This a small hack I wrote to bypass the "No previous installation!" error of Microchip s C18 Upgrade. Simply run C18Fake and select your C18 installation directory (create if necessary) and click Save button. A "fake" file will now be created into that directory, and this will fool the Upgrade into thinking you already have a former legit installation. After that you run the C18 Upgrade setup and select the same directory you created/selected before. Your installation will now begin installing the full C18 package.
标签： installation Microchip previous Upgrade
上传时间： 2014-01-21
上传用户：zhuyibin
You know the rudiments of the SQL query language, yet you feel you aren t taking full advantage of SQL s expressive power. You d like to learn how to do more work with SQL inside the database before pushing data across the network to your applications. You d like to take your SQL skills to the next level.
标签： advantage the rudiments you
上传时间： 2014-01-18
上传用户：xz85592677
You imagine? Right, there s more than one possibility, this time I ll give you tree. One for your private data, one for the common data in order to receive data from other applications like Excel, WinWord etc. and at last, I ll give you a handy-dandy class you can derive ANY MFC object from, to make it a drop target
标签： possibility imagine Right there
上传时间： 2013-12-21
上传用户：jichenxi0730
01背包问题题目 有N件物品和一个容量为V的背包。第i件物品的费用是c[i],价值是w[i]。求解将哪些物品装入背包...但它却是另一个重要的背包问题P02最简捷的解决方案,故学习只用一维数组解01背包问题是十分必要的。
上传时间： 2014-08-21
上传用户：金宜
卷积运算：取x(t)和h(t)的长度为nx,nh。平移量n=nh+nx-1，利用for和if语句实现倒序求和运算。外循环用一个for语句实现平移，通过在求和时取数组元素的顺序实现倒序求和
上传时间： 2013-12-20
上传用户：saharawalker
T-S模糊神经网络的matalb程序,可以用做端点检测
上传时间： 2016-02-14
上传用户：123啊
This program don t special because it just can run some number,and I hope I can be a menber on this website
标签： can program because special
上传时间： 2013-12-26
上传用户：netwolf
这是我心仪已久的一本书 顶级大师Stanley B Lippman J o s é e L a j o i e合著的
上传时间： 2014-01-01
上传用户：nanfeicui
输入一个数字N,找出i个连续自然数累加为N的所有和式(i>1)。
上传时间： 2016-02-29
上传用户：zhichenglu
数据结构 1、算法思路： 哈夫曼树算法：a）根据给定的n个权值{W1，W2… ，Wn }构成 n棵二叉树的集合F={T1，T2…，T n }，其中每棵二叉树T中只有一个带权为W i的根结点，其左右子树均空；b）在F中选取两棵根结点的权值最小的树作为左右子树构造一棵新的二叉树，且置新的二叉树的根结点的权值为其左、右子树上结点的权值之和；c）F中删除这两棵树，同时将新得到的二叉树加入F中； d）重复b)和c),直到F只含一棵树为止。
上传时间： 2016-03-05
上传用户：lacsx
