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bintree
一个关于二叉树应用的小程序,包括二叉树的建立，前中后遍历等操作(a program about bintree structure)
- 2009-05-17 16:54:29下载
- 积分：1
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maze
说明： 由数字1和0组成迷宫，0代表可走路径，程序会用坐标的形式给出每一步的走法(Composed by a maze of numbers 1 and 0, 0 to go the path, the program will use the coordinates given in the form of moves at each step)
- 2011-03-27 20:57:47下载
- 积分：1
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FreeTree
无环连通图G=(V, E)亦称为自由树T，其直径是树中所有顶点之间最短路径的最大值，设计一个算法求T的直径(Acyclic connected graph G = (V, E) is also known as free tree T, whose diameter is the tree of shortest paths between all vertices maximum design an algorithm for the diameter of T)
- 2013-10-29 16:31:47下载
- 积分：1
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48qqqq
主要用栈和树求解骑士问题，用C++语言编写的(The main stack and tree Cavalier Problem Solving with C++ languages)
- 2009-03-21 22:34:39下载
- 积分：1
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BinarySearchTree
在Visual Studio 2008运行成功 数据结构 二叉查找树的类(In Visual Studio 2008 running a successful binary search tree data structure classes)
- 2013-07-23 22:04:29下载
- 积分：1
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Calender
日历显示，查询1900年后的日历。每一屏显示一年的日历祥情。(Calender dispaling. Acess the calender detail of years after 1900.)
- 2011-12-18 14:57:15下载
- 积分：1
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BinaryTree
数据结构二叉树，及其遍历，镜像，输出等相关操作C++源码。(Binary tree data structure and its traversal, mirror, output the operation C++ source.)
- 2012-09-01 11:25:15下载
- 积分：1
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Work3.5
上海交通大学数据结构课程的课后作业系列3.5：仅供参考，最好还是自己去写(Shanghai Jiaotong University data structure course homework Series 3.5: for reference, it is best to write their own)
- 2013-12-27 21:00:03下载
- 积分：1
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polynomial
本程序实现一元多项式合并。本程序输入方式为按项输入，每输入一个项，会先提示输入项的幂，再提示输入项的系数。(This procedure combined to achieve a dollar polynomial. This procedure according to entry input methods for the input, one for each input item, it will first prompt entry of the power, and then prompts entry of the coefficient.)
- 2009-12-16 17:29:54下载
- 积分：1
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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
