搜索资源列表
TASTE_GDI_1
- 初学 GDI 绘图，初学者看看，大牛就别看了。-GDI graphics novice, the cow on the Do not.
tuling
- 《对图灵机与计算问题》22页的文章，一牛人关于计算本质的思考和讨论。-" Turing machine with the calculation of the problem" of article 22, one cow people consider about the nature of the calculation and discussion.
8549056tX7h
- J2ME手机游戏黑暗部落源代码 J2ME手机游戏黑暗部落源代码，运行于诺基亚手机，源码可以编译。-Plastic Cow Games has the rights to all graphics, and has licensed the rights to use the sound and music. Feel free to use any graphics in your own game or application excluding the pcg logo. If
howhtml
- 一个牛人用汇编写的浏览器源码 A cow with a compilation of source code written in the browser-A cow with a compilation of source code written in the browser
milk2
- 解决下面英文描述提到的问题.问题来源于USACO的培训题.-Three farmers rise at 5 am each morning and head for the barn to milk three cows. The first farmer begins milking his cow at time 300 (measured in seconds after 5 am) and ends at time 1000. The second farmer begins at ti
1
- 交替字符倒三角形.txt级数求和.txt密钥加密.txt密钥解密.txt5位以内的对称素数.txt母牛问题.txt-Inverted triangle alternate characters. Txt Sum. Txt key encryption. Txt decryption key. Txt5 symmetric spaces within the prime number. Txt cow problem. Txt
TheInternetWormProgram_AnAnalysis
- The Internet Worm Program - An alysis 普渡大学一牛人写的-The Internet Worm Program- An Analysis of Purdue University who wrote a cow
java
- 一个牛人给java初学者的建议,很深刻，建议仔细阅读-A cow were the recommendations to java beginners, it is profound, it is recommended to read
3278
- poj 3278 Catch That Cow 采用广度优先搜索实现-poj 3278 Catch That Cow using breadth-first search to achieve
XIAONIU
- 牛生牛的问题 说是：有一个农场有一头成年母牛，每三个月后一头小牛， 小牛一年后长大，长大后每三个月又可以生一头小牛， 如此循环，问10年后农场一共有多少牛?-The issue of beef cattle Health That is: there is a farm there is an adult cow, a calf every three months, One year after the calf grew u
queen
- 超牛的8皇后程序，整个程序只有短短的10行，而且是按标准格式，控制台上输出的格式也是符合美观条件的-8, Queen' s super-cow program, the whole process is only a short span of 10 lines, and is based on a standard format, the format of the console output is also in line with the conditions of appeara
eMule
- 超级牛叉的商业开源软件源代码，电驴的源代码-Super cow fork commercial open source software source code, source code eDonkey
Symbiannotice
- Symbian学习笔记，一牛人写的，很经典，从基础到深入，很适合Symbian学习者参考-Symbian study notes, written by a cow, it is classic, from basic to in depth, it is suitable for learners Symbian Reference
PhDthesis_ZhongGang
- 曲线曲面重建方法研究 ，浙大某牛博士论文 -Reconstruction of curves and surfaces, Zhejiang University doctoral dissertation of a cow
MemUtil
- 一个牛公司自己开发一个内存管理工具,能够很好的调试内存泄漏,及示例代码.-A cow' s own memory management to develop a tool for debugging memory leaks very good, and sample code.
Niuos-0.1.05.03-repot-0.08
- 国内的一位牛人开发的操作系统，感觉还不错，功能还挺全，所以上传上来分享给大伙-A domestic cow operating systems were developed, I feel good, function quite wide, so upload up to everyone to share
cmdshellv2
- 都是些老掉牙的技术了，AVP狂报，NOD32没反应。 呵呵~~ 适合新手看看，大牛就略了。 主要有 线程注入 文件释放 添加自启 一些常规手段-Are some of the old technology, AVP mad at, NOD32 did not react. Ha ha ~ ~ suitable for novice to see the big cow on the slightly. Thread into the main file to add the re
poj
- pku 3613 Cow Relays 题意：给一个无向图，求从起点s到终点e尽力n条边的路径最小值。 边的数量2<=t<=100(每条边可以重复遍历) 算法：利用dp求出 path[ l ] [ i ] [ k ]=MIN(path[ l ][ i ] [ k ] , path[ l ][ i ] [ j ] + path[ l ][ j ] [ k ]) l 表示的是2的指数，假设 i 到 k 要途径 2^l 条边，那么他可以被分为两段 途径2
Factory
- 设计模式知识,工厂方法,VS2005 .NET2.0 c# 依照cnblog网站上一巨牛人物写的,-Knowledge of design patterns, factory method, VS2005. NET2.0 c# Web site in accordance with a giant cow cnblog characters written
c1
- 对于元音字母（a e i o u）开头的英文单词结尾加上cow；相反，如果是辅音开头的英文单词，开头第一个字母移到末尾加ow-Bovine Latin The cows have heard that the pigs use a secret language called "Pig Latin" when they want to communicate with each other without Farmer John being able to understand what t