11组表达式求值

本程序是用c语言编写的在800位整数之间的乘积运算(This procedure is written in c the 800 integers in the product operation between)

链表L，创建公有成员函数Split（A,B ），创建2个新表A,B，使的A 中含有L中奇数位置元数，B中含L偶数位置元数 (Chain L, create public member function Split (A, B), to create two new tables A, B, A so containing L odd position of a few yuan, B-L even a few locations yuan)

使用三叉链表实现的哈夫曼树,统计inputfile1.txt中各字符的出现频率，并据此构造Huffman树，编制Huffman 码；根据已经得到的编码，对01形式的编码段进行译码。 具体的要求： 1．将给定字符文件编码，生成编码，输出每个字符出现的次数和编码； 2．将给定编码文件译码，生成字符，输出编码及其对应字符。 (Emergence of the frequency of each character in the trigeminal lists using the Huffman tree, statistics inputfile1.txt, and accordingly the Huffman tree structure, preparation of Huffman code according to the coding has been. 01 to form the code segment for decoding. Specific requirements: 1 coding the given character file, generating the encoding, and outputting the number and encoding of each character 2 will be given the encoding file decoding, generating the character, the output code and its corresponding character.)

参加数学建模的同学很可能会遇到这套题。本程序采用链表进行编写，很方便地解决了小白鼠问题。(Mathematical modeling of the students participating are likely to encounter this question, the program uses the list to solve the problem of mice, very convenient.)

电梯模拟，实现一个模拟电梯的环境，用C程序实现的，封装良好(elevator modify)

找出数组A中最长的递增序列的长度，使用动态规划算法，复杂度为O(N*N)。(Find the longest increasing sequence)

最优二叉搜索树问题具有最优子结构性质，即其任意一棵子树都是一棵独立的最优二叉搜索树，所以可以采用动态规划方法，并且在每次更新时保存子树的根。(Optimal binary search tree problem has optimal substructure property that any of its sub-tree are an independent optimal binary search tree, so you can use dynamic programming methods, and save each update subtree roots.)

经典的 数据结构问题 八皇后问题。问题是在8*8的棋盘上放置棋子，要求每条直线上不能出现两个或以上的棋子。(The classical data structure problem Eight queens problem.Question is placed on the 8* 8 board piece, requirements on each line cannot appear two or more pieces.)

实现折半查找，非递归形式，从而实现数据结构所要的结果。(To achieve binary search, the non-recursive form)

普利姆算法用来求解最小生成树问题很容易，使得树的最小权值最小。(Prim algorithm used to solve the minimum spanning tree problem is easy, the minimum weight is the lowest of the tree)