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Description: 以单片机为核心电路。让其在按键的作用下产生0到99的二进制数。用DAC0832作D/A转换器件。转换后的电压作为运放OP0的输入。经过运放扩展得到要求的电压。再经过具有很大电流输出能力的三端稳压lm317扩展电流。使电流也能达到要求的指标。同时单片机还起到驱动显示的功能。-core microprocessor circuit. Giving him the keys to the effects of 0-99 have a binary number. DAC0832 used for the D / A conversion device. After converting the voltage operational amplifier OP0 as input. After the expansion to be operational amplifier voltage requirements. After another great current output capacity of the three-Regulators lm317 current expansion. The electric current will also achieve the required targets. SCM also played while driving showed that the function.
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Size: 6947 |
Author: 万红军 |
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Description: 编写目的
A.由于原先图书管理采用的人工系统，浪费人力物力财力，图书管理系统可以改善这种情况，只需几台电脑和几个管理员老师就能轻松完成。
B.提高对新书入库、借阅图书以及学生信心管理的速度。
C.减少人力的投入的同时提高了信息处理的精度和准确度，在输入无误的基础上可保证数据的正确性。
D.改进了管理服务的质量，可由系统对入库图书进行自动分类、归类，学生信息整理、借阅情况登记入档。
E.改进人员的利用率，减少了前台操作人员，更多的人员可用于为借阅者服务、更新采购图书等其他事物，节省人力资源。
-prepared the original purpose of library management system using artificial waste of manpower, financial and material resources library management system to improve the situation, only a few computer teachers and several administrators can easily completed. B. improve the book stock and borrow books and the students confidence in the management of speed. C. reduce the input at the same time improve the information processing precision and accuracy, In relation to the importation can guarantee that on the basis of the accuracy of the data. D. Improvement of the quality of management services, the storage system can automatically library classification, classification, and student information collated Readers of registrations for stalls. E. improve the efficiency and reduce the prospects of
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Size: 12806 |
Author: 张三 |
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Description: 本系统是以凌阳SPCE061A单片机为核心控制器，具有电流可预置、可步进调整、输出的电流信号和预置的电流信号可同时显示的数控直流电流源。系统主要包括：凌阳SPCE061A精简开发板、键盘与显示电路、压控恒流源电路、校正电路、电源电路等。系统中通过键盘按键对电流值进行预置，凌阳SPCE061A单片机送出相应的数字信号，经过D/A转换、信号放大、电平转换、压控恒流源，再输出所需电流；实际输出的电流经过精密电阻变成取样电压信号，经高输入阻抗放大器、A/D转换器，将信号反馈到凌阳SPCE061A单片机中构成闭环控制；通过液晶显示器显示此信号的值。-the system is Sunplus SPCE061A as the core controller, with current can be preset. Stepping can be adjusted to the current output signal and the current signal Preferences can also showed NC DC current source. Systems include : Sunplus SPCE061A streamline the development board, keyboard and display circuit, voltage-controlled current source circuit, Correction Circuit, and the power circuit. System through the keyboard keys on the current values of Preferences, Sunplus SPCE061A send the corresponding digital signal through D / A conversion, signal amplification, level translators, voltage-controlled current source, and the required output current; actual output current through precision resistor voltage signal into sampling, After high input impedance amplifier, A / D converters. feedback
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Size: 71926 |
Author: panchuxin |
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Description: 在sco unix 下显示前n天或后n天日期的处理程序。
printf(\"功能： 时间戳与时间格式字符串的转换程序\\n\")
printf(\" -i 输入的参数为时间戳\\n\")
printf(\" -s 输入的参数为格式化时间\\n\")
printf(\" -t 输入的时间取当前系统时间\\n\")
printf(\" -x 输入的时间增加或减少的天数\\n\")
printf(\" -o 输出的时间为时间戳格式\\n\")
printf(\" -O 输出的时间为格式化时间\\n\")
printf(\" -h 帮助文件-查看[format]如何定义，例 ％s -h\\n\",procname)
printf(\"范例：1 取当前日期的前2天的时间显示 ％s -t -x -2\\n\",procname)
printf(\" 2 取20030101的前5天的时间显示 ％s -s 20030101000000 -x -5\\n\",procname)
printf(\" 3 取20031231的后5天的时间显示 ％s -s 20031231000000 -x 5 -O \\\"％％D ％％T\\\"\\n\",procname)
printf(\" 4 取时间戳为1089619417的时间显示 ％s -i 1089619417 \\n\",procname)
printf(\" 5 取时间戳为1089619417的后4天的时间显示 ％s -i 1089619417 -x 4 \\n\",procname)
-in sco unix under the former n n days or days after the date of processing. Printf ( "function : timestamp and time format string conversion program \\ n ") printf (" - i input parameters timestamp \\ n ") printf (" - s input parameters for formatting time \\ n ") printf (" - t input the time for the current system time \\ n ") printf (" - x importation time increase or decrease the number of days \\ n " ) printf ( "- o output format for timestamp \\ n") printf ( "- O when output time for the inter-formatted \\ n ") printf (" - h Help-View [format] definition , the cases% s-h \\ n ", procname) printf (" Examples : a date from the current two days before the time displayed% s-t-2 x \\ n ", procname) printf ( &q
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Size: 2560 |
Author: yux |
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Description: 本程序是D.R.J.OWEN主编的FINITE ELEMENTS IN PLASTICITYG一书中第八章例题的有限元程序，网上流传的这个版本的源码里有错误，我把它更正过来了，而且还添加了输入文本。绝对好东西啊。-D. R. J. OWEN editor of the PLA FINITE ELEMENTS IN STICITYG a book of the 8th chapter of excellence finite element program, the spread of the online version of the source there is an error, I see it corrected now, but also added to the input text. Ah absolute good things.
