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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.

    标签: receiving multipath Antenna signals

    上传时间: 2017-09-13

    上传用户: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.

    标签: notification synchronous detector circuit

    上传时间: 2017-09-18

    上传用户: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)

    标签: Polynomia following require Number

    上传时间: 2017-09-21

    上传用户:许小华

  • 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.

    标签: input potentiometer meaningful converted

    上传时间: 2017-09-25

    上传用户:iswlkje

  • input method tool using java

    input method tool using java

    标签: method input using java

    上传时间: 2013-12-20

    上传用户:阳光少年2016

  • lcd output test for Atmega128 toolkit.

    lcd output test for Atmega128 toolkit.

    标签: toolkit Atmega output test

    上传时间: 2017-09-27

    上传用户:leixinzhuo

  • Linux中Input输入子系统分析

    Linux中Input输入子系统分析(pdf)

    标签: linux

    上传时间: 2015-04-10

    上传用户:zsp26414609

  • 蛇形矩阵(规律)

    题目描述 蛇行矩阵 Problem 蛇形矩阵是由1开始的自然数依次排列成的一个矩阵上三角形。 输入 Input 本题有多组数据,每组数据由一个正整数N组成。(N不大于100)  输出 Output 对于每一组数据,输出一个N行的蛇形矩阵。两组输出之间不要额外的空行。  矩阵三角中同一行的数字用一个空格分开。行尾不要多余的空格。  样例输入 5 样例输出 1 3 6 10 15 2 5 9 14 4 8 13 7 12 11

    标签: 数字规律 数组 三角形

    上传时间: 2016-02-29

    上传用户:lwol2007

  • 运动会源代码

    #include <malloc.h>       #include <stdio.h>       #include <stdlib.h>       #include <string.h>       #define NULL 0      #define MaxSize 30          typedef struct athletestruct /*运动员*/     {         char name[20];          int score; /*分数*/         int range; /**/         int item; /*项目*/     }ATH;     typedef struct schoolstruct /*学校*/     {         int count; /*编号*/         int serial; /**/          int menscore; /*男选手分数*/         int womenscore; /*女选手分数*/         int totalscore; /*总分*/         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输入运动项目序号 (x<=%d):",ntsp);         scanf("%d",pafirst);         overgame = *pafirst;         if ( pafirst != phead )         {             for ( pasecond = phead ; pasecond < pafirst ; pasecond ++ )             {                 if ( overgame == *pasecond )                 {                     printf("\n这个项目已经存在请选择其他的数字\n");                     goto Is_Game_DoMain;                 }             }         }         pafirst = pafirst + 1;         if ( overgame > ntsp )         {             printf("\n项目不存在");             printf("\n请重新输入");             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输入序号 of the NO.%d (0<x<=%d): ",i,nsc);                 scanf("%d",&serial);         if ( serial > nsc )          {             printf("\n超过学校数目,请重新输入");             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继续输入运动项目(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号学校的结果成绩:",pfirst->serial);             printf("\n\n项目的数目\t学校的名字\t分数");             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 进入下一页");             getchar();             pfirst = pfirst->next;         }     //  clrscr();          printf("\n运动会结果:\n\n学校编号\t男运动员成绩\t女运动员成绩\t总分");         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按任意建结束");         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("文件已经成功保存\n");     }         void main()     {         system("cls");         printf("\n\t\t\t 运动会分数统计\n");         printf("输入学校数目 (x>= 5):");         scanf("%d",&nsc);          printf("输入男选手的项目(x<=20):");         scanf("%d",&msp);          printf("输入女选手项目(<=20):");         scanf("%d",&wsp);          ntsp = msp + wsp;                  phead = (int *)calloc(ntsp,sizeof(int));         pafirst = phead;         pasecond = phead;         input();         calculate();          output();         Save();     }             

    标签: 源代码

    上传时间: 2016-12-28

    上传用户:150501

  • AP2406技术手册

    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

    标签: 2406 AP 技术手册

    上传时间: 2017-02-23

    上传用户:w124141