 1 1. The fault is i > 0, it should be  i >= 0, because if the case is x = [0], y= 0, we expect = 1, but actually, we get -1; 2.  test: x = [], y = 1 does not execute the fault; 3.  test: x = [1,2,3], y = 3 executes the fault, but does not re…

I have met some errors in recent years, one of them which impress me most. It happend when I try to finish a web project with javascript for my web final project. In a web project, we often use 'button' to trigger some events, some times it is very s…

Ubuntu 14.04 安装 libpcap-1.1.1 & libpnet-1.1.4 & NMAC function lib 参考: NetMagic.org yacc: command not found libpcap-1.1.1 $ cd libpcap-1.1.1 $ ./configure Hint: 出现错误: configure: error: Your operating system's lex is insufficient to compile 原因是缺少fle…

Recently, I have a project, I use Unity3D to finish a visualization work, which is the final project for my visualization course. In this project, I try to visualization some frescoes' deterioration data to help the cultural workers to analiysis and…

I just want to say something about my java project that I did last year. Our task is to finish a linking game. It is a game that you should link all the same picture with specific rules. It is such an interesting game but creating the game can not be…

1. What is Quality? Quality means, “meeting requirements.” ..Whether or not the product or service does what the customer needs. In another word Quality is, “fit for use.” 2. Explain “Prevention” & “Detection”. ..Prevention means to prevent quality d…

title: Software Engineering-HW2 date: 2017-09-21 10:35:47 tags: HW --- 题目描述 从<构建之法>第一章的 "程序" 例子出发,完成一个能自动生成小学四则运算题目的命令行 "软件",满足以下需求: 1.参与运算的操作数(operands)除了100以内的整数以外,还要支持真分数的四则运算,例如:1/6 + 1/8 = 7/24.操作数必须随机生成. 2.运算符(operators)为 +…

title: Software Engineering-HW3 date: 2017-10-05 10:04:08 tags: HW --- 小组成员 264 李世钰 249 王成科 项目地址 https://github.com/lishiyuwhu/Homework/tree/master/SE/HW3 运行 dist目录下的DONE.exe 运行方法(环境PyQt5+py3.6.2) >python display_QWidget.py 本来想用pyinstaller打包成EXE, 但好像…

Preparation for MCM/ICM Writing -- by Chance Zhang $1^{st}ed$ key words: MCM/ICM, format, phrases, tools 目录 Preparation for MCM/ICM Writing Format The Basic Format Style and Layout Phrase bank Tools Codes Latex Table & Figure Finding Data Collaborati…

http://www.geekwire.com/2012/chart-pays-seattle-software-engineers/ Chart: Who pays the most in Seattle for software engineers BY JOHN COOK on February 2, 2012 at 3:14 pm     15 Comments Share  35 Tweet Share  19 Reddit Email FAQ: What to know for to…

4320: ShangHai2006 Homework Time Limit: 10 Sec Memory Limit: 128 MB Description 1:在人物集合 S 中加入一个新的程序员,其代号为 X,保证 X 在当前集合中不存在. 2:在当前的人物集合中询问程序员的mod Y 最小的值. (为什么统计这个?因为拯救 过世界的人太多了,只能取模) Input 第一行为用空格隔开的一个个正整数 N. 接下来有 N 行,若该行第一个字符为"A" ,则表示操作 1:若为&quo…

CentOS / Redhat : Configure CentOS as a Software Router with two interfaces   Linux can be easily configured to share an internet connection using iptables. All you need to have is, two network interface cards as follows: a) Your internal (LAN) netwo…

题目连接:http://acm.fzu.edu.cn/problem.php?pid=2137 题解: 枚举x位置,向左右延伸计算答案 如何计算答案:对字符串建立SA,那么对于想双延伸的长度L,假如有lcp(i-L,i+1)>=L那么就可以更新答案 复杂度  建立SA,LCP等nlogn,枚举X及向两边延伸26*n #include<iostream> #include<algorithm> #include<cstdio> #include<cstring…

