微机原理 汇编题一道，急求解答，高分求解答，要多少金币我可以追加

;用MASM5.0编译通过

data segment
T DB 45
A DB 0
B DB 0
C DB 0
D DB 0
E DB 0
F DB 0
N DB 45 DUP(5 DUP(0)) ;定义成绩存放单元
AVR DB 45 DUP(0) ;定义平均成绩存放单元
MESS DB 0dh,0ah

DB "Total $"
AA DB " : "
M1 DB " "
M2 DB " ",0DH,0AH,24H

data ends

code segment
main proc far
assume cs:code,ds:data

start:
PUSH DS
XOR AX,AX
PUSH AX

MOV AX,DATA
MOV DS,AX
PUSH DS
POP ES

;以下产生随机成绩
MOV CX,5*45
XOR AL,AL
OUT 43H,AL
IN AL,40H
MOV DI,OFFSET N
NEXT:
MOV BX,11
MOV AH,0
MUL BX
ADD AX,5463
MOV DX,0
MOV BX,50
DIV BX
ADD DL,50
MOV [DI],DL
INC DI
MOV AL,DL
LOOP NEXT

;以下求平均成绩

MOV DI,OFFSET AVR
MOV SI,OFFSET N
MOV CX,45

L0:
PUSH CX

MOV CX,5
MOV AX,0
L1:
MOV BH,0
MOV BL,[SI]
ADD AX,BX
INC SI
LOOP L1
MOV BL,5
DIV BL
MOV [DI],AL

POP CX
INC DI
LOOP L0

;以下统计成绩类型

MOV CX,45
MOV BX,OFFSET AVR

C_NEXT:
MOV AL,[BX]

CMP AL,90
JB C1
INC A
JMP C_END
C1:

CMP AL,80
JB C2
INC B
JMP C_END
C2:

CMP AL,70
JB C3
INC C
JMP C_END
C3:

CMP AL,66
JB C4
INC D
JMP C_END
C4:

CMP AL,60
JB C5
INC E
JMP C_END
C5:

INC F

C_END:
INC BX
LOOP C_NEXT

;以下显示结果
MOV DX,OFFSET MESS
MOV AH,09
INT 21H

MOV AL,T
CALL BIN_D

MOV AL,A
MOV AA,'A'
CALL BIN_D

MOV AL,B
INC AA
CALL BIN_D

MOV AL,C
INC AA
CALL BIN_D

MOV AL,D
INC AA
CALL BIN_D

MOV AL,E
INC AA
CALL BIN_D

MOV AL,F
INC AA
CALL BIN_D

RET

main endp
;=====================
;以下数字-ASCII码转换

BIN_D PROC NEAR
MOV AH,0
MOV BL,10D
DIV BL
CMP AL,0
JE NO10
ADD AX,3030H
MOV M1,AL
MOV M2,AH
JMP DISP11
NO10:
ADD AH,30H
MOV M1,AH
MOV AL,' '
MOV M2,AL

DISP11:
MOV DX,OFFSET AA
MOV AH,09
INT 21H

RET
BIN_D ENDP
;========================

code ends
end start

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