计量单位算法和数据结构分配1，
UoM Algorithms and Data Structures Assignment 1, (2023-10-08, C, 0KB, 下载0次)
为以太坊区块链实现BF跳过索引。
Implementation of BF skip indexes for the Ethereum blockchain. (2023-09-20, Jupyter Notebook, 0KB, 下载0次)
Probabilistic data structures in python <a href="http://pyprobables.readthedocs.io/en/latest/index.html" rel="nofollow">http://pyprobables.readthedocs.io/en/latest/index.html</a> , stars:97, update:2023-06-19 20:01:22
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其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，由于S较为窄小，车厢在S中不能调头或超车，这意味着任何一个在S中驻留的车厢能从S中驶出，当且仅当所有在它之后驶入S的车厢都已经从S中驶出（或者没有车厢在它之后驶入S）。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
Among them, a is the inlet, B is the outlet, and S is the transfer blind end. All railways are monorail unidirectional: the train can only travel from a to s, and then from s to B; it can also run directly from a to B without staying in S; in addition, due to the narrow size of S, cars can not turn around or overtake in s, which means that any car staying in s can move out of s if and only if all the cars that enter s after it have moved out of S (or No cars enter s after it. Because the cars can stay in s, the order in which they leave from end B may be different from that in which they enter from end a. However, the capacity of S is limited and no more than m cars can stay at the same time. (2020-11-23, Visual C++, 2KB, 下载0次)
教练员 A、B 和 C 将要从编号为1到n的队员中挑选自己的队员。为公平起见，每个教练都根据自己的喜好程度将队员排序；你负责根据以下规则为他们分配队员。
你拿到的数据是 a、b、c 三个数组，表示三个教练对队员的喜好程度排序，每个数组都是数字 1 到 n 的一个排列，下标越小表示教练越喜欢该队员（例如教练 A 最喜欢编号为 a[0] 的队员，其次是编号为 a[1] 的队员）。你的分组规则是，从还未被分配的队员中找一个教练A最喜欢的队员分到 A 组；然后，在未分配的队员中分配教练B最喜欢的队员到 B 组；然后是教练 C；再是教练 A、B......依次类推直到所有队员分配完毕。
现在队员 k 希望知道自己被分配给哪位教练，请你来告诉他。
Among them, a is the inlet, B is the outlet, and S is the transfer blind end. All railways are monorail unidirectional: the train can only travel from a to s, and then from s to B; it can also run directly from a to B without staying in S; in addition, due to the narrow size of S, cars can not turn around or overtake in s, which means that any car staying in s can move out of s if and only if all the cars that enter s after it have moved out of S (or No cars enter s after it. Because the cars can stay in s, the order in which they leave from end B may be different from that in which they enter from end a. However, the capacity of S is limited and no more than m cars can stay at the same time.
