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aes算法c(c++)_支持128、192和256位密钥
aes算法c(c++)_支持128、192和256位密钥 实现一个C++的AES算法类，可直接调用
- 2022-02-02 20:47:48下载
- 积分：1
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一维时间序列分行维数，利用计盒法计算得到的分形维数，希望对大家有所帮助！...
一维时间序列分行维数，利用计盒法计算得到的分形维数，希望对大家有所帮助！-Branch of one-dimensional time series dimension, using box method calculated the fractal dimension, and they hope to be helpful to everyone!
- 2022-02-07 12:30:37下载
- 积分：1
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分数阶统一混沌系统的离散算法
分数阶统一混沌系统的离散算法-This the program for fractional order Unified system, a simple but fast
method in time domain.-Fractional order unified chaotic system discrete algorithm-This the program for fractional order Unified system, a simple but fastmethod in time domain.
- 2022-01-24 10:22:58下载
- 积分：1
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模拟分时系统
模拟分时系统-Simulation of Time-Sharing System
- 2022-03-14 18:50:12下载
- 积分：1
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指纹识别源码
基于Gabor的指纹识别系统，有源码，有Demo，指纹图像预处理，Gabor增强，二值化，细化，特征点提取，模板保存与匹配。
- 2022-06-15 23:09:49下载
- 积分：1
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a full 3D simulation of electromagnetic waves with efficient absorbing boundary...
a full 3D simulation of electromagnetic waves with efficient absorbing boundary a full 3D simulation of electromagnetic waves with efficient absorbing boundary conditions SYNOPSIS- a full 3D simulation of electromagnetic waves with efficient absorbing boundary a full 3D simulation of electromagnetic waves with efficient absorbing boundary conditions SYNOPSIS
- 2022-05-25 07:53:42下载
- 积分：1
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Надоело
- 2022-03-23 05:42:02下载
- 积分：1
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军团问题， 某国家在一条战线上（一条直线）上部署了n 支军队。如果两支或两支以上军队相邻，它们将形成一个军团。如果一支军队不和任何一支军队相邻，那么它也是一个独...
军团问题， 某国家在一条战线上（一条直线）上部署了n 支军队。如果两支或两支以上军队相邻，它们将形成一个军团。如果一支军队不和任何一支军队相邻，那么它也是一个独立的团。给定一组军队在战线上左右两端的位置及编号，求出所有的军团
-Corps issues, a country in a front (a straight line) on n-support forces deployed. If two or more than two neighboring armies, they will form a regiment. If an army is not adjacent to any one army, then it is also an independent regiment. Given a group of troops on the front left and right ends of the location and number, find all of the Army
- 2022-02-28 17:22:05下载
- 积分：1
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fortran对称正定矩阵求逆
可以用来计算对称正定矩阵的逆矩阵，比通常的求逆程序的计算效率快很多，经过测试很好用，计算非常准确，使用方便，与大家共享。
- 2022-05-19 01:47:52下载
- 积分：1
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文化基因算法伪代码
memetic是一种高效的文化进化计算方法，该范例以Rosenbroek函数为例进行了演示，你也可以加入其它函数进行测试。-memetic cultural evolution is an efficient method of calculating the sample to Rosenbroek function as an example of the presentation, you can also add other functions to be tested.非原创，仅限分享。
- 2022-10-13 04:35:03下载
- 积分：1