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Size: 62427 |
Author: 栖木 |
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Description: 代入法的启发示搜索
我的代码实现是：按照自然语言各字母出现频率的大小从高到低（已经有人作国统计分析了）先生成一张字母出现频率统计表（A）--------(e),(t,a,o,i,n,s,h,r),(d,l),(c,u,m,w,f,g,y,p,b),(v,k,j,x,q,z)
，再对密文字母计算频率，并按频率从高到低生成一张输入密文字母的统计表（B），通过两张表的对应关系，不断用A中的字母去替换B中的字母，搜索不成功时就回退，在这里回朔是一个关键。
-generation into a search of inspiration I said a code is : According to the Natural Language alphabet frequency of the size of precedence (has been for the State Statistical Analysis), Mr. into an alphabet frequency tables (A )--------( e), (t, a, o, i, n, s, h r), (d, l), (c, u, m, w, f, g, y, p, b), (v, k, j, x, q, z), again close to the mother language calculated frequency and frequency input precedence generate a secret letter to the mother TAB (B), Table 2 by the corresponding relations, use of the letters A to B to replace the letters of the alphabet, when unsuccessful search regression, Here is a retrospective key.
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Size: 3971 |
Author: rtshen |
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Description: JSP 在线投票系统.附加数据库。附加D:\\tomcat5.0\\webapps\\chap17目录下的questionnaire_Data.MDF数据库文件。
运行。启动Tomcat，然后在浏览器中输入http://localhost:8080/chap17/admin/login.jsp。在文本框中输入用户名“admin”，密码“123”。
-JSP online voting system. Additional database. Additional D : \\ tomcat5.0 \\ webapps \\ directory under the chap17 questionnai re_Data.MDF database files. Operation. Start Tomcat, and then in the browser input http://localhost : 8080/chap17/admin/login.jsp. In the text box enter a user name of "admin" and the password "123."
