登录
首页 » Visual C++ » chess

chess

于 2009-06-01 发布 文件大小:170KB
0 83
下载积分: 1 下载次数: 1

代码说明:

  在N*N的棋盘上(1

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • Josephus(data_struct)
    数据结构中基于约瑟夫环问题,用C++方式实现的。。。。。(Data structure based on Josephus problem with C++ a manner. . . . .)
    2013-11-17 11:30:24下载
    积分:1
  • LCS
    the longest common serials
    2011-12-13 00:32:40下载
    积分:1
  • KMP
    说明:  KMP算法的具体实现,能够在O(n+m)时间内找出字串在父串中的位置,n为父串长度,m为字串长度(KMP in the proposed algorithm, can be O (n+ m) time to find the string in the parent string position, n the parent string length, m for the string length)
    2010-04-28 16:18:00下载
    积分:1
  • migong
    迷宫问题,利用随机数自动生成迷宫,并自动进行探索。迷宫起点及终点由用户自己设定。(Maze problem, the use of random number automatically generated maze, and automatically explore. The starting point and end point of the maze is set by the user. 双语对照 )
    2021-04-26 16:48:45下载
    积分:1
  • FinalExam
    C语言期末考试题,含答案。数组、链表、函数等的使用和C编程。(failed to translate)
    2011-11-07 20:07:25下载
    积分:1
  • jose
    约瑟夫环(约瑟夫问题)是一个数学的应用问题:已知n个人(以编号1,2,3...n分别表示)围坐在一张圆桌周围。从编号为k的人开始报数,数到m的那个人出列;他的下一个人又从1开始报数,数到m的那个人又出列;依此规律重复下去,直到圆桌周围的人全部出列。通常解决这类问题时我们把编号从0~n-1,最后结果+1即为原问题的解。 (Josephus (Josephus problem) is the application of a mathematical problem: Given n individuals (with numbers 1,2,3 ... n respectively) sitting around a round table. From the number of people gettin k, number of the m man out of the line he s the next person and a number of gettin number to m the man was out of the line and so the law is repeated until the round table were all out of the column. We numbered 0 ~ n-1, the final result is the original problem solution+1 usually solve these problems.)
    2015-01-09 17:28:41下载
    积分:1
  • test
    使用vc编写的一个动态链接库,测试动态链接库的例程(Vc prepared using a dynamic link library, dynamic link library routine testing)
    2010-05-28 23:27:09下载
    积分:1
  • maze
    迷宫求解一般采用“穷举法”,逐一沿顺时针方向查找相邻块(一共四块-东(右)、南(下),西(左)、北(上))是否可通,即该相邻块既是通道块,且不在当前路径上。用一个栈来记录已走过的路径栈是限定仅在表尾(top)进行插入或删除操作的线性表。(Maze solving generally use the " exhaustive" one by one to find adjacent blocks in a clockwise direction (a total of four- East (right), South (down), West (left), North (on)) whether to pass, that the both channel blocks adjacent blocks, and not in the current path. Use a stack to record the path traversed stack is limited only in the tail (top) to insert or delete a linear form.)
    2013-12-13 17:55:45下载
    积分:1
  • yeschengxu
    这是数据结构很重要的程序,主要是课后题的答案,和代码,功能很强大(the data struct of the book)
    2013-12-22 23:25:14下载
    积分:1
  • Library
    数据结构课程设计 数据结构B+树 B+ tree Library(B+ tree Library)
    2009-05-18 14:07:25下载
    积分:1
  • 696524资源总数
  • 103939会员总数
  • 12今日下载