气相色谱greatcode 1.140完成C / C++代码的源代码的缩进杂波，诉..

简介： Visual Basic代码库 是一个集收集，管理Visual Basic代码于一体的软件。它本身带有100多个源码，同时用户可以自己添加，删除自己的源码，不用打开Visval Basic，你就可以查找和浏览全部源代码。同时，它支持在Visual Basic中打开源代码。 Visual Basic代码库 一定会使您在编程学习之路上走得更好。-Synopsis: Visual the Basic code storehouse is one collection collection, manages Visual the Basic code to one body software. Its itself has more than 100 sources codes, simultaneously the user may own increase, deletes oneself the source code, with does not open Visval Basic, you may search and glance over the complete source code. At the same time, it supports in Visual Basic opens the source code. Visual the Basic code storehouse certainly can enable you to walk even better road in on the programming study.

笔迹识别笔迹识别-handwriting recognition handwriting recognition

一个组态王机械手控制的实例教程。版本为6.53-A robot controlled KingView tutorial examples. Version 6.53

Space time block codes

数码管控件.rar，有事例。-digital pipe controls. Rar, stories.

排序算法，使用方便-algorithm, easy to use

基于javaweb基础知识实现的学生信息管理系统,包含对学生信息的增删查改,以及模糊查询,用之前需连接数据库,在配置文件c3p0-config.xml里设置要连接的数据库

本系统实现了VC++环境中对WORD的写入和读取操作，希望对大家有所帮助-The system achieved a VC++ environment on WORD write and read operations, want to help everyone

还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不断地循环 e 次之后，很多情况下都可以循环出原文，只是计算量过余多一些罢了。不过有不少情况下，根本都无须循环 e 次，不过对于1024位的 n 级别来说，e 也是一个相当大的数值，所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考，如有不实，请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected

火车车厢重排问题:给定一个任意的车厢排列次序。重新排列车厢，使其按照从1到n的次序排列。规定重排时只能从入轨到缓冲铁轨，或者从缓冲铁轨到出轨。-Train cars rearrangement problem: given an arbitrary compartment order of merit. Rearrange the cars, so that from 1 to n according to the order. Rearrangement can only be provided to the buffer from the orbit track, or tracks from the buffer to the derailment.