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KPCA故障检测
说明: 基于KPCA的故障诊断matlab代码。包括t^2统计量和q统计量(Fault Diagnosis Matlab Code Based on KPCA)
- 2021-04-12 11:18:57下载
- 积分:1
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c++ 图书管理系统源码(含数据库)
c++ 图书管理系统源码(含数据库)
- 2019-10-02下载
- 积分:1
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sqlite-source-3_6_6_2
说明: sqlite的最新源码
This ZIP archive contains preprocessed C code for the SQLite library as individual source files. It is strongly recommended that the amalgamation above be used instead of this package(sqlite latest source This ZIP archive contains preprocessed C code for the SQLite library as individual source files. It is strongly recommended that the amalgamation above be used instead of this package)
- 2008-11-29 16:13:56下载
- 积分:1
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PrintTheSchedule_DataStructure_Lab4
数据结构试验-打印输出计算机本科专业4年每学期的课表;C++实现;图的邻接表存储结构以及拓扑排序的基本思想(Data structure test- printout computer 4 years of undergraduate per semester timetables C++ achieve basic idea of the adjacency table storage structure and topological sorting)
- 2012-11-24 23:30:41下载
- 积分:1
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oracleOCI
ORACLE接口超过20个字干尼玛啊,傻逼论坛(oracle hand)
- 2013-11-05 10:40:16下载
- 积分:1
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Access数据库的存储和读取图片
Access数据库的存储和读取图片
Access数据库的存储和读取图片
Access数据库的存储和读取图片
Access数据库的存储和读取图片
- 2023-03-23 07:25:04下载
- 积分:1
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vulution-transformation
C++经典数值算法源码, 包括线性代数方程组的求解, 数学变换与滤波等(C classical numerical algorithm source code, including the solution of linear algebraic equations, mathematical transformation and filtering, etc.)
- 2018-11-08 17:00:38下载
- 积分:1
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outline
哈希表函数,纯C语言编写的哈希表计算函数(Hash table functions, written in pure C language function to calculate the hash table)
- 2013-12-02 14:42:14下载
- 积分:1
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SubSonic.Core orm框架示例源码(基于Mssql版本)
修正提交数据库有失败时也返回成功的BUG 修正脏数据提交bug 修正无法赋值整数0的BUG
- 2019-04-14下载
- 积分:1
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test
某公司收到若干报价,然后报价由低到高进行排序。设最低报价为F,最高报价为G。n=0.2*(G-F)
设A,B,C,D,E五个等距区间并取:
A=[F,F+n) B=[F+n,F+2n) C=[F+2n,F+3n) D=[F+3n,F+4n) E=[F+4n,G)
所有的报价都按各自的大小,分别列入上面五个区间。各自区间内的最小报价为该区间的代表报价。如果某区间内没有报价 则以小于该区间的且与该区间相邻的区间内的最高报价代表该区间报价,如果该区间与与之相邻的较小区间内都没有报价则该区间不参加最后计算。取各区间的代表值, 求出其平均值 做为评标基准报价
(A company received a number of quotations, and quotations from low to high order. Minimum price for the F, the highest offer for the G. n = 0.2* (GF) Let A, B, C, D, E, and five equidistant intervals obtained: A = [F, F+ n) B = [F+ n, F+2 n) C = [F+ 2n, F+3 n) D = [F+3 n, F+4 n) E = [F+4 n, G) All quotes are by their size, five were included in the above range. The minimum offer their own range representative of the range quoted. If a range does not offer less than the range of places and range of adjacent intervals with the highest offer within the offer on behalf of the interval, if the interval and the adjacent smaller range do not offer the range not to participate the final calculation. Take the representative value of each interval, calculate the average price as a benchmark evaluation)
- 2011-11-10 22:44:43下载
- 积分:1