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ConsoleApp
aop基础源码,适合初学者进行学习。适合初学者进行学习。(aop foundation source)
- 2013-05-25 12:08:44下载
- 积分：1
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I2C_Labview
Labview 访问2线制数据i2C，值得研究(Labview access I2C)
- 2009-04-14 10:37:50下载
- 积分：1
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C Primer Plus 第6版 非扫描版 中文版
C Primer Plus 第6版 非扫描版 中文版(C Primer Plus The sixth Edtion)
- 2020-06-18 21:00:02下载
- 积分：1
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CtrlIPClt
网络方面关于ip的得到与获取方面的代码，哈哈哈 (ip on the network and get access to the area code, Ha Ha Ha)
- 2006-11-14 11:08:37下载
- 积分：1
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zoj1094
zoj094
Matrix multiplication problem is a typical example of dynamical programming.
Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.
For example, let A be a 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix.
There are two different strategies to compute A*B*C, namely (A*B)*C and A*(B*C).
The first one takes 15000 elementary multiplications, but the second one only 3500.
Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy.
(Matrix multiplication problem is a typical example of dynamical programming.
Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.
For example, let A be a 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix.
There are two different strategies to compute A*B*C, namely (A*B)*C and A*(B*C).
The first one takes 15000 elementary multiplications, but the second one only 3500.
Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy.
)
- 2012-06-10 10:38:53下载
- 积分：1
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shop_manager
设计一个程序，对商铺信息管理，商铺信息包括：商铺编号，商铺名，信誉度（0-5），（商品名称1，价格1，销量1），（商品名称2，价格2，销量2），（商品名称3，价格3，销量3）…。
商品名称包括(毛巾，牙刷，牙膏，肥皂，洗发水，沐浴露等6种以上商品)，每个商铺具有其中事先确定若干商品及价格，由文件输入，销量初始为0。(Designing a program to manage shop information, which includes: shop number, shop name, reputation (0-5), (commodity name 1, price 1, sales volume 1), (commodity name 2, price 2, sales volume 2), (commodity name 3, price 3, sales volume 3)....
Commodity names include (towels, toothbrushes, toothpaste, soap, shampoo, bath lotion, etc.) more than 6 kinds of goods, each shop has a number of goods and prices in advance, input by documents, initial sales of 0.)
- 2019-02-06 19:05:42下载
- 积分：1
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SServer
这个demo是一个twain协议实现的demo(This demo is a demo implemented by Twain protocol)
- 2020-06-19 22:00:01下载
- 积分：1
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不同进程间通讯
说明： 实现vc下不同进程之间的通讯：Copyrecv接收进程，Copysend发送进程(achieve vc different process of communication between : Copyrecv receiving process, this process Copysend)
- 2020-08-23 14:07:54下载
- 积分：1
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electromotor1.6
VC （MODBUS协议）上位机程序VC （MODBUS协议）上位机程序(VC （MODBUS）)
- 2013-10-10 10:55:46下载
- 积分：1
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LOL防封
说明： LOL防封，可以参考研究，不能直接用要更新(Lol is sealed, research-ready, not ready to be updated)
- 2020-03-02 12:04:05下载
- 积分：1
