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Input-Output

  • Smart Antenna system receiving three white input multipath signals from two sources (3 each) and dis

    Smart Antenna system receiving three white input multipath signals from two sources (3 each) and distinguishing each multipath.

    標簽: receiving multipath Antenna signals

    上傳時間: 2017-09-13

    上傳用戶:youlongjian0

  • This circuit is a nice edge detector that gives you synchronous notification of edges on your input

    This circuit is a nice edge detector that gives you synchronous notification of edges on your input signal. There s no excuse for not doing this it s a tiny circuit in just five lines of Verilog.

    標簽: notification synchronous detector circuit

    上傳時間: 2017-09-18

    上傳用戶:xieguodong1234

  • These codes require an ASCII input file interp.dat of the following form: N: Number of Polynomia

    These codes require an ASCII input file interp.dat of the following form: N: Number of Polynomial Interpolation Points (Small) First Sample (x1,y1) Second Sample (x2,y2) ... Nth Sample (xN,yN) N1: Number of Error Evaluation Points (Large) First Sample (x1,y1) Second Sample (x2,y2) ... N1th Sample (xN1,yN1)

    標簽: Polynomia following require Number

    上傳時間: 2017-09-21

    上傳用戶:許小華

  • Use the on-board potentiometer to simulate an analog input,this input converted into a meaningful di

    Use the on-board potentiometer to simulate an analog input,this input converted into a meaningful digital value in a microcontroller register, and use a group of LED to display this value.

    標簽: input potentiometer meaningful converted

    上傳時間: 2017-09-25

    上傳用戶:iswlkje

  • input method tool using java

    input method tool using java

    標簽: method input using java

    上傳時間: 2013-12-20

    上傳用戶:陽光少年2016

  • lcd output test for Atmega128 toolkit.

    lcd output test for Atmega128 toolkit.

    標簽: toolkit Atmega output test

    上傳時間: 2017-09-27

    上傳用戶:leixinzhuo

  • Linux中Input輸入子系統分析

    Linux中Input輸入子系統分析(pdf)

    標簽: linux

    上傳時間: 2015-04-10

    上傳用戶:zsp26414609

  • 蛇形矩陣(規律)

    題目描述 蛇行矩陣 Problem 蛇形矩陣是由1開始的自然數依次排列成的一個矩陣上三角形。 輸入 Input 本題有多組數據,每組數據由一個正整數N組成。(N不大于100)  輸出 Output 對于每一組數據,輸出一個N行的蛇形矩陣。兩組輸出之間不要額外的空行。  矩陣三角中同一行的數字用一個空格分開。行尾不要多余的空格。  樣例輸入 5 樣例輸出 1 3 6 10 15 2 5 9 14 4 8 13 7 12 11

    標簽: 數字規律 數組 三角形

    上傳時間: 2016-02-29

    上傳用戶:lwol2007

  • 運動會源代碼

    #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

  • AP2406技術手冊

    The AP2406 is a 1.5Mhz constant frequency, slope compensated current mode PWM step-down converter. The device integrates a main switch and a synchronous rectifier for high efficiency without an external Schottky diode. It is ideal for powering portable equipment that runs from a single cell lithium-Ion (Li+) battery. The AP2406 can supply 600mA of load current from a 2.5V to 5.5V input voltage. The output voltage can be regulated as low as 0.6V. The AP2406 can also run at 100% duty cycle for low dropout operation, extending battery life in portable system. Idle mode operation at light loads provides very low output ripple voltage for noise sensitive applications. The AP2406 is offered in a low profile (1mm) 5-pin, thin SOT package, and is available in an adjustable version and fixed output voltage of 1.2V, 1.5V and 1.8V

    標簽: 2406 AP 技術手冊

    上傳時間: 2017-02-23

    上傳用戶:w124141

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