[Windows编程] XpMenu3.1
OFFICE 2000/97 菜单
OFFICE 2000/97 menu (2004-11-01, C++ Builder, 99KB, 下载76次)
[其他嵌入式/单片机内容] 3328F-china-ptv
ALI M3328F A1 uploader
ALI M3328F A1 uploader (2012-08-03, C++ Builder, 491KB, 下载29次)
[驱动编程] 24c01-24c16驱动程序
//24c01-24c16读写驱动程序， sbit a0=ACC^0 //定义ACC的位，利用ACC操作速度最快 sbit a1=ACC^1 sbit a2=ACC^2 sbit a3=ACC^3 sbit a4=ACC^4 sbit a5=ACC^5 sbit a6=ACC^6 sbit a7=ACC^7
//24c01- 24c16 literacy drivers, ACC sbit a0 = ^ 0// ACC definition of the position, using the speed of the fastest ACC sbit ACC a1 = ^ 1 = a2 sbit ACC sbit a3 ^ 2 ^ 3 = ACC sbit a4 = ^ ACC a5 = 4 sbit ACC sbit a6 ^ 5 ^ 6 = ACC sbit a7 ACC = ^ 7 (2005-07-04, C++ Builder, 1KB, 下载24次)
[数学计算] 2012.4.13
DLT645 检验和、+/-33计算小软件，附带源码。
DLT645 inspection and+/-33 computing software, with source code. (2012-07-07, C++ Builder, 489KB, 下载22次)
[处理器开发] TEST
MCU是msp430g2553，用timerA0控制LED的闪烁，闪烁速度由快到慢，循环执行
#include "msp430g2553.h"
unsigned char J,K
void main( void )
{
//unsigned char i
// Stop watchdog timer to prevent time out reset
WDTCTL = WDTPW+ WDTHOLD
P1DIR |= BIT6
/*BCSCTL1 &= ~BIT7
do
{
IFG1 &= ~OFIFG
for(i=0xff i>0 i--)
}
while(IFG1&OFIFG)
BCSCTL2 |= SELS+ DIVS_3 */
CCTL0 |= CCIE
CCR0=65000
TACTL |= TASSEL_2+ MC_1 //+ ID_3
_EINT()
LPM0
}
#pragma vector=TIMER0_A0_VECTOR
__interrupt void key(void)
{
P1OUT ^= BIT6
J++
if(J==16)
{
J=0
K++
if(K==1)
{
TACTL &=~(ID0+ ID1)
TACTL |= ID_0
}
if(K==2)
{
TACTL &=~(ID0+ ID1)
TACTL |= ID_1
}
if(K==3)
{
TACTL &=~(ID0+ ID1)
TACTL |= ID_2
}
if(K==4)
{
TACTL &=~(ID0+ ID1)
TACTL |= ID_3
K=0
}
}
}
(2012-07-09, C++ Builder, 21KB, 下载19次)
[单片机开发] upp
The software is owned by TI and/or its suppliers, and is protected under applicable copyright laws.
The software is owned by TI and/or its suppliers, and is protected under applicable copyright laws. (2012-12-25, C++ Builder, 6KB, 下载18次)
[数学计算] 2005102611013310480
最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50 (2005-11-28, C++ Builder, 6KB, 下载18次)
[网络编程] forumbot-a1.0
用于 各种 论坛批量注册。。Bulk Registration for a variety of forums. .
Bulk Registration for a variety of forums. . (2011-08-25, C++ Builder, 721KB, 下载17次)
[其他嵌入式/单片机内容] Echolink-China-Ali3328F
Echolink Ali 3328F A1 bin file
Echolink Ali 3328F A1 bin file (2012-08-03, C++ Builder, 497KB, 下载16次)
[单片机开发] JD9161BA-CTC3.97TN-9bit-MIPI_OK-(1)
This is a 9161+ over 3.97 of the project, MIPI dual channel, there is a need to please download, thank you
This is a 9161+ over 3.97 of the project, MIPI dual channel, there is a need to please download, thank you (2016-10-27, C++ Builder, 292KB, 下载12次)
[数值算法/人工智能] SAT
SAT问题也称为合取范式的可满足问题，一个合取范式形如：A1∧A2∧…∧An，子句Ai（1≤i≤n）形如：a1∨a2∨…∨ak，其中，ai称为文字，为某一布尔变量或该布尔变量的非。SAT问题是指：是否存在一组对所有布尔变量的赋值（TRUE或FALSE），使得整个合取范式取值为真。
SAT problems are also called conjunctive normal form of the satisfiability problem, a conjunctive normal form such as: A1 lambda A2 lambda ... A An, clause Ai (1 i n ) shaped like: A1 V A2 v ... On AK, wherein, AI is known as the text, a boolean variable or the boolean variable non. SAT problem refers to: the existence of a group of all boolean variable assignment ( TRUE or FALSE ), making the entire conjunctive normal form a value of true.
