代码游戏蛇

最新noi2099解题报告.包含源程序.标注以及解答.-Latest noi2099 problem-solving report. Contains source. Callout and answer.

以高效率的方法查找一个数组中的第二大数，数组中的数可以相同.同时对各种可能发生的错误返回相应得错误代码。-find second max number in a high effective method.The number in the array could be the same.meanwhile ,the program return special error code

求矩阵特征值、解方程组 -seeking matrix eigenvalue solution equations for matrix eigenvalue equations Solutions

经典八数码问题,回溯法的解决-eight digital classic, backtracking solution

马的遍历问题。在中国象棋棋盘上，对任一位置上放置的一个马，均能选择合适的路线，使得该棋子能够按象棋的规则不重复的走过棋盘上的每一个位置。-Traversal problem horses. In Chinese chess chessboard, for any one location to place a horse, able to choose the right line, making the chess pieces according to chess rules can not repeat the past chessboard for each location.

1.[问题描述] 编写递归算法，计算二叉树中叶子结点的数目 [输入] 按照先序序列的顺序输入该结点的内容。其输入abd eh cf i g . [输出] 按中序序列输出，输出的结果为；dbheaficg并计算出二叉树中叶子结点的数目为4 [存储结构] 采用二叉表存储 [算法的基本思想] 采用递归方法建立和遍历二叉树。首先建立二叉树的根结点，然后建立其左右子树，直到空子树为止，中序遍历二叉树时，先遍厉左子树，后遍厉右子树，最后访问根结点。根据左右子树的最后一个结点计算出二叉树中叶子结点的数目。 程序如下： #include #include #include"stdlib.h" -1. [Description of the issue] to prepare recursive algorithm, Binary calculation leaves the number of nodes [imported] in accordance with the first order input sequence in the order of the node contents. Input abd eh i g cf. [Output] by the order sequence output, the results of the output; dbheaficg calculated Binary leaf node to the number four [storage structure] Table 2 forks storage [ The basic idea algorithm] recursive method and traverse binary tree. First established binary tree root node, and then to build their son around the tree, the tree until the loopholes, which preorder binary tree, Li first times the left sub-tree, right af

这是一个用于计算最小生成树的程序。当人们逐个输入每条边的权，再点击生成，便能得到这个图的最小生成树。-This is a used to calculate the minimum spanning tree procedure. When people enter one by one each side of the right, then click generate, will be able to get this graph minimum spanning tree.

这个医院管理系统管理病人信息、 人员信息、 商店和药品、 计费并生成报告。这个复杂的应用程序与后端数据库服务器进行通信和管理医院物流相关的所有信息。

问题描述： 给定n个石子，其重量分别a1,a2,a3...an,要求将其划分成m份，每一份的划分费 义为这份石了中最大重量与最小重量的差的平方。总划分费用等丁m份划分费用之和。 编程任务 对于给定的n个石子，求一种划分方案，使得总划分费用最小。-Description of the problem: Given n-stones, its weight, respectively a1, a2, a3 ... an, asked that it be divided into m copies, each of the division of costs defined as the weight of the stone of the largest and the smallest weight difference the square. Small m the total costs were divided into the sum of the cost. Programming tasks for a given n-stones, seek a kind of division, making the total cost of the smallest division.

pb做出的outlook界面框架,可以直接用在管理信息系统中来使用.-pb"s outlook interface framework, can be directly used in the management information systems to use it.