资源列表
200553225018296
- MPEG-2视频编解码源代码(VC版)，以实践过-MPEG-2 video codec source code (VC version), off to practice
bp_feel_classification
- 此为BP感知神经分类器,自己写的.有须要的请下载,很不错的.-this perception of BP neural classifier himself wrote it. The need to please download, very good.
xvidcore-1.1.2
- mpeg-4的编解码源码，是xvid格式的，在官方网站下的-mpeg-4 codecs to the source, xvid format, in the official website
BPSK_radar
- 此为雷达辐射源信号BPSK的仿真信号,自己编写的,已调试通过,贡献出来.可放心使用.-sources such as radar signal simulation BPSK signals, their preparation, testing has passed, contribute. use can be assured.
hPSOption
- 一个国外学者编辑的粒子群算法，对于初学者和其他人员均有很大的帮助作用-a foreign scholar edit the PSO for beginners and other staff have been very helpful
Huffman_C
- 用c实现Huffman编码,基于概率统计,压缩率较高-c achieve with Huffman coding, based on statistical probability, the higher compression rate
C_AI
- C语言写的神经网络源程序 是学习C语言和人工智能很好得例子-write C language source of neural network learning C language and a good example of artificial intelligence
humantree
- 这就是所谓的哈夫曼编码的代码,我自己编的,编的不好还多请教,这是个学生级别的代码!嘿嘿!-This is the so-called Huffman coding code, I developed the addendum to the more bad advice, This is a student-level code! laughter!
Projectjohu123006
- crc任意位生成多项式 任意位运算 自适应算法 循环冗余校验码(CRC，Cyclic Redundancy Code)是采用多项式的 编码方式，这种方法把要发送的数据看成是一个多项式的系数 ，数据为bn-1bn-2…b1b0 (其中为0或1)，则其对应的多项式为： bn-1Xn-1+bn-2Xn-2+…+b１X+b０ 例如：数据“10010101”可以写为多项式 X７+X４+X２+1。 循环冗余校验CRC 循环冗余校验方法的原理如下：
huiwen333
- 这是一个关于回文数的判断算法!只要输入的数据形式是回文形式,将输出该回文数是回文模式!-This is a palindrome number on the judgment algorithm! As long as the form of data input is palindrome form Palindrome will output the model number is palindrome!
n1qn3-lib
- 最优化算法，应用有限内存拟牛顿方法（Limited Memory (variable-storage)quasi-Newton method）求解高维最优化问题，使用更多的内存将使算法更有效。-optimization algorithm, Application limited memory quasi-Newton method (Limited Memory (variable-stora ge) quasi--Newton method) for high-dimensional opt
CrossPointNo
- CrossPointNo=53 ％％％输入图中节点的总数目 对已知的边进行赋值，注意：有向图的Cost(i,j)=Cost(j,i)-CrossPointNo = 53%%% imported map nodes to the total number of known side Fu value, attention : to map Cost (i, j) = Cost (j, i)