-
一个类CARD牌的小程序,可以作到抛砖引玉吧
一个类CARD牌的小程序,可以作到抛砖引玉吧-CARD licensing a class of small procedures, you can definitely do it
- 2022-09-20 13:00:03下载
- 积分:1
-
windows核心编程源程序
windows核心编程源程序-windows core programming source
- 2023-08-26 21:25:03下载
- 积分:1
-
拦截iofcalldriver,发送按键的用户
Intercept IofCallDriver, send pressed key to user-mode parser. C-Intercept IofCallDriver, send pressed key to user-mode parser. C++
- 2023-01-03 08:45:03下载
- 积分:1
-
基于C++实现的斗地主单机游戏
基于C++实现的斗地主单机游戏,开发工具VC++2005,没有使用MFC,使用的是WinMain实现,支持简单的人工智能。
- 2022-06-21 08:13:15下载
- 积分:1
-
计算器的c++编程程序,源代码,经过老师的验证
计算器的c++编程程序,源代码,经过老师的验证-Calculator c++ programming procedures, source code, through the teacher certification
- 2022-03-06 16:53:45下载
- 积分:1
-
Spring+MyBatis基础框架
省去每次都要搭载框架的烦烦恼
- 2023-08-10 00:50:02下载
- 积分:1
-
不我不描述
no i have no describe
- 2022-07-15 12:06:47下载
- 积分:1
-
风险投资编写的日历和即时表
vc编写日历和即时钟表-vc prepared calendar and instant watches
- 2022-01-28 13:50:22下载
- 积分:1
-
0-1背包问题
给定N中物品和一个背包。物品i的重量是Wi,其价值位Vi ,背包的容量为C。问应该如何选择装入背包的物品,使得转入背包的物品的总价值为最大。
在选择物品的时候,对每种物品i只有两种选择,即装入背包或不装入背包。不能讲物品i装入多次,也不能只装入物品的一部分。因此,该问题被称为0-1背包问题。该算法中,矩阵c的大小为(m+1)×(n+1),物体的重量、价值和解向量大小都等于物体个数n,故该算法的空间复杂度为O(nm)。对物体重量、价值的初始化(算法实现略)所需时间都为n,解向量和矩阵第0行初始化时间为n,矩阵第0列初始化时间为m,对矩阵c的计算所需时间为n×m,解向量X的确定时间为n,故整个算法的时间复杂度为O(nm)。
- 2022-01-25 23:07:45下载
- 积分:1
-
During the code development phase I detected some variables and actions that are...
During the code development phase I detected some variables and actions that are general for all the resizable controls and for this reason I decided to move them in a general interface, IResizeControl, from which all the resizable controls are to inherit. Technically, IResizeControl is not an interface because it doesn t have any pure virtual functions, but I still consider it an interface because it doesn t make any sense to declare IResizeControl objects. In fact, I am preventing the creation of IResizeControl objects by declaring the IResizeControl s constructor and destructor as protected. The IResizeControl class
- 2022-07-01 21:38:20下载
- 积分:1