我在诺姆·尼桑（Noam Nisan）和西蒙·肖肯（Shimon Schocken）的《计算系统的要素》（the Elements of Computing Systems）一书和《从南德到俄罗斯方块》（From Nand to Tetris）课程中的项目实施。[在制品]
My implementations of the projects in the book "The Elements of Computing Systems" and the course "From Nand to Tetris" by Noam Nisan and Shimon Schocken. [WIP] (2024-03-24, Assembly, 0KB, 下载0次)
我在8086年用nasm做了一个俄罗斯方块游戏。试一试，我希望你会喜欢它。它还包括音乐，并给它一个星球大战的触摸。
i have made a Tetris game in 8086 using nasm .Try it out , i hope you will like it. It also includes music in it and have given it a star wars touch as well. (2024-02-25, Assembly, 0KB, 下载0次)
该存储库包含我对迈克尔·西科尔斯基（Michael Sikorski）和安德鲁·霍尼格（Andrew Honig）的《实用恶意软件分析》（Practical malware Analysis）一书的恶意软件实验室的写作。
This repository contains the my writeups to the malware labs of the book "Practical Malware Analysis" by Michael Sikorski and Andrew Honig. (2024-02-20, Assembly, 0KB, 下载0次)
powell算法设计
鲍威尔法又称方向加速法，它由Powell于1964年提出，是利用共轭方向可以加快收敛速度的性质形成的一种搜索方法。该方法不需要对目标函数进行求导，当目标函数的导数不连续的时候也能应用，因此，鲍威尔算法是一种十分有效的直接搜索法。
Powell法可用于求解一般无约束优化问题，对于维数n<20的目标函数求优化问题，此法可获得较满意的结果。
不同于其他的直接法，Powell法有一套完整的理论体系，故其计算效率高于其他直接法。该方法使用一维搜索，而不是跳跃的探测步。同时，Powell法的搜索方向不一定为下降方向。
Powell method can be used to solve general unconstrained optimization problems. For the optimization of dimension n<20 objective function, this method can obtain satisfactory results.
Unlike other direct methods, the Powell method has a complete set of theoretical systems, so its computational efficiency is higher than other direct methods. The method uses one-dimensional search instead of jumping detection steps. At the same time, the search direction of the Powell method is not necessarily for the downward direction. (2017-07-14, Visual C++, 135KB, 下载1次)
已知两个整数变量 A 和B，范围为0-100
（1）、若两个数中有一个是奇数，则将奇数存入A 中，偶数存入B 中。
（2）、若两个数均为奇数，则两数分别加1，并存回原变量。
（3）、若两个数均为偶数，则两个变量不变。
Two integer variables A and B, which range of 0-100 (1) if there is an odd number, store the odd number in A, and even in B. (2) If the two numbers are odd, the two numbers should stored back to the original variables, after add 1. (3) If the two numbers are even, then the two variables do not change. (2013-05-14, Asm, 1KB, 下载4次)
以GRADE为首地址的10个字的数组中保存有学生成绩。建立一个10个字的RNAK 数组，并根据GRADE中的学生成绩将学生名次填入RANK数组中（提示：一个学生的名次等于成绩高于等于该学生的人数加1）。
Saved in an array of 10-word address GRADE headed student achievement. The establishment of a 10-word RNAK array, and student achievement in the GRADE the student ranking populate RANK array (Hint: a student ranking equal to score higher than or equal to the number of students plus 1). (2012-11-04, Asm, 1KB, 下载33次)
根据哈夫曼算法，建立哈夫曼树时，可以将哈夫曼树定义为一个结构型的一维数组HuffTree，保存哈夫曼树中各结点的信息，每个结点包括：权值、左孩子、右孩子、双亲
Huffman algorithm Huffman tree is created, the Huffman tree is defined as a structure of one-dimensional array HuffTree, save the information of Huffman tree nodes, each node including: the right valueleft child and right child, parents (2012-05-13, Visual C++, 1KB, 下载3次)
已知BUF1,BUF2字节单元中各存放一个无符号数，编写程序完成下述功能：
若两个数中一个是奇数一个是偶数，则奇数存入BUF1单元，偶数存入BUF2单元。
若两个数均为奇数，则分别加1后存回原单元
若两个数均为偶数，则不做任何操作。
The known BUF1 and BUF2 byte unit in the store a unsigned number, write a program to complete the following functions: if two number one is odd number one is even, the odd number in BUF1 unit, even in BUF2 unit. If the two Numbers are an odd number, add 1 respectively after the unit back to save two Numbers for all if even, do not do any operation. (2012-04-13, Asm, 1KB, 下载3次)
北京迪文智能液晶屏驱动程序及其范例，是开发智能串口液晶屏很好的资料。范例俄罗斯方块游戏。-
Beijing Di Wenzhi to LCD driver and the example of the development of intelligent serial LCD screen very good information. Example Russian square game.- (2011-06-16, Asm, 1069KB, 下载173次)
程序，把数据段中AAA标号地址中连续10个数（5，6，7，8，9，A，B，C，D，E，）读出，把每个数加2后再存入到数据段中BBB开始的标号地址中去，并把BBB数据显示出来；
Program, the data segment in 10 consecutive AAA label address the number of (5,6,7,8,9, A, B, C, D, E,) read out, and then add 2 to each number stored in the BBB data segment to the beginning of the address label, and display the BBB data
(2010-10-26, DOS, 531KB, 下载3次)
需要一个八位LED数码管显示器用于显示时间（分，秒）；显示的初值在启动实验箱运行程序后输入；按任意键后时间开始走动，每隔一秒改变一次显示值（每60秒为一分钟，60分一小时，24小时后复位）；能同时显示时分秒共六位；秒钟为60时，能自动复位到零，并使分钟加一；分钟为60时，能自动复位到零，并使小时加一；小时为24时时，自动复位到零。LED数码管显示器循环显示时分秒的动态值。现显示过程中，按A键可重新置初值；按B键暂停时间，再按一次从当前时间开始计时；D、E、F键分别清除小时、分钟、秒钟对应两位。
