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Description: 基于stm32f103的ad7880采集程序-#include"sys.h"
#define CS_AD1 PDout(15)
#define CONVST_AD1 PDout(14)
#define RD_AD1 PDout(13)
#define BUSY_AD1 PDin(12)
#define CS_AD2 PGout(15)
#define CONVST_AD2 PGout(14)
#define RD_AD2 PGout(13)
#define BUSY_AD2 PGin(12)
void IO_init_AD(void)
int Read_dat_AD1(void)
int Convert_AD1(void)
int Read_dat_AD2(void)
int Convert_AD2(void)
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Size: 1024 |
Author: zyang |
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Description: Test float to int extends Dx Test Case Source Code for Linux v2.13.6.
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Size: 1024 |
Author: sulingso |
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Description: 1.写一个程序提示用户键入某一货币数量(以分为单位)，该程序应计算出如何用最少量的1元，50分，10分和1分硬币得到该数目。如195分最少可以用一个1元硬币，1个50分硬币，4个10分硬币和5个1分硬币组成。
注意：可以不用数组。
2.写一个程序可以对两个字符串进行测试，得知第一个字符串是否包含在第二个字符串中。如字符串”PEN”包含在字符串“INDEPENDENT”中。
可以利用字符串类的方法
boolean regionMatches(int toffset, String other, int ooffset, int len)
关于本方法的说明可以在网上（如ORACLED的网站）查到，
(b)改进程序，当第一个字符串包含在第二个字符串中时，显示第一个字符串在第二个字符串中的起始位置。
(c)改进程序，即使当用户输入顺序错误时(第一个是长字符串，第二个是短字符串)，能自动在长字符串中搜寻短字符串。
同花色的13张扑克牌，按照顺序A,2,3……10 J Q K排成一圆圈，从A开始，总是按照同一方向数。每数到第10张，便拿出来，再从下一个开始数，如此继续下去，问：最后剩下的那张扑克牌是哪张?
-1. Write a program that prompts the user to type in a certain currency amount (in divided into unit), the program should calculate the how to use the least amount of 1 yuan, 50 points, 10 points and penny to get the number 1. Such as 195 points can be used at least a $1 COINS, a 50 cent coin, four 10 points of COINS and five 1 cent coin.
Note: you can need not an array.
2. Write a program to test two strings, learned that the first string is included in the second string. Such as the string "PEN" included in the string "INDEPENDENT".
Can the method of using string class
Boolean regionMatches (int toffset, String other, int ooffset, int len)
For an explanation of this method can (such as ORACLED site) on the Internet,
(b) improvement program, when the first string contains the second string, showed the first word
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Author: 杨阳 |
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Description: 程序1】
题目：古典问题：有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月 后每个月又生一对兔子，假如兔子都不死，问每个月的兔子总数为多少？ 1.程序分析： 兔子的规律为数列1,1,2,3,5,8,13,21.... 2.程序源代码： main() { long f1,f2 int i f1=f2=1
for(i=1 i<=20 i++)
{ printf(" 12ld 12ld",f1,f2)
if(i 2==0) printf("\n") /*控制输出，每行四个*/ f1=f1+f2 /*前两个月加起来赋值给第三个月*/ f2=f1+f2 /*前两个月加起来赋值给第三个月*/ } }
-Program 1] Topic: classical problem: a pair of rabbits, from the first three months after birth month, a pair of rabbits are born, grow a small rabbit to the first month and three months after the birth of a pair rabbit, hare, if not death, the total number of rabbits asked how much per month? 1 Program analysis: the law of the rabbit number of columns 1,1,2,3,5,8,13,21 .... 2 source code:. Main () {longf1, f2 inti f1 = f2 = 1 for (i = 1 i < = 20 i++) {printf (" 12ld 12ld" , f1, f2) if (i 2 == 0) printf (" \ n" ) /* Control output per line four*/f1 = f1+f2 /* add up to two months before the assignment to a third Month*/f2 = f1+f2 /* two months before the first three months of assignment to add* /}} =============== ======================
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Author: 刘源 |
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Description: Test rsub int lit8 extends DxTestCase for Linux v2.13.6.
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Size: 1024 |
Author: hequnpao |
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Description: Test rem int extends DxTestCase for Linux v2.13.6.
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Size: 1024 |
Author: lerbuzao |
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Description: Test long to int extends DxTestCase Source Code for Linux v2.13.6.
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Size: 1024 |
Author: fingrengtiu |
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Description: Test shr int extends DxTestCase Source Code for Linux v2.13.6.
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Size: 1024 |
Author: qeisuijen |
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Description: read file tgt int stats for Linux v2.13.6.