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Size: 823045 |
Author: 刘卡卡 |
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Description: 波兰表达式
以下是几个标准的表达式：
5 * 2 + -3
5 * (2 + -3)
5 + ((-4 * -5) + (((5 + (6 - 2)) * 7 + ((4 + 2) * (3 - 1))))
与之等价的波兰表达式为
+ * 5 2 -3
* 5 + 2 -3
+ 5 + * -4 -5 + * + 5 - 6 2 7 * + 4 2 - 3 1
在普通的表达式中，符号是位于运算对象之间的，而在波兰表达式中，运算符号位于参与运算的对象之前。
波兰式在计算中的价值在于它不需要使用括号，之所以如此是由于波兰式的操作符的先后顺序是明确的。
如果我们用 P 表示波兰表达式，用 O 表示操作符，用 D 表示数字，则可以将波兰表达式定义为 P = O P P 或 P = D。
任务
编写程序计算波兰表达式的值。
输入
输入第一行是一个整数，表示输入文件中共有几个波兰式。
以后每行一个表达式。每个表达式长度不超过 80 个字符，且每个表达式仅包含数字 (取值范围是[-10000, 10000]) 和二元操作符 + 、- 和 * 。任何表达式或子表达式的值都在 [-1000000, 1000000] 范围内。每个数字与操作符之间，至少有一个空格作为间隔。(负号是整数值的一部分，不是操作符，并且和整数值之间没有空格。)
输出
对每个表达式输出其值。 -Poland expression Following is the expression of several criteria : 5 * 2 3 * (2 -3) 5 ((-4 -5 *) (((5 (6-2)) * 7 ((2) * ( 3 - 1)))) which are equivalent to the expression of Poland 5 2 -3 * * 5 2 -3 5 -4 -5 * * 5 - 6 2 7 * 4 2 - 3 1 expression in general, the symbolic object is located between the operator and expression in Poland, Operators participating in Operation symbols at the target before. Poland-in the calculation of the value is that it does not require the use of brackets, The reason is because the Polish operator in the order is clear. If we use P said that Poland expression, O said the operator, and D. said figures Poland can be defined as the expression of P = O or P P P = D. task of preparing Poland calculated the value of the expression. input importation of the first trip
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Size: 1077 |
Author: zheng |
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Description: TLV1544与TMS320VC5402通过串行口连接，此时，A／D转换芯片作为从设备，DSP提供帧同步和输入／输出时钟信号。TLV1544与DSP之间数据交换的时序图如图3所示。
开始时， 为高电平（芯片处于非激活状态），DATA IN和I/OCLK无效，DATAOUT处于高阻状态。当串行接口使CS变低（激活），芯片开始工作，I／OCLK和DATAIN能使DATA OUT不再处于高阻状态。DSP通过I／OCLK引脚提供输入／输出时钟8序列，当由DSP提供的帧同步脉冲到来后，芯片从DATA IN接收4 b通道选择地址，同时从DATAOUT送出的前一次转换的结果，由DSP串行接收。I／OCLK接收DSP送出的输入序列长度为10～16个时钟周期。前4个有效时钟周期，将从DATAIN输入的4 b输入数据装载到输入数据寄存器，选择所需的模拟通道。接下来的6个时钟周期提供模拟输入采样的控制时间。模拟输入的采样在前10个I／O时钟序列后停止。第10个时钟沿（确切的I／O时钟边缘，即上升沿或下降沿，取决于操作的模式选择）将EOC变低，转换开始。
-TLV1544 with TMS320VC5402 through serial port connectivity, at this time, A / D conversion chip as from the equipment, to provide frame synchronization DSP and input / output clock signal. TLV1544 DSP and data exchange between the chronology of the map is shown in figure 3. At the beginning of the margin (in chip-activated), and I DATA IN / OCLK invalid, DATAOUT at high resistance state. When the serial interface CS change low (activator), the chip start work, I / OCLK and DATAIN can DATA OUT is no longer in a state of high resistance. DSP through I / OCLK pin provide input / output clock 8 sequence, When the DSP from the frame synchronization pulse, the chip from the DATA IN receive four channels to choose b address, DATAOUT sent from the same time the previous conversion results from the
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Size: 1207 |
Author: john |
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Description: MSC 1210 A/D Conversion for 1 input signals (-2.5V ... +2.5V)
Inputs pairs AIN0-AIN1 read in an interrupt service routine-MSC 1210 A / D Conversion for an input signals (-2.5V ... 2.5V) Inputs pairs compute-con read in an interrupt service routine
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Size: 2180 |
Author: jacky |
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Description: 操作系统课程设计
资源管理系统
safe() //用银行家算法判定系统是否安全
do{
printf(\"\\n*****输入要进行的操作 1：分配资源 2： 修改资源 3:显示资源 4：离开*****\")
scanf(\"％d\",&choice) -courses on operating system design resource management system safe () / / algorithm with bankers determine whether the safety system do (printf ( "\\ n ***** input to the operation of a : 2 distribution of resources : 3 amend resources : resources four shows : left *****") Scanf ( "% d",
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Size: 2481 |
Author: 叶风 |