Need an 8 LED digital display to display time (tube) second, The show started in the initial input to run the program after experimental box, Press any key after every time, a change that value (once every 60 seconds for a minute, 60 points, 24 hours after an hour reset) When can display of six seconds, For 60 seconds, can be reset to zero, and make a minute, For 60 minutes, can be reset to zero, and make a hour, 24 hours, automatic reset to zero. LED digital display monitor cycle when tube of dynamic value determined. Now display process, according to A key to offset the initial, According to B key stoppage time, click again from the start time, D, E, F key respectively, minutes and seconds clear hour corresponding two. (2009-12-27, Asm, 585KB, 下载5次)
利用DAC0832产生三角波，程序如下：
本程序中产生的电压的最大幅度受DAC0832的基准电压限制
产生的的三角波周期受累加器A的控制，可以通过改变每次加的数值来改变周期，但连续性绘变差
也可以改变延时时间来实现
DAC0832 generated using the triangular wave, as follows: This procedure generated by the voltage of the most significant restrictions on the reference voltage of the DAC0832 generated by the triangular wave cycle control accumulator A can be added by changing the value every time to change the cycle, However, the continuity of paint deterioration also can change the delay time to achieve (2009-07-29, Others, 32KB, 下载16次)
求积分,可以计算被积函数高震荡的一维积分
D01AKF IS AN ADAPTIVE INTEGRATOR, ESPECIALLY SUITED TO
C STRONGLY OSCILLATORY NON-SINGULAR INTEGRANDS, WHICH CALCULATES
C AN APPROXIMATION TO THE INTEGRAL OF A FUNCTION OVER A FINITE
C INTERVAL (A,B)
C
C D01AKF ITSELF IS ESSENTIALLY A DUMMY ROUTINE WHOSE FUNCTION IS TO
C PARTITION THE WORK ARRAYS WORK AND IWORK FOR USE BY D01AKV.
C WORK IS PARTITIONED INTO 4 ARRAYS EACH OF SIZE LIMIT, WHERE
C LIMIT = MIN(LWORK/4, LIWORK).
C IWORK IS A SINGLE ARRAY IN D01AKV OF SIZE LIMIT. (2009-06-20, Fortran, 72KB, 下载7次)
可以调节分和小时（只能加），不知道在硬件上可不可以实现上电复位，反正在仿真的时候不行，所以设置了一个清零端
在这里附上我的报告（里面有详细的设计原理及过程）和原文件，还望大家指点，交流：
It can adjust the hours and hours (only plus), do not know the hardware can achieve the power-on reset and, at any time in the simulation will not work, so set up a zero-side in my report is attached here (there are detailed design Principle and process) and the original document, we hope that pointing, exchange: (2009-01-04, Asm, 1KB, 下载2次)
利用BIOS和DOS的中断来设计一个计算器，要求在屏幕上显示一个主菜单，提示用户输入相应的数字键，分别执行加、减、乘、除四种计算功能和结束程序的功能。
Use BIOS and DOS interrupt to design a calculator, the requirements on the screen displays a main menu that prompts the user to input the corresponding number keys, respectively, the implementation of add, subtract, multiply, with the exception of four procedures for calculating the function and the end of the function. (2008-11-22, DOS, 3KB, 下载5次)
简单的计算器
1.键盘输入两个10进制数据（1到4位），选择实现加、减、乘、除运算，并选择以2进制、8进制、10进制、16进制方式输入。
要求：
程序界面尽量方便使用并简洁。
Simple calculator 1. Keyboard input two 10-band data (1-4), select the realization of add, subtract, multiply, with the exception of computing, and choose to 2 M, 8 M, 10 M, 16 M enter. Requirements: easy-to-use interface and simple as possible. (2008-06-24, DOS, 3KB, 下载12次)
農夫過河，用verilog寫的，動物種類加農夫共有4種，農夫一次只能帶一種動物過河，但是雞跟貓不能一起，豆不能跟雞一起
The farmer across the river, using Verilog written animal species plus a total of four kinds of farmers, farmers can only bring an animal to cross the river, but the chicken together with the cat should not, should not with the chicken with beans (2008-03-10, Others, 138KB, 下载4次)
里面收录了19个经典汇编案例例！电子时钟，动画！图形变换！
inside contains 19 classic case compilation cases! Electronic clock, animation! Graphics transform! (2005-12-06, Asm, 252KB, 下载487次)
随机产生10道互不相同的20以内的加减法（加减数及和差均<=20）待用户答完题后，显示用户的得分、评语及正确答案。并带彩色，计时器和系统日期和时间。这是我的汇编课程设计！
randomly generated 10 incompatible with the 20 within the addition and subtraction (plus or minus a few and the poor and have lt; = 20) to be wound that users, showed scores of users, reviews and the correct answer. Bring color, and a timer system date and time. This is a compilation of curriculum design! (2005-01-20, Asm, 5KB, 下载36次)
从键盘输入两个数字(0~7)之间 （1）用16进制表示两个数的和 （2）用四位二进制表示两个数的差，结果一定>=0 （3）用两位十进制表示两个数的积 exemple:
keyboard input from the two figures (0 to 7) between (a) a 16-band, said the number two and (2) with four two binary said the number of poor, the results would certainly gt; = 0 (3) with two decimal number two said the plot example : (2005-01-05, WINDOWS, 1KB, 下载5次)