Suppose that a train is composed of N cars numbered {A1, A2,..., an}. The dispatcher would like to know whether, according to the above traffic rules, these cars can be rearranged in the order of {1,2,..., n} and then drive out from end B. (2020-11-23, Visual C++, 1KB, 下载0次)
Shannon编码步骤：
1、将信源符号按概率从大到小的顺序排列，
p(a1)≥ p(a2)≥…≥ p(an)
2、确定满足下列不等式的整数Ki，
－log2 p(ai) ≤ Ki < 1－log2 p(ai)
3、令P(a1)=0，用Pi表示第i个码字的累加概率，
Pi=sum_{k=1}^{i-1} p(ak)
4、将Pi用二进制表示,并取小数点后Ki位作为符号ai的编码。
Shannon coding steps:
1. The source symbols are arranged in order of probability from large to small
2. Determine the integer ki satisfying the following inequality
3. Let P(A1)= 0, and PI denotes the cumulative probability of the i-th codeword,
4. PI is represented in binary system,and Ki bit after decimal point is taken as the code of symbol AI. (2020-09-06, C/C++, 1KB, 下载0次)
AVL树有不同的版本。通过初始化一个空的AVL树来启动您的程序。程序接受一行作为输入。输入行包含n个“修改移动”，用空格分隔(1≤n≤100)。可用的
修改举措
?Aint(字符A后跟一个1到100之间的int值):A3表示将值3插入AVL树中。如果3已经在树中，什么都不做。
?Dint(字符D后跟1到100之间的int值):D3表示将值3删除到AVL树中。如果3不在树中，什么都不做。
然后，在输入之后正好有一个完成步骤(PRE或POST或IN):如果完成步骤是PRE，那么应该按预先顺序打印出树(在当前情况下)。如果树为空，则打印为空。否则，打印用空格分隔的值。POST和IN的处理方式类似。
不需要担心无效的输入。样本输入1:A1 A2 A3 IN
示例输出1:1 2 3
样本输入2:A1 A2 A3 PRE
样本输出2:2 1 3样本输入3:A1 D1 POST样本输出3:空
There are different versions of AVL trees. Start your program by initializing an empty AVL tree. The program accepts one line as input. The input line contains n "modify moves" separated by Spaces. The available Modification measures (2020-08-25, C/C++, 1KB, 下载0次)
A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
train traffic problem (2018-12-24, C/C++, 1KB, 下载2次)
给定n个矩阵构成的一个链<A1,A2,…,An>其中i 1,2,…,n,矩阵Ai的维数为pi-1pi,对乘积A1
A chain of given n Matrices (2018-09-05, C++, 3KB, 下载0次)
对顺序表元素就地逆置，将线性表（a1,a2,……,an）逆置为（an,an-1,……,a1）。
The sequential table element is placed in place, and the linear table (A1, A2,... An) inversed (an, an-1,... A1). (2018-01-13, C/C++, 489KB, 下载1次)
对一个数组进行摇摆排序使a1<a2>a3<a4>a5
sort the array in wiggle way (2016-03-11, C++, 1KB, 下载1次)
现有喜串定义如下。a 与b 互为喜串需满足以下两个条件之一： 1. a 和b 相同。 2. 将a 分成a1 与a2 两个等长串，b 分成b1 与b2 两个等长串，其子串需满足以下两个 条件之一： a) a1与b1互为喜串且a2 与b2 互为喜串。 b) a1与b2互为喜串且a2 与b1 互为喜串。 任务： 判断给定的两个串是否互为喜串
Hi existing string defined below. a and b mutually hi string must meet one of the following two conditions: 1. a and b are the same. 2. Place a a1 and a2 into two equal length strings, b b1 and b2 into two equal length strings, which substring must meet one of two conditions: a) a1 and a2 and b1 string hi to each other and b2 mutual hi string. b) a1 and a2 and b2 each other hi string hi string and b1 each other. Task: to determine whether a given two strings string hi to each other (2015-11-17, Visual C++, 1KB, 下载3次)
实验二-矩阵链乘
Description
给定n个矩阵A1，A2，…，An，其中，Ai与Aj+1是可乘的，i=1，2，…，n-1。
你的任务是要确定矩阵连乘的运算次序，使计算这n个矩阵的连乘积A1A2…An时总的元素乘法次数达到最少。
例如：3个矩阵A1，A2，A3，阶分别为10×100、100×5、5×50，计算连乘积A1A2A3时按（A1A2）A3所需的元素乘法次数达到最少，为7500次。
MatrixChain dp (2015-09-25, Java, 1KB, 下载1次)
建立一个关于学生姓名及其若干课程成绩的数据处理系统，在满足基本的数据存储功能以外，还需要向用户提供添加新学生信息、删除已有学生信息、按某一门成绩对学生信息进行排序、按要求查询学生信息、输出学生信息到文件等功能。从而熟悉已学过的各种数据结构，并掌握它们各自的特点与适用情形，同时熟悉已学过的各种数据处理算法，特别是查询、排序、添加单元、删除单元，以及遍历等等，学会根据实际需要选择数据结构和算法设计。
