用MATLAB編寫的一個猜數字游戲 就是隨機生成4位數，你要在有限的步數內猜出。 用法：輸入4位數字后按’ok’鍵，可看到當前結果。 按answer鍵查看正確答案 按reset鍵重新開始 還有很多不成熟的地方，發上來想和大家一起商量商量 主要問題： 1、步數限制功能沒有完成 2、應當在顯示區保留顯示以前的所有步驟，但目前只能顯示當前步驟 3、按鍵次數沒有保留，無法判斷你已經猜了幾次 一個小bug：若輸入3位數，則會自動把第一位置0
上傳時間： 2014-01-13
上傳用戶：王者A
This is svd source code implementation in c++ , the zip file contain the ap file ,more detail ,please see the ap library adapted for c++ in the zip file.also there is faq to answer your question.
標簽： file implementation the contain
上傳時間： 2013-12-27
上傳用戶：aappkkee
#include <malloc.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define NULL 0 #define MaxSize 30 typedef struct athletestruct /*運動員*/ { char name[20]; int score; /*分數*/ int range; /**/ int item; /*項目*/ }ATH; typedef struct schoolstruct /*學校*/ { int count; /*編號*/ int serial; /**/ int menscore; /*男選手分數*/ int womenscore; /*女選手分數*/ int totalscore; /*總分*/ ATH athlete[MaxSize]; /**/ struct schoolstruct *next; }SCH; int nsc,msp,wsp; int ntsp; int i,j; int overgame; int serial,range; int n; SCH *head,*pfirst,*psecond; int *phead=NULL,*pafirst=NULL,*pasecond=NULL; void create(); void input () { char answer; head = (SCH *)malloc(sizeof(SCH)); /**/ head->next = NULL; pfirst = head; answer = 'y'; while ( answer == 'y' ) { Is_Game_DoMain: printf("\nGET Top 5 when odd\nGET Top 3 when even"); printf("\n輸入運動項目序號 (x<=%d):",ntsp); scanf("%d",pafirst); overgame = *pafirst; if ( pafirst != phead ) { for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ ) { if ( overgame == *pasecond ) { printf("\n這個項目已經存在請選擇其他的數字\n"); goto Is_Game_DoMain; } } } pafirst = pafirst + 1; if ( overgame > ntsp ) { printf("\n項目不存在"); printf("\n請重新輸入"); goto Is_Game_DoMain; } switch ( overgame%2 ) { case 0: n = 3;break; case 1: n = 5;break; } for ( i = 1 ; i <= n ; i++ ) { Is_Serial_DoMain: printf("\n輸入序號 of the NO.%d (0<x<=%d): ",i,nsc); scanf("%d",&serial); if ( serial > nsc ) { printf("\n超過學校數目,請重新輸入"); goto Is_Serial_DoMain; } if ( head->next == NULL ) { create(); } psecond = head->next ; while ( psecond != NULL ) { if ( psecond->serial == serial ) { pfirst = psecond; pfirst->count = pfirst->count + 1; goto Store_Data; } else { psecond = psecond->next; } } create(); Store_Data: pfirst->athlete[pfirst->count].item = overgame; pfirst->athlete[pfirst->count].range = i; pfirst->serial = serial; printf("Input name:) : "); scanf("%s",pfirst->athlete[pfirst->count].name); } printf("\n繼續輸入運動項目(y&n)？"); answer = getchar(); printf("\n"); } } void calculate() /**/ { pfirst = head->next; while ( pfirst->next != NULL ) { for (i=1;i<=pfirst->count;i++) { if ( pfirst->athlete[i].item % 2 == 0 ) { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 5;break; case 2:pfirst->athlete[i].score = 3;break; case 3:pfirst->athlete[i].score = 2;break; } } else { switch (pfirst->athlete[i].range) { case 1:pfirst->athlete[i].score = 7;break; case 2:pfirst->athlete[i].score = 5;break; case 3:pfirst->athlete[i].score = 3;break; case 4:pfirst->athlete[i].score = 2;break; case 5:pfirst->athlete[i].score = 1;break; } } if ( pfirst->athlete[i].item <=msp ) { pfirst->menscore = pfirst->menscore + pfirst->athlete[i].score; } else { pfirst->womenscore = pfirst->womenscore + pfirst->athlete[i].score; } } pfirst->totalscore = pfirst->menscore + pfirst->womenscore; pfirst = pfirst->next; } } void output() { pfirst = head->next; psecond = head->next; while ( pfirst->next != NULL ) { // clrscr(); printf("\n第%d號學校的結果成績:",pfirst->serial); printf("\n\n項目的數目\t學校的名字\t分數"); for (i=1;i<=ntsp;i++) { for (j=1;j<=pfirst->count;j++) { if ( pfirst->athlete[j].item == i ) { printf("\n %d\t\t\t\t\t\t%s\n %d",i,pfirst->athlete[j].name,pfirst->athlete[j].score);break; } } } printf("\n\n\n\t\t\t\t\t\t按任意建 進入下一頁"); getchar(); pfirst = pfirst->next; } // clrscr(); printf("\n運動會結果:\n\n學校編號\t男運動員成績\t女運動員成績\t總分"); pfirst = head->next; while ( pfirst->next != NULL ) { printf("\n %d\t\t %d\t\t %d\t\t %d",pfirst->serial,pfirst->menscore,pfirst->womenscore,pfirst->totalscore); pfirst = pfirst->next; } printf("\n\n\n\t\t\t\t\t\t\t按任意建結束"); getchar(); } void create() { pfirst = (struct schoolstruct *)malloc(sizeof(struct schoolstruct)); pfirst->next = head->next ; head->next = pfirst ; pfirst->count = 1; pfirst->menscore = 0; pfirst->womenscore = 0; pfirst->totalscore = 0; } void Save() {FILE *fp; if((fp = fopen("school.dat","wb"))==NULL) {printf("can't open school.dat\n"); fclose(fp); return; } fwrite(pfirst,sizeof(SCH),10,fp); fclose(fp); printf("文件已經成功保存\n"); } void main() { system("cls"); printf("\n\t\t\t 運動會分數統計\n"); printf("輸入學校數目 (x>= 5):"); scanf("%d",&nsc); printf("輸入男選手的項目(x<=20):"); scanf("%d",&msp); printf("輸入女選手項目(<=20):"); scanf("%d",&wsp); ntsp = msp + wsp; phead = (int *)calloc(ntsp,sizeof(int)); pafirst = phead; pasecond = phead; input(); calculate(); output(); Save(); }
標簽： 源代碼
上傳時間： 2016-12-28
上傳用戶：150501
