1. Subprogram design
Complete exercise 6.9.
The scores of 10 students are 76, 69, 84, 90, 73, 88, 99, 63100 and 80 respectively. Trial compilation
The number of people with 60-69 points, 70-79 points, 80-89 points, 90-99 points and 100 points is calculated by subprogram respectively
Store in units S6, S7, S8, S9, and S10.

DSEG SEGMENT
GRADE DB 76,69,84,90,73,88,99,63,100,80,100      ;Points stored in GRADE Storage unit
len dw $-GRADE
SX DB 5 DUP(0)                               ;sx It is a fraction temporary accumulation unit
S1 DB 0                                      ;S1 Store 60~69 Number of people in fractions
S2 DB 0                                      ;S2 Store 70~79 Number of people in fractions
S3 DB 0                                      ;S3 Store 80~89 Number of people in fractions
S4 DB 0                                      ;S4 Store 90~99 Number of people in fractions
S5 DB 0                                      ;S5 Deposit 100 points
NOTICE DB 'INPUT A NUMBER',0DH,0AH,'$'
DSEG ENDS
CSEG SEGMENT
ASSUME DS:DSEG,CS:CSEG                      ;Segment and code segment initialization
STA:MOV AX,DSEG
    MOV DS,AX
    MOV DI,OFFSET GRADE                     ;DI Point to the selected fraction unit address                        
    MOV CX,len
    LEA SI,SX                               ;SI For operation of temporary accumulation unit
AGIN:MOV AL,[DI]
    XOR BX,BX
    CALL COMP
    ;MOV AH,02H                                    
    ;MOV DL,BL
    ;ADD DL,30H
   ; INT 21H     ;Display the fraction segment of the temporary unit to be operated on the screen BL Content representation of
    INC BYTE PTR[SI+BX]    ;Add 1 to the unit corresponding to the fraction segment where the score is
    INC DI
    LOOP AGIN
    XOR BX,BX              ;The statistics content of the fraction segment of the temporary accumulation unit
    MOV AL,[SX+BX]
    MOV S1,AL
    INC BX
    MOV AL,[SX+BX]
    MOV S2,AL
    INC BX
    MOV AL,[SX+BX]
    MOV S3,AL
    INC BX
    MOV AL,[SX+BX]
    MOV S4,AL
    INC BX
    MOV AL,[SX+BX]
    MOV S5,AL               ;Deposit separately S1,S2,S3,S4,S5
   ; MOV AH,02H
   ; MOV DL,0AH
   ; INT 21H
   ; MOV AH,09H
   ; LEA DX,NOTICE
  ;  INT 21H
   ; MOV AH,08H             ;Enter the unit number of the fraction segment where you want to display the statistics, 1~5 Between
    ;INT 21H
   ; SUB AL,30H                          
  ;  XOR BX,BX
   ; MOV BL,AL
   ; MOV AH,02H
  ;  MOV DL,[SX+BX-1]
  ;  ADD DL,30H
  ;  INT 21H                ;Display the number of people in the corresponding fraction segment according to the entered unit number
    MOV AH,4CH
    INT 21H
;Entry conditions, AL,BX;AL Store compared scores, BX Return fraction segment value
;Subroutine, which is used to compare the score segment and give the value of the segment to BX
COMP  PROC
    CMP AL,69
    JA C1
    JMP EXI
    C1:CMP AL,79
    JA C2
    INC BX
    JMP EXI
    C2:CMP AL,89
    JA C3
    ADD BX,2
    JMP EXI
 C3:CMP AL,99
    JA C4
    ADD BX,3
    JMP EXI
 C4:ADD BX,4
  EXI:RET
    COMP ENDP
CSEG ENDS
    END STA

2. Recursive subprogram design
Complete exercise 6.13.
Given a positive number N ≥ 1 stored in NUM unit, a recursive subroutine is developed to calculate FIB (N), and
Store the results in the RESULT cell.
Fibonacci number is defined as follows:
FIB(1)= 1
FIB(2)= 1
FIB(n)= FIB(n-2)FIB(n-1) n>2

3 3. Advanced assembly language technology
Complete exercise 7.11.
Try to write a program to complete the following functions: if the length of a string named X is greater than 5, the following refers to
Order will be compiled 10 times.
ADD AX, AX
(requirement: use conditional assembly, generate LST file during compilation, and view macro expansion)

DATAS SEGMENT
    ;Enter segment code here  
    string db '1234567'
    len equ $-string   ;Calculate string length
DATAS ENDS

STACKS SEGMENT
    ;Enter stack segment code here
STACKS ENDS

CODES SEGMENT
    ASSUME CS:CODES,DS:DATAS,SS:STACKS
START:
    MOV AX,DATAS
    MOV DS,AX
    ;Code snippet code here
    
    mov ax,1
    
    IF len GT 5
    REPT 10
    ADD ax,ax
    ENDM
    ENDIF
    
    MOV AH,4CH
    INT 21H
CODES ENDS
    END START

The result is 2^10 D=400H