题目链接:http://acm.fzu.edu.cn/problem.php?pid=1914 题意: 给出一个数列,如果它的前i(1<=i<=n)项和都是正的,那么这个数列是正的,问这个数列的这n种变换里, A(0): a1,a2,-,an-1,an A(1): a2,a3,-,an,a1 - A(n-2): an-1,an,-,an-3,an-2 A(n-1): an,a1,-,an-2,an-1 问有多少个变换里,所以前缀和都是正整数. 思路:因为变换是a[n]后面接着a[1]所以我们把…

Windows应用软件 Windows Applications (TBU) 全本推荐微软Windows 10操作系统 Quanben recommends Microsoft Windows 10 开发工具 Development Tools 软件名称Name of Application 版本认证Version 说明Description 适用平台System 评级Rating 官网或安全资源链接Offcial/Download Link Git for Windows 当前版本Curren…

 FZU 2105  Digits Count Time Limit:10000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Practice Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are intege…

原文作者:kkun 原文地址:http://www.cnblogs.com/kkun/archive/2011/07/06/agile_software_development.html 敏捷是什么 敏捷开发是一种软件开发方法,基于迭代和增量开发,通过自组织,跨团队,沟通协作完成开发工作. 敏捷宣言的诞生 2001年2月11日到13日,17位软件开发领域的领军人物聚集在美国犹他州的滑雪胜地雪鸟(SnowBird)雪场.经过两天的讨论,“敏捷”(Agile)这个词为全体聚会者所接受,用以概括一套全…

First we should know some basic conceptions about network: 1.Every PC is supposed to have its own IP,So we can connent other's PC by WAN.That's just like a ID of netwrok world.   2.But if  every PC has its own IP,Then IPs of this world will be insuff…

Atitit GRASP(General Responsibility Assignment Software Patterns),中文名称为"通用职责分配软件模式" 1. GRASP (职责分配原则)1 1.1. GRASP设计模式 1 1.2. 9个基本的OO设计原则或基本的设计构件.这9个设计模式分别是:创建者(Creator).信息专家(Information Expert).低耦合(Low Coupling).控制器(Controller).高内聚(High Cohesion…

用Java客户端程序访问Java Web服务器时出错: java.net.SocketException: Software caused connection abort: socket write error at java.net.SocketOutputStream.socketWrite0(Native Method) at java.net.SocketOutputStream.socketWrite(Unknown Source) at java.net.SocketOutputS…

最近在整一台华硕笔记本,大概有5年寿命了吧,质量还行,由于系统出了问题,打算用自制U盘WinPE进去修复一下.按照个人经验,在主板设置里启用了USB启动选项,并且设置USB启动顺序为第一个,可是进系统居然报错,如下图所示. Windows failed to start. A Recent hardware or software change might be the cause. To fix the problem: 1. Insert your windows installation…

软件开发流程(Software development process)即软件设计思路和方法的一般过程,包括设计软件的功能和实现的算法和方法.软件的总体结构设计和模块设计.编程和调试.程序联调和测试以及编写.提交程序.1需求调研分析 ●相关系统分析员和用户初步了解需求,然后用WORD列出要开发的系统的大功能模块,每个大功能模块有哪些小功能模块,对于有些需求比较明确相关的界面时,在这一步里面可以初步定义好少量的界面. ●系统分析员深入了解和分析需求,根据自己的经验和需求用WORD或相关的工具再做出…

原题:http://acm.fzu.edu.cn/problem.php?pid=2112 首先是,票上没有提到的点是不需要去的. 然后我们先考虑这个图有几个连通分量,我们可以用一个并查集来维护,假设有n个连通分量,我们就需要n-1条边把他们连起来. 最后对于每个联通分量来说,我们要使它能一次走完,就是要求他是否满足欧拉通路,也就是这个联通分量中至多有2个度为奇数的点,每多出2个度为奇数的点,就多需要一条边(因为单个连通分量的所有点的度数之和为偶数,所以不可能存在奇数个奇数度数的点). #inc…

FZU 2107 Hua Rong Dao Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice  Description Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisl…

 FZU 2112 Tickets Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Practice Description You have won a collection of tickets on luxury cruisers. Each ticket can be used only once, but can be used in either direction betwee…

 FZU 2102   Solve equation Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description You are given two positive integers A and B in Base C. For the equation: A=k*B+d We know there always existing many non-nega…

 FZU 2110  Star Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u  Practice Description Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sk…