这个用C++编写的PUBG Steam Cheat源代码提供了一组全面的功能，旨在增强玩家在游戏中的体验。
This PUBG Steam Cheat Source Code, written in C++, offers a comprehensive set of features designed to enhance the players experience in the game. (2024-04-17, C++, 0KB, 下载0次)
“该项目的特点是游戏2048的Python实现，包括使用Q-learning训练的代理。该代码允许手动游戏以及训练代理自主玩游戏。使用这个现成的实现探索游戏动态和强化学习过程。”
"This project features a Python implementation of the game 2048, including an agent trained using Q-learning. The code allows for manual gameplay as well as training the agent to autonomously play the game. Explore the game dynamics and reinforcement learning process with this ready-to-use implementation." (2024-02-18, Jupyter Notebook, 0KB, 下载0次)
如果您正在寻找免费的编码练习，以准备您的下一次工作面试，我们已经为您提供了覆盖。本节包含数百个编码挑战，测试您的算法、数据结构和数学知识。它也有许多带回家的项目，你可以用来加强你的技能，或添加到你的投资组合。
If you re looking for free coding exercises to prepare for your next job interview, we ve got you covered. This section contains hundreds of coding challenges that test your knowledge of algorithms, data structures, and mathematics. It also has a number of take-home projects you can use to strengthen your skills, or add to your portfolio. (2024-01-08, Others, 0KB, 下载0次)
用Java开发的交互式扣篮比赛模拟，实现各种设计模式（策略、工厂方法、单例）来建模玩家交互、判断评估和评分逻辑。该项目展示了面向对象的编程技能、用于扣篮评估的复杂算法和全面的用户交互。
An interactive Dunk Contest simulation developed in Java, implementing various design patterns (Strategy, Factory Method, Singleton) to model player interactions, judge evaluations, and scoring logic. This project showcases object-oriented programming skills, intricate algorithms for dunk evaluation, and comprehensive user interaction. (2024-01-06, Java, 0KB, 下载0次)
游戏规则: 1:玩家=888, 机器人=robot; 2:每个玩家5张牌: 3:牌有1<2,2<3,3<1,每张牌数量未知 4:比大小,(1)比拼牌归位,(2)胜者可交换点数以内任意牌,(3)败者获得胜者手中一张牌并弃一张牌 5:两人以上胜利条件:手牌先清空;平局:手牌均相...
Game rules: 1: player=888, robot=robot; 2: Each player has 5 cards: 3: there are 1<2,2<3,3<1 cards, the number of cards per card is unknown, 4: the size of the game, (1) the game card returns, (2) the winner can exchange any card within the number of points, (3) the loser gains a card in the winner s hand and discards a card, 5: the victory condition for more than two people: the hand card is cleared first; Draw: hand is even (2021-06-24, Python, 0KB, 下载0次)
Jogo educativo de matemática Paraensino基础教育。Treina raciocício rápido，modo单人游戏（com levels e explica o de conteúdo），modo多人游戏，新玩家综合排名
Jogo educativo de matemática para ensino fundamental e médio. Treina raciocício rápido, modo single player (com levels e explica o de conteúdo), modo multiplayer, compra de novos personagens e ranking (2019-05-16, C#, 0KB, 下载0次)
HDT Damage Counter是炉石牌堆跟踪器（HDT）的插件，通过提供玩家在当前游戏状态下可以处理的最小可能伤害的视觉表示，增强您的炉石牌游戏体验。
HDT Damage Counter is a plugin for Hearthstone Deck Tracker (HDT) that enhances your Hearthstone gameplay experience by providing a visual representation of the minimum possible damage that can be dealt by players in the current game state. (2023-05-24, C#, 0KB, 下载0次)
这是一个压缩，这个github只是用于pltform下载的重定向，请在Synapse.exe链接处下载平台，下载并玩得开心，代码可能会在短时间内返回
This is a compression, this github is simply redirect for download of the pltform, please download the platform at the link that says Synapse.exe, download and have fun, code might come back for short periods of time (2018-04-06, C#, 0KB, 下载0次)
用javascript克隆了松鸡游戏，开发了一个AI来玩这个游戏，并增加了一些有趣的难度。基线算法是神经进化，它在几代之后向前进化一组馈电。
Clone of flappy bird game in javascript, developed an AI to play this game and added some interesting difficulty. The baseline algorithm is neuro-evolution which evolves a set of feed foward after a number of generations. (2019-07-18, JavaScript, 0KB, 下载0次)
将经典和量子tic tac toe实现为2人游戏和玩家对计算机游戏，具有获胜概率和策略，作为第一年年底在ISI加尔各答的夏季实习的一部分。
Implementation of classical and quantum tic tac toe as 2 player and player vs computer games, with winning probabilities and strategies, as part of summer internship at ISI Kolkata at the end of first year. (2022-06-08, C, 0KB, 下载0次)
2022年：在机器学习中，有几个子类别。在这里，我们将关注一种接近于称为强化学习的试错法的学习类型。该程序（随后将被称为代理或系统）执行一系列操作，在一段时间后，状态rea的质量将被提高...
