data-structure

在远距离数据通信中，为确保高效而无差错地传送数据，必须对数据进行校验即差错控制。循环冗余校验CRC(Cyclic Redundancy Check/Code)是对一个传送数据块进行校验，是一种高效的差错控制方法。-in long-distance data communication, in order to ensure efficient without error to transmit data, the data must check that the error control. Cyclic Redundancy Check (Cyclic Redundancy Check/Code) is to block transmission of data over a check out, is a highly efficient error control methods.

查看transfert表或EXCEL工作表以访问和检索

环形链表结点出排-circular linked list node wait-

vfp编写的简单计数器.可以轻松实现计数,清零等功能 -Probe preparation of simple counters. Easily count, reset function

this water jug problem in c-this is water jug problem in c++

c 的一些经典算法,满好的,适合初学者. 也可以当作小程序看,对初学者会很有帮助-c some of the classic algorithms, over a good, suitable for beginners. can also be read as a small program, it would be helpful for beginners

迷宫算法（maze） /* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination. */ -/* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination.

检验SQL语法正确性-test SQL syntax correctness

在一台机器上只能处理n个作业，每个作业均可在单位时间内完成，又假定每个作业都有一个截止期限，当且仅当作业i在它的截止期限之前完成时，可以获得一定的效益。 给定作业序列，要求一种作业排序算法，使获得的总效益值最大。 -On a single machine can handle only n-jobs, each job can be completed within the unit, but also assumes that each job has a deadline, if and only if job i in its completed before the deadline, they can to receive some benefits. For a given job sequence, requires a job sorting algorithm, so that the greatest value of total benefits received.

单循环存储结构实现约瑟夫环,n个人从1到n编码,然后围坐一圈,每人持有一密码-one-cycle storage structure to achieve Josephus, n individuals from 1 to n coding, and then sitting around a circle, each holding a password