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agreement h261 coding yards inside lib_261.h provided a detailed interface and a...
h261协议的h261编码码,lib_261.h里面提供了详细的接口和说明.-agreement h261 coding yards inside lib_261.h provided a detailed interface and annotations.
- 2022-07-01 22:46:26下载
- 积分:1
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经典算法,快来下,
经典算法,快来下,-classical algorithm, breaking under ,
- 2022-02-01 09:16:28下载
- 积分:1
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关于一些算法的例子,我看着还不错,有不足的地方希望大家指导...
关于一些算法的例子,我看着还不错,有不足的地方希望大家指导-Algorithm on some examples, I looked pretty good, has the shortage of places I hope to guide U.S.
- 2022-11-02 15:55:03下载
- 积分:1
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Pseudo
伪随机数的产生程序 有m序列还有同余序列等等 用与嵌入信息位置的选择或密钥的制作-Pseudo-random number generation process has m sequences are congruent with and embedded in sequence and so the choice of location information or the production of key
- 2023-07-31 12:45:03下载
- 积分:1
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pvml源代码,提供下载,共同学习,共同进步
pvml源代码,提供下载,共同学习,共同进步-pvml source code, available for download, a common learning and common progress
- 2022-08-02 12:35:37下载
- 积分:1
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based on the SDK image acquisition procedures
基于SDK的图像采集程序-based on the SDK image acquisition procedures
- 2022-01-26 05:04:55下载
- 积分:1
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SPIHT的一个Matlab程序,经过改进的。2006年编写的。
SPIHT的一个Matlab程序,经过改进的。2006年编写的。-SPIHT, a Matlab procedure improved. Prepared in 2006.
- 2022-02-05 05:37:29下载
- 积分:1
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奎因
Quine-McCluskey算法
- 2022-08-07 15:00:09下载
- 积分:1
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Base64编码解码。
Base64 encode decode.
- 2022-02-20 07:51:12下载
- 积分:1
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0_1背包问题使用一个给定n种去的动态规划解法…
0_1背包问题用动态规划解给定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。-0_1 knapsack problem using dynamic programming solution of a given n types of 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.
- 2023-04-12 03:25:02下载
- 积分:1