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Size: 5120 |
Author: yuexoncao |
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Description: TM1620,2.#include<reg52.h>
13.#include<intrins.h>
14.#define nop _nop_()
15.#define uchar unsigned char
16.#define uint unsigned int -TM1620,AT89S52 2.#include<reg52.h>
13.#include<intrins.h>
14.#define nop _nop_()
15.#define uchar unsigned char
16.#define uint unsigned int
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Size: 2048 |
Author: david hu |
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Description: 1. 编写筛选查找素数函数：
void sieve(bool isPrime[], int n)
其中isPrime[n]为一个布尔型数组，n为数组大小。由于2是第一个素数，所以设置isPrime[0]和isPrime[1]的值为false，并设置其余的元素初值为true。然后对从4到n－1的每一个i，判断i是否能够被2整除，如果i能够被2整除，则设置isPrime[i]为false。对从2到n/2的每一个可能的数值重复以上处理，当操作结束时，所有值为true的isPrime[i]所对应的i就是素数，它们将是从筛子中掉下来的。
测试程序：
输入：一个整数,表示n
输出：1行，1到n之间的所有素数，两个素数之间用空格隔开，最后一个素数输出后，输出一个空格和空行，输出参见函数Disp的代码。
例如，当输入20时，输出如下：
2 3 5 7 11 13 17 19
-1. Preparation of screening to find primes function: void sieve (bool isPrime [], int n) where isPrime [n] is a Boolean array, n is the array size. Because 2 is the first prime number, so set isPrime [0] and isPrime [1] the value is false, and set the rest of the elements of the initial value to true. Then 4 to n-1 every i, i determine whether it can be divisible by 2, if i can be divisible by 2, set isPrime [i] is false. For every possible value 2 to n/2 Repeat the above process, when the end of the operation, all the value of true isPrime [i] corresponding to the i is a prime number, they will be falling the sieve come. Test Procedure: Input: An integer representing the n output: 1 line, all the prime numbers between 1 and n, separated by a space between two prime number, a prime number after the last output, the output of a blank and blank lines, see the output function Disp code. For example, when the input 20, the output is as follows: 235711131719
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Size: 117760 |
Author: 光 |
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Description: cvmx ciu2 en iox int mem for Linux v2.13.6.
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Size: 5120 |
Author: yinvjlp |
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Description: sidus v[箭头]~指标，能很清晰的显示出交易信号，适合初学外汇交易者试用学习。
安装路径：MT4\MQL4\Indicators-
Generated by EX4-TO-MQ4 decompiler V4.0.224.1 []
Website: http://purebeam.biz
E-mail : purebeam@gmail.com
#property copyright Sidus
#property link http://www.metaquotes.net/
#property indicator_chart_window
#property indicator_buffers 4
#property indicator_color1 Yellow
#property indicator_color2 Red
#property indicator_color3 Lime
#property indicator_color4 Fuchsia
extern int FastEMA = 8
extern int SlowEMA = 13
extern int RSIPeriod = 17
extern bool Alerts = FALSE
double g_ibuf_92[]
double g_ibuf_96[]
double g_ibuf_100[]
double g_ibuf_104[]
int gi_108 = 0
int gi_112 = 0
double gd_116 = 0.0
double gd_unused_124 = 0.0
int init() {
SetIndexStyle(0, DRAW_LINE)
SetIndexBuffer(0, g_ibuf_92)
SetIndexStyle(1, DRAW_LINE)
SetIndexBuffer(1, g_ibuf_96)
SetIndexStyle(2, DRAW_ARROW, STYLE_DASH, 2)
SetIndexArrow(2, 233)
SetIndexBuffer(2, g_ibuf_100)
SetIndexEmptyValue(2, 0.0)
SetIndexStyle(3,
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Size: 1024 |
Author: chenxiao |
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Description: struct string { int last, max char p string1, string2 Device Driver for linux v2.13.6.
-struct string { int last, max char p string1, string2 Device Driver for linux v2.13.6.
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Size: 1024 |
Author: puyaivn |
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Description: 10.编写一个程序，从键盘输入一批学生的成绩信息，每个学生的成绩信息包括：学号、姓名以及8门课的成绩。然后按照平均成绩由高到低顺序输出学生的学号、姓名以及平均成绩。
11.把在链表中插入一个新结点的操作写成一个函数：
bool insert(Node *&h,int a,int pos)
其中，h为表头指针，a为要插入的结点的值，pos（≥0）表示插入位置。当pos为0时表示在表头插入；否则，表示在第pos个结点的后面插入。操作成功返回true，否则返回false。
12.把在链表中删除一个结点的操作写成一个函数：
bool remove(Node *&h,int &a, int pos)
其中，h为表头指针，a用于存放删除的结点的值，pos（>0）表示删除结点的位置。操作成功返回true，否则返回false。
13.编写一个程序，首先建立两个集合（从键盘输入集合的元素），然后计算这两个集合的交集、并集以及差集，最后输出计算结果。要求用链表实现集合的表示。
14.名表查找中折半查找的递归函数。-10. Write a program, enter the number of student achievement information the keyboard, each student achievement information includes: Student ID, name, and 8 units results. Then according to the average score of students in descending order of the output number, name and grade point average. 11. The insert a new node in the linked list operation is written as a function: bool insert (Node* & h, int a, int pos) wherein, h is the header pointer, a is the node you want to insert a value, pos (≥0) indicates insertion position. When pos 0 are shown in the table header insert otherwise, indicating that after the first pos-node insertion. Operation successful return true, otherwise returns false. 12. to delete a node in the linked list operation is written as a function: bool remove (Node* & h, int & a, int pos) wherein, h is table pointer, a used to store deleted node value, pos (> 0) indicates the position of deleted nodes. Operation successful return true, otherwise re
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Author: xing |
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Description: 超级刷单王外部参数详解：
extern int Profit = 70;//目标盈利点数，注意该EA按5位数报价的
extern int gi_maxbuy = 6;//买单最大单数， 该EA理论上可以操作的上限为13，但是13单很吓人的！
extern int gi_maxsell = 6;//卖单的……
extern int gi_addstep = 50;//1-6单加码间距基数； 加码间距= 基数×系数的单量次方；不过是否加仓还取决于上一根柱子的阴阳，也就是说实际的间距只要超过这个数就行，不一定等于这个数。
extern int Prosadka = 1500;//7-max 加仓间距基点数参考
extern int Povtor = 1;//下单错误后休眠分钟数
extern bool gi_lot_Personalized = FALSE;//是否人为设定第一单手数；若否，系统自动(warehouse still depends on the Yin and Yang of the previous column, that is to say, the actual spacing only needs to exceed this number, which is not necessarily equal to)
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Size: 100352 |
Author: like8 |
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