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Dan Saks关于const的所有用法
Dan Saks关于const的所有用法-Dan Saks talks about the usage of const
- 2022-03-11 23:02:03下载
- 积分:1
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学习MFC的好例子3
学习MFC的好例子3-learning a good example of 3
- 2022-06-21 08:08:08下载
- 积分:1
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Service Pack 一手搞定.mht Windows 瘦身.mht 创建可引导光盘更完美.mht 装Win Me
Service Pack 一手搞定.mht Windows 瘦身.mht 创建可引导光盘更完美.mht 装Win Me-2000免输序列号.htm-Service Pack get first-hand. Mht Windows thin. Mht creation may be cited I. perfect CD. mht installed Win Me-2000-lost serial number. htm
- 2022-11-02 12:45:03下载
- 积分:1
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书上载的源码summax,求和的最大值,用c++实现,要的可以看啊看...
书上载的源码summax,求和的最大值,用c++实现,要的可以看啊看-The source book contains summax, the maximum sum, and c++ Realize, it is necessary to look ah can look
- 2022-01-26 01:50:15下载
- 积分:1
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ad7705的中文资料很好用的
ad7705的中文资料很好用的 -ad7705的中文资料很好用的 ad7705的中文资料很好用的
- 2022-02-02 04:36:45下载
- 积分:1
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basics of operating system
basics of operating system
- 2022-11-22 17:30:04下载
- 积分:1
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Lunar transfer function (C language version), can be converted to the Lunar cale...
农历转换函数(C语言版),可将公历转换为农历-Lunar transfer function (C language version), can be converted to the Lunar calendar
- 2022-07-19 19:40:43下载
- 积分:1
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《Junit In Action》是一本很好的书,它讲了如何在J2EE架构中构建单元测试,以及如何设计...
《Junit In Action》是一本很好的书,它讲了如何在J2EE架构中构建单元测试,以及如何设计-"Junit In Action" is a very good book, which stresses how the J2EE framework Construction unit testing, and how to design
- 2022-02-12 20:05:02下载
- 积分:1
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java数据库编程源码,包含了所有文章的示例
java数据库编程源码,包含了所有文章的示例-source, the article contains examples of a
- 2022-02-02 05:13:57下载
- 积分:1
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背包问题动态规划算法JAVA
给定n种物品和一背包。物品i的重量是wi,其价值为vi,背包的容量为c。问应如何选择装入背包中的物品,使得装入背包中物品的总价...
背包问题动态规划算法JAVA
给定n种物品和一背包。物品i的重量是wi,其价值为vi,背包的容量为c。问应如何选择装入背包中的物品,使得装入背包中物品的总价值最大?
在选择装入背包的物品时,对每种物品i只有两种选择,即装入背包或不装入背包。不能将物品i装入背包多次,也不能只装入部分的物品i。因此,该问题称为0-1背包问题。-Knapsack problem dynamic programming algorithm for a given n species JAVA goods and one backpack. Item i is the weight of wi, the value for vi, knapsack capacity c. Question should be how to choose the items into the backpack, the backpack into the total value of the largest items? In the selection of the items into the backpack, for each of the items i have only two choices, that is loaded or not loaded backpack backpack. Items should not be many times i load backpack can not only part of the goods loaded i. Therefore, the problem known as the 0-1 knapsack problem.
- 2023-04-22 23:20:03下载
- 积分:1