The establishment of a student' s name and the number of courses on the results of the data processing system, other than the satisfaction of basic data storage capabilities, you need to add new users to provide student information, delete the existing student information, according to the results of a door sort of student information , the required student information queries, student information to the output files and other functions. So familiar with has learned the various data structures, and master their characteristics and exclusions, while familiar already learned the various data processing algorithms, especially the query, sort, add unit, remove the unit, and traversal, etc., learn the actual need to the data structure and algorithm design based on. (2015-05-19, Visual C++, 5574KB, 下载3次)
用图的相关算法，解决清华园旅游问题，包括计算游览完不包括教学楼的所有景点需要的次数，找出最长的最短路径最小的点，计算铺设能到达所有景点的光纤的最小长度及路径。通过实验掌握图的表示方法和算法。
With related algorithms map, solve Tsinghua tourism issues, including the calculation does not include the number of times after visiting all the attractions of teaching building needs to find the shortest path to the smallest longest point, calculated to reach the minimum length of the laying of optical fiber and all the attractions path. Through experiments grasp graph represents the methods and algorithms. (2015-05-19, Visual C++, 7918KB, 下载3次)
对单链表实现就地逆置,即利用原表的存储空间将线性表（a1,a2,…,an）逆置为（an,an-1,…,a1）。输入数据：文本文件input.txt
文件分2行：
第1行是单链表的数据元素个数n，n<20。如果n<1或n>19则视为输入数据出错。
第2行是单链表的n个数据元素，每个元素都是-999～999之间的十进制整数，元素之间用空格分开。
输出数据：文件output.txt
文件只有1行：
第1行是完成逆置之后的单链表n个元素的十进制整数，每个元素之间用空格分开。
如果输入数据出错，或者运行过程出错、问题无解等，则第1行输出字符串“ERROR”。
Single linked list implementation situ reverse position, namely the use of the storage space of the original tables linear table (a1, a2, ..., an) inverse set (an, an-1, ..., a1). Input data: text file input.txt file sub 2 lines: Line 1 is the number of a single list of data elements n, n < 20. If n < 1 or n> 19 is considered to input data error. The second line is a single-chain n data elements, and each element is a decimal integer between-999 to 999, separated by a space between the elements. Output data: file output.txt file has only one line: Line 1 is set after the completion of the reverse single linked list of n decimal integer elements, separated by a space between each element. If the input data error, or an error during the operation, the problem has no solution, etc., then the first line of the output string " ERROR" . (2014-05-08, Visual C++, 1KB, 下载6次)
平衡二叉树生成 输入任意个节点 如 2 6 8 0为显示 可自动生成平衡二叉树 bf为平衡因子 h给深度 可插入删除 计算转动次数
Balanced binary tree to generate arbitrary input nodes as 2680 to show a balanced binary tree can be automatically generated for the balance factor bf depth h to calculate the number of rotations can be inserted deleted (2014-01-15, Visual C++, 3KB, 下载3次)
现有4堆石子,两个人轮流取石子,他们有n种可能的取法,取法表示从第1堆取A1个石子,从第2堆中取B1个,第3堆取C1个,第4堆取D1个.如果取的时候某一堆的石子数比所要取的石子数要少,则这种取法是不可行的.取到最后没有可行取法的人就算输了.
现给出4堆石子的石子数目以及n种取法,请问如果两个人都采用最优取法,先取的人是赢还是输.