(2011-12-25, C++ Builder, 564KB, 下载12次)
[DSP编程] Adc_T1_correct
AD采样频率为10K，序列发生器SEQ1和SEQ2级联成一个16通道的序列发生器，
* 采样模式采用顺序采样。利用通用定时器T1的周期中断来触发AD转换。
ADCINA0和ADCINB0为参考电平，对ADCINA0,ADCINA1，ADCINB0，ADCINB1连续10次采样，然后分别排序，滤波，取均值，用A0，B0的值计算增益和偏移量，用来校正A1，B1
AD sampling frequency of 10K, the sequence generator SEQ1 and SEQ2 cascaded into a 16-channel sequencer,* sampling mode with sequential sampling. The use of general-purpose timer interrupt to trigger the cycle T1 AD converter. ADCINA0 and ADCINB0 the reference level, the ADCINA0, ADCINA1, ADCINB0, ADCINB1 10 consecutive samples, then were sorting, filtering, taking the mean, with A0, B0 the value gain and offset used to correct the A1, B1 (2011-09-02, C++ Builder, 299KB, 下载10次)
[通讯编程] a1.tar
parser in C++~~~~~~~~~~~~
parser in C++~~~~~~~~~~~~ (2009-02-19, C++ Builder, 120KB, 下载7次)
[通讯编程] Clock
MOTO E6手机上的一个桌面时钟，可以始终显示在桌面，我参照QT示例编写并移植到E6上的，非常漂亮。
MOTO E6 a desktop clock on your phone, you can always appear on the desktop, I refer to examples of written and ported to QT on the E6, very beautiful. (2011-04-10, C++ Builder, 83KB, 下载7次)
[通讯编程] ScreenLocker
这个是MOTO E6上用的屏幕锁 我以前编写的，可以自动锁屏，在屏幕上显示时间。
This is MOTO E6 with the screen lock on my previous writing, you can automatically lock screen, the screen time. (2011-04-10, C++ Builder, 109KB, 下载6次)
[matlab编程] primegen-0.97.tar
This program is a Prime number generator. It generates prime numbers very fast. it uses many algorithms to speed up things and uses cache.
This program is a Prime number generator. It generates prime numbers very fast. it uses many algorithms to speed up things and uses cache. (2010-11-20, C++ Builder, 31KB, 下载6次)
[数值算法/人工智能] mcm_assign1
这是美国数学建模大赛97年的a题一种解决方法是利用计算机仿真的方法
This is the American Mathematical Modeling Contest in a 97 that a solution is to use computer simulation methods (2007-05-10, C++ Builder, 4KB, 下载6次)
[RFID编程] Firmware_cv
This file contains code about UHF RFID Using AS399x chip
This file contains code about UHF RFID Using AS399x chip (2014-12-10, C++ Builder, 469KB, 下载6次)
[处理器开发] TLC549
当/CS变为低电平后， TLC549芯片被选中， 同时前次转换结果的最高有效位MSB （A7）自 DATA OUT 端输出，接着要求自 I/O CLOCK 端输入8个外部时钟信号，前7个 I/O CLOCK信号的作用，是配合 TLC549 输出前次转换结果的 A6-A0 位，并为本次转换做准备：在第4个 I/O CLOCK 信号由高至低的跳变之后，片内采样/保持电路对输入模拟量采样开始，第8个 I/O CLOCK 信号的下降沿使片内采样/保持电路进入保持状态并启动 A/D开始转换。转换时间为 36 个系统时钟周期，最大为 17us。直到 A/D转换完成前的这段时间内，TLC549 的控制逻辑要求：或者/CS保持高电平，或者 I/O CLOCK 时钟端保持36个系统时钟周期的低电平。由此可见，在自 TLC549的 I/O CLOCK 端输入8个外部时钟信号期间需要完成以下工作：读入前次A/D转换结果；对本次转换的输入模拟信号采样并保持；启动本次 A/D转换开始。
TLC549 is TI company a low price, high performance 8 bit A/D converter, it takes 8 switched capacitor successive approximation method to realize A/D conversion, the conversion speed is less than 17us, the maximum conversion rate of 40000HZ, 4MHZ typical internal system clock, the power of 3V to 6V. It is convenient to use three line serial interface mode and a variety of microprocessor is connected, a variety of inexpensive measure and control system. (2011-11-15, C++ Builder, 262KB, 下载5次)
[Windows编程] copy
拷贝构造函数的应用
哈工大的实验课作业
注释完整
得分97
copy construction (2010-09-04, C++ Builder, 151KB, 下载4次)