Compilation of ten small problem

1, a character string input from the keyboard (not longer than 30), counting the number of digital strings Africa, and the statistical results displayed on the screen, implemented EXE format.

data segment
  str db 30,?,30 dup(?);30是限制个数，？存储实际个数
  count db 0
data ends
code segment
  assume cs:code,ds:data
  main proc far
start:
    push ds
    mov ax,0
    push ax

    mov ax,data
    mov ds,ax
    
    lea dx,str
    mov ah,0ah
    int 21h
    
    mov cl,str+1
    
    mov ch,0
    mov si,2
  L3:
    cmp str[si],30h
    jae L1
    inc count
    jmp L2
  L1:
    cmp str[si],39h
    jbe L2
    inc count
  L2:
    inc si
    loop L3

    mov ah,02h
    mov dl,0dh
    int 21H 
    mov ah,02h
    mov dl,0ah
    int 21H 

    lea bx,count
    mov ax,[bx]
或
mov al,count
mov ah,00h
    mov bl,10
    div bl;余数在ah，商在al
    mov dx,ax
    add dx,3030h;转化成十\个位对应的ASCII码
    mov ah,2
    int 21h
    mov dl,dh
    mov ah,2
    int 21h 

    mov ah,4ch
    int 21h

code ends
end start

2, count the number of 16-bit binary number a 1, and the results are displayed in hexadecimal on the screen, with the COM format achieved.

code segment
  org 100h
  assume cs:code
  main proc near
  start:
    mov bx,0a39h
    mov si,0
    mov cx,16
  next:
    shr bx,1
    jnc l1
    inc si
  l1:
    loop next
    mov dx,si
    add dx,30h
    cmp dl,3ah ;看1是否超过十个
    jb l2
    add dl,7
  l2:   
    mov ah,02h
    int 21h
    mov ax,4c00h
    int 21h
    main endp
code ends
end start

3, the keyboard input from a two decimal numbers, and seeking thereof, and outputs the result in decimal form.

code segment
  assume cs:code
start:

  mov ah,01
  int 21h
  mov bl,al;存储在al中
  mov ah,01
  int 21h
  mov bh,al
  sub bx,3030h
  add bl,bh
  
  mov al,bl
  mov ah,00h
  mov bl,10
  div bl
  
  add ax,3030h
  push ax;保护ax中数据
  
  mov dl,al
  mov ah,2
  int 21h
  
  pop ax

  mov dl,ah
  mov ah,2
  int 21h
  
  mov ah,4ch
  int 21h

code ends
end start

4, the keyboard input from a decimal digit, the number corresponding to the number displayed on the screen.

For example, input 3, the screen will display "333."

code segment
assume cs:code
start:
  mov ah,01h
  int 21h
  
  mov bl,al
  
  sub al,30h
  mov cl,al
  mov ch,00h
  
  mov ah,02h
  mov dl,0dh
  int 21h
  
  mov ah,02h
  mov dl,0ah
  int 21h
  
L1:
  mov dl,bl
  mov ah,02h
  int 21h
  loop L1
  
  mov ah,4ch
  int 21h
 
code ends
end start

5, find all odd and less than 100, the variable X stored in the word.

data segment
  x dw ?
data ends
code segment
  assume cs:code,ds:data
  main proc far
start:
    mov ax,data
    mov ds,ax
        mov ax,0
        mov bx,1
        mov cx,50
    L1: add ax,bx
        add bx,2
        loop L1
mov x,ax
mov ah,04h
int 21h
        main endp
code ends
end start

6, the BX number in binary form is displayed on the screen.

code segment
  assume cs:code
  start:
    mov bx,1234h
    mov cx,16
  l1:
    mov dl,30h
    shl bx,1
    jnc l2
    inc dl
  l2:
    mov ah,02h
    int 21h
    loop l1
    
    mov ah,4ch
    int 21h
code ends
end start

7, stored in a byte array X 0 ~ F 16 hexadecimal numbers, these numbers set displayed on the screen in hexadecimal form.

data segment
  x db 0,1,2,3,4,5,6,7,8,9,0ah,0bh,0ch,0dh,0eh,0fh
data ends

code segment
  assume cs:code,ds:data
  start:
    mov ax,data
    mov ds,ax
    
    mov cx,16
    mov si,0
    
  l1:
    mov dl,x[si]
    add dl,30h
    
    cmp dl,39h
    jbe l2
    add dl,7
  l2:
    mov ah,02h
    int 21h
    
    inc si
    loop l1
    
    mov ah,4ch
    int 21h
code ends
end start

8, selected from the byte array contains an array of unsigned 10 in a minimum number stored in the variable MIN, and the number displayed in decimal form.

data segment 
  array db 33,57,65,62,90,69,85,56,53,39
  min db ?
data ends
stack segment
  dw 10 dup(?)
stack ends
code segment 
  assume cs:code,ds:data,ss:stack
  start:
    mov ax,data
    mov ds,ax
    
    mov al,255
    mov cx,10
    mov si,0
    mov di,0
  l1:
    cmp al,array[si]
    jbe l2
    mov al,array[si]
  l2:
    inc si
    loop l1
    
    mov min,al
    mov ah,0

  l3:
    mov bl,10
    div bl
    mov dl,ah
    push dx
    inc di
    mov ah,0
    cmp ax,0
    jz l4
    loop l3
  l4:
    mov cx,di
  
  l5:
    pop dx
    add dl,30h
    
    mov ah,02h
    int 21h
    loop l5
    
    mov ah,4ch
    int 21h
    
code ends
end start

9, the start address is stored is provided in the storage space of a string STRING (a string which has been stored in memory without having to enter, and the string length is not more than 99), the number of characters in the string count of "A", and the results are displayed on the screen.

data segment 
  string db 'aaaaaaaaaasbbbbbcccy'
  len dw $-string
  result db 0
data ends
code segment
  assume cs:code,ds:data
  
  start:
    mov ax,data
    mov ds,ax
    mov di,0
    mov cx,len
    lea si,string
    
  l1:
    lodsb
    cmp al,'a'
    jnz l2
    inc result;直接加即可
  l2:
    loop l1
    
    mov al,result;直接将result传递给al
    mov ah,00h
  l3:
    mov bl,10
    div bl
    mov dl,ah
    push dx
    inc di
    mov ah,0
    cmp ax,0
    jz l4
    loop l3
  l4:
    mov cx,di
  l5:
    pop dx
    add dl,30h
    mov ah,02h
    int 21h
    loop l5
    
    mov ah,4ch
    int 21h
code ends
end start   

10, comparing the two strings of equal length, if the same, then the output Match !, If different, the output No Match!

data segment
  str1 db 'computer'
  len1 dw $-str1
  str2 db 'computer'
  mess1 db 'MATCH$'
  mess2 db 'NO MATCH$'
data ends

code segment
  assume cs:code,ds:data
  start:
    mov ax,data
    mov ds,ax
    mov es,ax
    lea si,str1
    lea di,str2
    cld
    mov cx,len1 
    repe cmpsb
    jz l1
    lea dx,mess2
    jmp l2
  l1:
    lea dx,mess1
  l2:
    mov ah,09h
    int 21h
    
    mov ah,4ch
    int 21h
    
code ends
end start