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说明：  nuaa1108 多少连续的1 解题报告(Problem solving report )

说明：  数据结构：Huffman编码。哈夫曼编码是哈夫曼树的一个应用。哈夫曼编码应用广泛，如JPEG中就应用了哈夫曼编码。 哈夫曼树又称最优二叉树，是一种带权路径长度最短的二叉树。(Data structure: Huffman coding. Huffman coding is an application of the Huffman tree. Huffman coding is widely used, such as JPEG in the application of the Huffman coding. Huffman tree, also known as the best binary tree is a weighted shortest path length binary tree.)

堆栈实现C++,以VC++环境为准——逍遥大洋至上(The stack of the C++, VC++ environment will prevail)

二叉树的操作，涉及到三种顺序的遍历，树的凹入表示法，转化为一维数组储存，节点的计数，以及中序线索化。(Binary operation, involving three kinds of the order of traversal, tree concave representation into a one-dimensional array of storage, the node count, as well as in the threaded sequence.)

一个很好的银行家算法，能运行(A good banker s algorithm, will be able to run)

本代码提供一种非常简单且快速的哈夫曼编码方法。 (this code provide a very simple and rapid method of Huffman coding.)

单链表逆置的源码~提供了源码和源码说明~简单实用，适用于课堂作业程度(Reverse single linked list provides a set of source code source code and source code ~ ~ simple and practical instructions for classroom work degree)

北大ACM1702题 转换为3进制 很经典(NULL)

数据结构，利用霍夫曼树实现文本数据的压缩。(Huffman data compression and decompression.)

数字根算法，用最简单的代码计算数字根。如果把一个大数的各位数字相加得到一个和，再把这个和的各位数字相加又得一个和，再继续作数字和，直到最后的数字和是个位数为止，这最后的数称为最初那个数的“数字根”。(Digital root algorithm, with the most simple code to calculate the digital root. If you figure in a large numbers is the sum of a, then this and you figure the sum is again a, and then continue for digital and until the final figure and the single digits, the last number called initially that the number of "digital root".)