Four existing heap of stones, two people take turns stones, they have emulated n possible, take the A1 emulated said a stone from a pile of rubble taken from the second B1, 3rd stacker C1, 4th heap Take a D1 and if it' s a pile of stones take a few less than the number of stones to take it, then this is not feasible emulated Take to the last man, even if there is no feasible emulated lost. heap of stones now gives 4 n kinds of stones as well as the number of emulated, what if two people are using the optimal emulated, first take the people win or lose. (2013-11-26, C/C++, 1KB, 下载1次)
基因串是由一串有限长度的基因所组成的，其中每一个基因都可以用26个英文大写字母中的一个来表示，不同的字母表示不同的基因类型。一个单独的基因可以生长成为一对新的基因，而可能成长的规则是通过一个有限的成长规则集所决定的。每一个成长的规则可以用三个大写英文字母A1A2A3来描述，这个规则的意思是基因A1可以成长为一对基因A2A3。
用大写字母S来表示一类称作超级基因的基因，因为每一个基因串都是由一串超级基因根据给出的规则所成长出来的。
请写一个程序，读入有限条成长的规则和一些我们想要得到的基因串，然后对于每个基因串，判断它是否可以由一个有限长度的超级基因串成长得出。如果可以，给出可成长为该基因串的最短超级基因串的长度。
Gene by a gene string is a string that consists of finite length, wherein each gene can be used in the 26 capital letters to indicate a different type of letters different genes. A single gene may be grown to be a new gene, the rule is likely to grow by a limited set of rules determined by the growth. Each rule can be used to grow three capital letters A1A2A3 to describe the meaning of the rule A1 gene is a gene can grow A2A3. Uppercase letter S to represent a class of genes called super-gene as a gene sequence are each composed of a string of super-gene grown out according to the given rules. Please write a program that reads the limited growth of the rules and want to get some of our gene cluster, and then for each gene cluster, to determine whether it is the growth of super-gene cluster consists of a finite length can be derived. If you can, given the length of the shortest that can grow super-gene cluster of genes strings. (2013-11-26, C/C++, 1KB, 下载1次)
计算每副牌的分值：每副牌有个原始大小(即排除对子，顺子，金花，顺金，筒子的大小），再
每张牌的分值为一个2位数，不足2位的补前导0，例如 A ：14，‘10’：10，’2‘：’02‘，’k‘：13，’7‘：07
将3张牌按点数大小排序(从大到小)，凑成一个6位数。例如’A27 ：140702，‘829’：090802，‘JK8’：131108，‘2A10’：141002
例外，对于对子要将对子的位数放在前两位(后面会看到为什么这么做）。例如‘779’：070709，‘7A7’：070714，‘A33’：030314
现在的分值是一个6位数，将对子设为一个原始值加上10*100000的值，现在为一个7位数。例如‘779’：1070709，‘7A7’：1070714，‘A33’：1030314
对于顺子，将结果加上20*100000.。例如‘345’：2050403，‘QKA’：2141312，‘23A’：2140302
对于金花，将结果加上30*100000。例如‘Spade K,Spade 6,Spade J ：3131106
因为顺金的时候其实是金花和顺子的和，所以顺金应该是50*10000。 例如‘Spade 7,Spade 6,Spade 8 ：5080706
对于筒子，将结果加上60*100000。例如’666‘：6060606，’JJJ‘：6111111
3“ 比较两幅牌的大小(用所计算的分值来比较）
就这么简单！！
Scores calculated for each of the cards : Each of the cards have the original size ( ie excluding pairs, straights , Jinhua, Shun-Jin , cheese the size ) , and then
The score of each card a 2-digit , less than two s complement leading 0, for example A : 14, 10 : 10, 2 : 02 , k : 13, 7 : 07
The three cards in descending order according to points ( descending ) , make up a six digits. For example A27 : 140702, 829 : 090802, JK8 : 131108, 2 A10 : 141002
Exception to the pair on the first two pairs of digits ( you ll see later how to do so ) . For example 779 : 070709, 7 A7 : 070714, A33 : 030314
Now the score is a six-digit , it will set an original value plus the sub- 10* 100000 value , is now a seven digits. For example 779 : 1070709, 7 A7 : 1070714, A33 : 1030314
For the straight , the result plus 20* 100000 . For example 345 : 2050403, QK (2013-11-13, C#, 1KB, 下载5次)