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Description: 4.asm…… 响铃程序，输入一个数字字符N，响铃N次。（完成）ysk3.asm ……显示一个星型倒三角。m1.asm ………编程将键盘输入的8位无符号二进制数转化为十六进制数和十进制数，并输出结果form.asm ……采用子程序编程按以下三种格式（██，◣，◥）打印九九乘法表：（完成）char.asm ……小写字母a b c d ……x y z的ASCII码分别为61H 62H 63H 64H……78H 79H 7AH, 而大写字母A B C D ….X Y Z的ASCII码分别为41H 42H 43H 44H …58H 59H 5AH， 使用串处理指令编程从键盘输入16个字符（大小写字母及其它字母均有）， 存入以BUF1开始的一片存储区中，并将其传送到以BUF2开始的一片存储区中， 在传送是将其中的小写字母均改为大写字母，并将第一个小写字母在串中的位置 （距串头BUF1的相对位移量）以十六进制形式输出。（完成）-4.asm ... ... beep procedures, the importation of a number of characters N, N-beep. (Completed) ysk3.asm ... shows a Star inverted triangle. ... ... M1.asm programming to the keyboard input of eight unsigned binary number into a hexadecimal number and decimal number, and the output form.asm ... using subroutine program by the following three formats ( , TT, Short) Print Jiujiuchengfabiao : ( completed) char.asm ... lowercase letters a b c d ... x y z ASCII respectively 61H 62H 63H 64H 78H 79H ... 7AH and capital letters A B C D .... X Y Z ASCII respectively 41H 42H 43H 44H 58H 59H L1.5AH ..., the use of string processing programming instructions from the keyboard 16 characters (letters and other case-sensitive alphanumeric both), credited to BUF1 started a storage area and its transmission
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Size: 2893 |
Author: 冯萍 |
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Description: 提供的程序是供四相步进电机使用。本实验使用的步进电机用直流+12V电压，电机线圈由A、B、C、D四相组成。 2. 驱动方式为四相单四拍方式,各线圈通电顺序如下表。表中首先向A相线圈输入驱动电流，接着向B，C，D线圈通电，最后又返回到A相线圈驱动，按这种顺序轮流切换，电机轴按顺时针方向旋转。若通电顺序相反，则电机轴按逆时针方向旋转。-provide for the procedures of the four-phase stepper motor use. This experiment used the stepper motor 12V DC voltage electrical coils A, B, C, D phase. 2. Four-driven approach to making four single-phase, the coil electrifying sequence the following table. Table A first phase coil input drive current, then to B, C, D coil electricity and eventually return to the A-phase coil driven by the rotation order of this switch, the electrical axis by rotating clockwise direction. If electricity in reverse order, the motor shaft by the anti-clockwise rotation.
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Size: 69496 |
Author: 蔡伟 |
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Description: LCD-7279的经过调试多次已经能够实现在指定的位置显示特定的数据。
程序lcd1.c实现基本的功能，即：键盘输入0、1、2、3、4、5、6、7、8、9、a\\b\\c\\d\\e\\f 在液晶的指定位置显示实现满屏或半屏显示点阵和字符，调入一幅图画的代码进行显示；-LCD-7279 after the repeated testing has been able to achieve the specified location-specific data. Lcd1.c procedures to achieve the basic functions, namely : the keyboard input 0, ,36.2. 8,9. a \\ b \\ c \\ d \\ e \\ f in the designated location LCD realize Manping or semi-Screen Display and characters, transferred to a picture of the code display;
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Size: 1191 |
Author: 吴新明 |
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Description: 我们一个小组，课程大作业中做的简单的输入法，有记忆功能。支持双拼。开发环境visual stdio 2005-a group courses have been operating so simple input method, and memory function. D. support. Visual development environment stdio 2005
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Size: 8712820 |
Author: 冀映辉 |
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Description: 1、 ：编写并调试一个模拟的进程调度程序，采用“最高优先数优先”调度算法对进程进行调度。 “最高优先数优先调度算法的基本思想是把CPU分配给就绪队列中优先数最高的进程。尝试静态优先数与动态优先数两种方法：
a) 静态优先数是指优先数在整个进程运行期间不再改变。优先数可以在数据输入时指定，也可以根据到达顺序、运行时间确定。
b) 动态优先数是指进程的优先数在创建进程时可以给定一个初始值，并且可以按一定原则修改优先数。例如进程获得一次CPU后就将其优先数减少1。或者进程等待的时间超过某一时限时增加其优先数的值。
2、 编写并调试一个模拟的进程调度程序，模拟实现多级反馈队列调度算法。
3、 编写并调试一个模拟的进程调度程序，模拟实现最低松弛度优先算法。
4、 程序与报告要求：
a) 对上述要求1、2、3，至少要完成一项，鼓励尝试多种算法。
b) 输出结果要尽量详细清晰，能够反映调度后队列变化，PCB内部变化。
c) 可以选择在Windows或Linux环境下编写、运行程序
d) 鼓励使用不同的开发工具在不同平台环境上进行开发比较。
e) 在实验报告中，一方面可以对实验结果进行分析，一方面可以对各种算法进行比较，分析它们的优劣，说明各种算法适用于哪些情况下的调度。
-1 : Prepare a simulation and debugging process scheduling procedures, "Priority number of the highest priority" to the process of scheduling algorithms for scheduling. "Highest priority priority scheduling algorithm for the basic idea is to place the CPU allocated to the priority queue highest process. Taste Examination several static and dynamic priority priority number two methods : a) static priority number is priority number in operation during the entire process will not change. priority number in the designated input data, can be reached under the order, running time determine. b) dynamic priority number refers to several priorities for the process of the creation process can be given an initial value, and according to the principle of amending certain priority number.