YEAR 2022 : In machine learning there are several subcategories. We will focus here on a type of learning that is close to the trial and error method called reinforcement learning. The program (which will subsequently be called either agent or system) performs a series of actions and after a certain period of time, the quality of the state (2022-07-08, C, 0KB, 下载0次)
给定含有n个元素的多重集合S，每个元素在S中出现的次数称为该元素的重数。多重集S中重数最大的元素称为 众数。例如，S={1，2，2，2，3，5}。多重集S的众数是2，其重数为3。 求众数方法很多，现要求你用分治算法来试一试，并分析其效率。 编程任务：对于给定的由n个自然数组成的多重集S，采用分治算法编程计算S的众数及其重数。 输入格式 第1行多重集S中元素个数n；接下来的一行为集合S，有n个自然数。( n < 100000 ) 输出格式 结果输出：输出2个数，第1个数为众数，第2个为其重数。 当有多个同样重数的众数，优先输出数值更小的众数。 输入样例 6 1 2 2 2 3 5 输出样例 2 3
Given a multiple set S containing n elements, the number of occurrences of each element in S is called the multiplicity of that element. The element with the largest number in the multiple set S is called the mode number. For example, S={1, 2, 2, 2, 3, 5}. The multiple set S has a modulus of 2 and a multiplicity of 3. There are many ways to find the mode number. Now you need to try the divide and conquer algorithm and analyze its efficiency. Programming tasks: For a given multiple set S consisting of n natural numbers, divide and conquer algorithm is used to program and calculate the modulus and multiplicity of S. (2019-01-07, C/C++, 7043KB, 下载0次)
若矩阵Am×n中的某个元素aij是第i行中的最小值，同时又是第j列中的最大值，则称此元素为该矩阵中的一个马鞍点。假设以二维数组存储矩阵Am×n ，试编写求出矩阵中所有马鞍点的算法。
If the Am× n matrix of a certain element in the AIJ line I in minimum value, at the same time it is the j column in the maximum, then the elements of the matrix in a saddle point. The hypothesis with two-dimensional array storage matrix Am× n, try to write out all the saddle point algorithm for matrix. (2012-11-05, Visual C++, 150KB, 下载2次)
实现贝叶斯概率计算过程。假设进行n次试生产，计算最后得到后验概率，即可信度。样本数据集个数和观测值从文件中读入，计算过程封装为动态链接库，以.dll形式提供，
主程序提供界面输入先验分布值、边缘分布值，显示计算结果。
Bayesian probability calculation process. Assuming that the n-th trial production, and calculate the final posterior probability that credibility. The number of sample data sets and observations read from the file, the calculation process packaged as a dynamic link library. Dll form, the main program interface input prior distribution value, value of marginal distribution, display calculation results. (2012-10-11, Visual C++, 1289KB, 下载22次)
FORTRAN，亦译为福传，是英文“FORmula TRANslator”的缩写，译为“公式翻译器”，它是世界上最早出现的计算机高级程序设计语言，广泛应用于科学和工程计算领域。FORTRAN语言以其特有的功能在数值、科学和工程计算领域发挥着重要作用。
FORTRAN, also translated as evangelization, in English "FORmula TRANslator" acronym, translated as "formula translator", which is the world s earliest high-level computer programming language, widely used in scientific and engineering computing. FORTRAN language-specific features in its value, scientific and engineering computing plays an important role. (2011-12-21, Fortran, 2020KB, 下载4次)
给定两个 4×4 方格矩阵组成的图形 A 和 B ，每个方格的颜色为黑色或白色。图形 A 和图形 B 的黑白方块总数相等。
通过交换相邻方格的颜色。试设计一个算法，计算最少需要多少步变换，才能将图形 A 变换为图形 B
Given two matrices 4 × 4 grid graph A and B, the color of each box is black or white. A graphic black and white box and graphics B equal to the total. Through the exchange of the color of the adjacent square. Try to design an algorithm to calculate how many steps need to transform at least, to transform the graphic A graphic B (2010-10-07, Visual C++, 3KB, 下载9次)
已知某哈希表的装载因子小于1，哈希函数
H(key)为关键字(标识符)的第一个字母在字母表中
的序号，处理冲突的方法为线性探测开放定址法。
试编写一个按第一个字母的顺序输出哈希表中所有
关键字的算法。
Known load factor of a hash table is less than 1, the hash function H (key) for the keywords (identifiers) of the first letter of the alphabet in the serial number, dealing with the conflict detection for the linear open addressing method . Try to write a letter by the first order of output of the hash table algorithm for all keywords. (2010-05-28, C/C++, 1KB, 下载22次)
给某个玩家100元的资本，让他不停押注直到输光，计算需要赌博多少次？
如果把次数放在数组ruinLength[]中，进行1000次实验后，看看破产的最大次数、
最小次数和平均次数分别是多少？
100 to a player' s capital, so that he kept until the bet输光to calculate how many times the need for gambling? If the number on the array ruinLength [] in experiments carried out after 1000 to see the largest number of bankruptcy, the minimum number and the average number of how many were? (2009-04-17, C/C++, 1KB, 下载4次)
函数再现机构设计
试设计一曲柄摇杆机构，再现函数
要求：
输入构件的转角范围180°，输出构件摆角范围30°，即：
当输入构件从a转至a+90时，输出构件从b转至b+30
当输入构件从a+90转至a+180时，输出构件从b+30转至b
Function reproduce the design of mechanism design test one crank-rocker mechanism, reproduction function requirements: the corner of the scope of the importation of component 180 °, the output component swinging angle range 30 °, namely: When the input from a component to a+ 90, the output components from the b Go to b+ 30, when the importation of components from a+ 90 to a+ 180, the output component from b+ 30 to b (2008-07-12, Others, 4KB, 下载6次)
试写一算法,实现顺序表的就地逆置,即利用原表的存储空间将线性表(a1,a2,...,an)逆置为(an,an-1,...,a1).
try to write an algorithm to achieve the order form in situ reverse home, namely the use of the original table of linear storage space (a1, a2 ,..., an) inverse home (an, an-1 ,..., a1). (2005-06-24, C/C++, 1KB, 下载17次)