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rabit
- android 平台的兔子跳铃游戏.游戏中兔子随着鼠标的位置移动，当触碰到铃铛是就会跳跃-the rabbit jump bell android platform game in which the position of the rabbit with the movement of the mouse, when it will jump to touch the bell is
HD6102
- Example of use an HD61202 graphic lcd writen in assembler for Rabbit core.
pg12864lrs-ann-h
- Example using PG12864 graphic lcd for rabbit core module.
acm4
- 在一座山上有n个山洞（每个洞从0~n-1分别编号），这山上有一只兔子躲在山洞里，有一只狼，从编号为0的山洞开始入洞，每隔m个洞，再次入洞抓兔子，现在问，兔子能否避免这场灾难？-The queue data with n cave in a mountain (each hole from the 0~n-1 number), this mountain has a rabbit hid in the cave, there is a wolf, start putting 0 cave from
Curso
- Examples for rabbit 4720 and dynamic C
vbjttlsrc
- 由鸡兔同笼问题编写的vb程序，貌似游戏但又不是游戏，又像是一个急转弯，但可确定是一个数字问题，输入鸡兔总只数和鸡兔总脚数，可进行计算，有兴趣的下载参考。-From chicken rabbit cage questions prepared vb program, but looks like the game is not a game, but also like a sharp turn, but can determine a number of issues, just enter t
pig
- 运行程序，打开魔法兔子主窗体。在主窗体中左侧是功能菜单，右侧是功能菜单的相应内容，上部是广告区域，下部是状态信息-Run the program, open the main form of magic rabbit. In the main form on the left is the function menu, the right is the corresponding content of the function menu, the upper is advertising are
HD6102
- Example of use an HD61202 graphic lcd writen in assembler for Rabbit core.
rabbitCantBeCatched
- 抓不住的兔子小游戏，实现对小兔子的简单操作-Rabbit catch small game, simple operation bunny
2014053082053
- 程序1】 题目：古典问题：有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月 后每个月又生一对兔子，假如兔子都不死，问每个月的兔子总数为多少？ 1.程序分析： 兔子的规律为数列1,1,2,3,5,8,13,21.... 2.程序源代码： main() { long f1,f2 int i f1=f2=1
Game
- 一个兔子过河的游戏，悬崖2岸兔子要过去，你移动横条帮助兔子跳到对面。-A rabbit across the river of the game, go over the cliff two shore rabbit, rabbit jump to help you move the bar opposite.
FINAL
- 这是一个用c语言编写的小游戏——兔子过河，有着较为华丽的图形界面，和完善的功能，包含暂停、继续、设置难度等级、查看排行榜、查看游戏说明等功能。-This is a language with a small c game- rabbit across the river, with more gorgeous graphical interface, and improved features, including pause, resume, set the difficulty level,
sim_langtu
- 首先计算狼的初始位置到兔子洞穴的直线距离： 由于狼奔跑的速度是兔子速度的两倍，兔子跑60码的时间狼可以跑120码。如果狼沿直线奔向兔窝，应该是可以追上兔子的。但是，有人推导出狼在追赶兔子过程中的运动曲线为 根据曲线方程，当 时， 。也就是说，在没有兔窝的情况下兔子一直往北跑，在跑到大约66码处将被狼追上。由此可知，在有兔窝时狼是追赶不上兔子的。 用计算机模拟的方法也可以得到同样的结论。取时间步长为1m，随时间步长的增加，考虑这一系统中的各个元素（狼和兔子）所处的位置变
GraspConey
- 抓不住的兔子，鼠标化身一只金色的小手，用户可以通过移动鼠标来抓取兔子。-Catch the rabbit, mouse incarnation of a golden little hand, the user can grab the rabbit by moving the mouse....
rabbit
- 纯AS3的项目，一个高完成度类似与跑酷的DEMO。可以用FLASH直接发布！-Pure AS3 project, similar and with a high degree of completion of parkour DEMO. Can be issued directly by the FLASH!
FINAL
- 这是一个用c语言编写的小游戏——兔子过河，有着较为华丽的图形界面，和完善的功能，包含暂停、继续、设置难度等级、查看排行榜、查看游戏说明等功能。-This is a language with a small c game- rabbit across the river, with more gorgeous graphical interface, and improved features, including pause, resume, set the difficulty level,
Gallery
- 类似塔防游戏的兔子别跑android手机游戏-Rabbit tower defense games like Hold android phone games
zhandanren
- j2me写的炸弹人源码，很小以前在FC机子上玩的炸弹人，就是那个炸兔子的。-The bombs were to write j2me source code, a small machine on the previous play FC bomb people is that deep-fried rabbit.
source
- 兔子产子的一种递归算法实现，供各位参考。-A recursive algorithm implementations rabbit birth, for your reference.
4
- opengl绘制桌子，兔子，茶壶，通过键盘控制各个物体的显示位置，用键盘切换显示列表和非列表显示方式-opengl drawing table, rabbit, teapot, through the keyboard to control the display position of each object, lists, and non-list display mode using the keyboard to switch the display