A major societal challenge for the decades to come will be the delivery of effective medical services while at the same time curbing the growing cost of healthcare. It is expected that new concepts-particularly electronically assisted healthcare will provide an answer. This will include new devices, new medical services as well as networking. On the device side, impressive innovation has been made possible by micro- and nanoelectronics or CMOS Integrated Circuits. Even higher accuracy and smaller form factor combined with reduced cost and increased convenience of use are enabled by incorporation of CMOS IC design in the realization of biomedical systems. The compact hearing aid devices and current pacemakers are good examples of how CMOS ICs bring about these new functionalities and services in the medical field. Apart from these existing applications, many researchers are trying to develop new bio-medical solutions such as Artificial Retina, Deep Brain Stimulation, and Wearable Healthcare Systems. These are possible by combining the recent advances of bio-medical technology with low power CMOS IC technology.
上傳時間： 2017-02-06
上傳用戶：linyj
Many times I have been asked to explain “ briefl y ” how SDH, SONET, and the OTN “ exactly ” work. The questions came mainly from new colleagues, stu- dents, and users of these technologies, personally or via the usenet newsgroup comp.dcom.sdh - sonet. I could have referred them to the standards documents, but to provide a more consistent and clear answer I decided to write this pocket guide. The objective of this book is that it can be used both as an introduction as well as a reference guide to these technologies and their spe- cifi c standards documents.
標簽： Generation Transport Optical ComSoc Guide
上傳時間： 2020-05-27
上傳用戶：shancjb
There are few technologies that have had a more profound effect on people’s lives than mobile communications. As recently as twenty years ago no one had a mobile phone, while today 1.4 billion men, women and children depend on them. This now exceeds the number of landline users, where it took the preceding one hundred years to reach the 1 billion mark. The ability to make mobile voice calls turns out to be the answer to a deeply felt need across different cultures who simply want to communicate.
標簽： Communications Wireless Mobile and
上傳時間： 2020-05-30
上傳用戶：shancjb
This effort started as an answer to the numerous questions the authors have repeatedly had to answer about electrostatic discharge (ESD) protection and input/output (1/0) designs. In the past no comprehensive book existed suffi- ciently covering these areas, and these topics were rarely taught in engineering schools. Thus first-time I/O and ESD protection designers have had consider- able trouble getting started. This book is in part an answer to such needs.
上傳時間： 2020-06-05
上傳用戶：shancjb
I would like to thank you for purchasing the second edition of “Industrial Network Security,” especially if you are one of the many supporters of the first edition. When the second edition was announced, many people asked me, “why a second edition?” and even more followed that up with, “and why a coauthor?” These ques- tions are harder to answer than you would think.
標簽： Industrial Security Network
上傳時間： 2020-06-07
上傳用戶：shancjb
We are in the era of ubiquitous computing in which the use and development of Radio Frequency Iden- tification (RFID) is becoming more widespread. RFID systems have three main components: readers, tags, and database. An RFID tag is composed of a small microchip, limited logical functionality, and an antenna. Most common tags are passive and harvest energy from a nearby RFID reader. This energy is used both to energize the chip and send the answer back to the reader request. The tag provides a unique identifier (or an anonymized version of that), which allows the unequivocal identification of the tag holder (i.e. person, animal, or items).
上傳時間： 2020-06-08
上傳用戶：shancjb
搶答器是一種智力競賽常用的器件,搶答器的設計方法千差萬別,文章利用常用的數字電子器件,設計了八路搶答器電路的設計、仿真及實現的全過程,提出兩種可行的設計方案:方案1采用74ls373實現電路鎖存,74ls148實現電路編碼,74ls74及數碼管實現電路顯示;方案二采用CD4511BCN和LMC555CM集成電路及數碼管實現搶答器的控制和顯示。本文設計用的器件簡單,容易理解,適用于初學電子技術的人員。answer scrambler is a common device in intelligence competition, and its design methods vary greatly. This paper designs the whole process of design, simulation and Realization of the circuit of eight-way answer scrambler by using common digital electronic devices, and puts forward two feasible design schemes: scheme 1 uses 74 ls373 to realize circuit latching, 74 ls148 to realize circuit coding,74 ls74 and digital tube to realize circuit. The second scheme uses CD4511 BCN, LMC555 CM integrated circuit and digital tube to control and display the answerer. The device designed in this paper is simple and easy to understand, and it is suitable for the beginners of electronic technology.
標簽： 搶答器
上傳時間： 2022-04-05
上傳用戶：