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Size: 365976 |
Author: huiting_liu |
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Description: 图的遍历的演示（c 语言 数据结构课程设计题）
/*定义图*/
typedef struct{
int V[M]
int R[M][M]
int vexnum
}Graph
/*创建图*/
void creatgraph(Graph *g,int n)
{
int i,j,r1,r2
g->vexnum=n
/*顶点用i表示*/
for(i=1 i<=n i++)
{
g->V[i]=i
}
/*初始化R*/
for(i=1 i<=n i++)
for(j=1 j<=n j++)
{
g->R[i][j]=0
}
/*输入R*/
printf(\"Please input R(0,0 END):\\n\")
scanf(\"％d,％d\",&r1,&r2)
while(r1!=0&&r2!=0)
{
g->R[r1][r2]=1
g->R[r2][r1]=1
scanf(\"％d,％d\",&r1,&r2)
}
}
-graph traversal of the display (c language curriculum design data structures that) / * definition of the map * / typedef s truct V (int int [M] R [M] [M]) int vexnum Graph / * create map * / void creatgraph (Graph * g, int n) (int i, j, r1, r2 g -
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Size: 1405 |
Author: 吉庆 |
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Description: 学会对文件的记录锁定，及解锁。#include <stdio.h>
#include <unistd.h>
#include <fcntl.h>
int main()
{
int fd
int i
struct {
char name[20]
uint ID
int age
} myrec
fd =open(\"name\", O_RDWR|O_CREAT, 0755)
if (fd == -1) return -1
printf(\"Input your name:\") scanf(\"％s\", myrec.name)
printf(\"Inpute your ID :\") scanf(\"％d\", &myrec.ID)
printf(\"Input your age :\") scanf(\"％d\", &myrec.age)
lseek(fd, 0,SEEK_END)
lockf(fd, 1, 0)
write(fd, (void *)&myrec, sizeof(myrec))
lockf(fd, 0 ,0)
return 0
}
执行命令cc lock.c –o lock.out
Chmod +x lock.out
./lock.out
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Size: 9297 |
Author: 华羿 |
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Description: c++冒泡排序#include \"stdio.h\"
void sort()
{
int i,count=0,a[20]
int k=20
printf(\"please input 20 defferent numbers?\\n\")
for(i=0 i<20 i++)
scanf(\" ％d\\a\",&a[i])
while(count<20)
{
for(i=0 i<k i++)
{
if(a[i] > a[i+1])
{
int temp
temp=a[i]
a[i]=a[i+1]
a[i+1]=temp
}
}
count=count+1
}
for(i=0 i<k i++)
printf(\"％d\",a[i])
}
void main()
{
sort()
}
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Size: 623 |
Author: 李仅 |
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Description: Problem D:合唱队形
Time Limit:1000MS Memory Limit:65536K
Total Submit:1237 Accepted:437
Language: not limited
Description
N位同学站成一排，音乐老师要请其中的(N-K)位同学出列，使得剩下的K位同学排成合唱队形。
合唱队形是指这样的一种队形：设K位同学从左到右依次编号为1，2…，K，他们的身高分别为T1，T2，…，TK， 则他们的身高满足T1 < T2 < ...< Ti > Ti+1 > … >TK(1<=i<=K)。
你的任务是，已知所有N位同学的身高，计算最少需要几位同学出列，可以使得剩下的同学排成合唱队形。
Input
输入包含若干个测试用例。
对于每个测试用例，输入第一行是一个整数N(2<=N<=100)，表示同学的总数。第二行有N个整数，用空格分隔，第i个整数Ti(130<=Ti<=230)是第i位同学的身高(厘米)。当输入同学总数N为0时表示输入结束。
Output
对于每个测试案例，输出包括一行，这一行只包含一个整数，就是最少需要几位同学出列。
Sample Input
8
186 186 150 200 160 130 197 220
3
150 130 140
0
Sample Output
4
1
Platform: |
Size: 969 |
Author: 罗伯晓 |
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