匯編:統(tǒng)計輸入字符串內(nèi)的數(shù)字個數(shù)(最多三位數(shù)0—256)

;This is made by wq 
;編程從鍵盤讀入不超過256個字符的字符串;
;編程統(tǒng)計其中數(shù)字的個數(shù)，并將結(jié)果在屏幕上輸出。
;經(jīng)進(jìn)一步完善(能輸出三位數(shù)啦)


DATAS SEGMENT
    sum    db 00h    ;累加器
    string db 256 dup(0dh);存放字符
    tital  db  'This program is made by wq.',0dh,0ah
           db  'It is used to calculater the amout of the number in the string ',0dh,0ah,'$'
    tital1 db  'PLease input the string: ','$'
    overs  db  'Amout of the nember is : ','$'
    tital2 db  'Do you want to have a try again(if yes,input "y" else "n") : ',0dh,0ah, '$'
    tital3 db  'Welcome to use this program again','$'
    kongzi db 0dh,0ah,'$'
    result db 256 dup(?)  
DATAS ENDS
CODES SEGMENT
    ASSUME CS:CODES,DS:DATAS
START:
    MOV AX,DATAS
    MOV DS,AX
  
    mov ax,offset tital;輸出抬頭
    mov dx,ax
    mov ah,09h
    int 21h
main:
       lea dx,kongzi
       mov ah,09h
       int 21h
       lea dx,tital1   ;輸出抬頭1
       mov ah,09h
       int 21h
  
    mov si,0
    mov ax,offset string
    mov dx,ax
again:           ;輸入字符
    mov ah,01h
    int 21h
    cmp al,0dh
    jz then
    mov string[si],al
    inc si
    jmp again    ;輸入完畢
then:          ;傳輸
    mov di,0    
    mov si,0
    mov bl,0
exchang:              ;判斷字符的性質(zhì)
     lea dx,string
     cmp string[si],0dh   
     jz over
     cmp string[si],30h
     jnb next
     inc si
    dec cx
    jmp exchang
next:                     
    mov al,string[si]
    cmp string[si],39h
    jbe calculator
    inc si
    dec cx
    jmp exchang
 
calculator:
     lea dx,result
     mov result[di],al
    inc si
    inc sum
    inc di
    dec cx
    jmp exchang  
over:                   ;處理并輸出數(shù)字個數(shù)
   lea dx,overs
   mov ah,09h
   int 21h
  mov ax,0000h
  mov al,sum[0]         ;判斷是否為三位數(shù)
  cmp al,64h
  jae san
  mov bl,10
  div bl
  mov dx,ax
  add dx,3030h
   mov ah,02h
   int 21h
   mov dl,dh
   mov ah,02h
    int 21h
    jmp a
san: mov bl,10          ;如果是三位數(shù)則多處理一位
     div bl
     mov cl,ah          ;把第一位給cl
     add cl,30h
     mov ah,00h
     div bl              ;第二次除十
     mov dx,ax      
     add dx,3030h
     mov ah,02h
     int 21h
     mov dl,dh
     mov ah,02h
     int 21h
     mov dl,cl
     mov ah,02h
     int 21h   
   
a:  lea dx,kongzi    
    mov ah,09h            ;判斷是否循環(huán)操作
    int 21h
    lea dx,tital2
    mov ah,09h
    int 21h
    mov ah,01h
    int 21h
  
    cmp al,'y'
    je main
    lea dx,tital3
    mov ah,09h
    int 21h
       
    MOV AH,4CH
    INT 21H
CODES ENDS
    END START