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支持分布式网络化制造的智能数控系统的研究
支持分布式网络化制造的智能数控系统的研究-support distributed network of intelligent manufacturing system of NC
- 2022-02-12 01:22:59下载
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In the learning process of digital signal processing algorithms using VC General...
在学习数字信号处理算法程序中用VC编写的几个通用算法程序。
卷积计算/DFT与FFT实现/-In the learning process of digital signal processing algorithms using VC General Algorithm for the preparation of several procedures. Convolution calculation/DFT and FFT realization /
- 2022-07-13 20:31:16下载
- 积分:1
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用Borland C写的B
用Borland C写的B-Tree算法- B-Tree algorithm writes which with Borland C
- 2023-02-02 02:20:04下载
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径向基神经网络源代码
this is an example demonstrating the Radial Basis Function,if you select a RBF that supports it (Gausian, or 1st or 3rd order polyharmonic spline), this also calculates a line integral between twopoints.
- 2022-02-14 22:25:54下载
- 积分:1
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大数运算库
大数运算库-Treasury operations of large numbers
- 2022-02-12 09:54:57下载
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多边形剪裁的源码
多边形被一个多边形被剪裁的源码,剪裁是建模中比较重的操作之一,也是很常用的工具。多边形被一个多边形被剪裁的源码,剪裁是建模中比较重的操作之一,也是很常用的工具。
- 2023-03-09 01:35:04下载
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ACM1087 realize the title of the request, the document provided in the form of t...
实现了ACM1087题中的要求,以文件的方式提供测试数据。-ACM1087 realize the title of the request, the document provided in the form of test data.
- 2022-06-27 12:01:22下载
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mime64是什么:mime64是一rfc1341 MIME base64编码描述为其铺…
WHAT MIME64 IS: MIME64 is an encoding described in RFC1341 as MIME base64.Its purpose is to encode binary files into ASCII so that they may be passedthrough e-mail gates. In this regard, MIME64 is similar to UUENCODE.Although most binaries these days are transmitted using UUENCODE, Ihave seen a few using MIME64, and I have had requests from friends thatI decode MIME64 files that have fallen into their hands. As long assome MIME64 continues to exist, a package such as this one is usefulto have.
- 2022-03-22 13:30:49下载
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Mahout在大数据的应用
该资料包括大数据内应用的的所有算,支持单机、伪分布、全分布等环境运行,主要应用推荐产品、分类产品。
- 2022-04-22 03:47:25下载
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矩阵运算
// 矩阵乘法: C=AB, A[m×v], B[v×n], C[m×n] ... for row-major matrices
// 允许A,B,C 带入相同参数.
double* matr_x(double*C, const double*Ao, const double*Bo, const int m,const int v, const int n){
if(!Ao || !Bo) return 0; if(!C) C=new double[m*n];
double* A=new double[m*v], * B=new double[v*n];
memcpy(A,Ao,m*v*sizeof(double)); memcpy(B,Bo,v*n*sizeof(double));
请点击左侧文件开始预览 !预览只提供20%的代码片段,完整代码需下载后查看 加载中 侵权举报
- 2022-04-07 01:39:24下载
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