Question

微机原理题

看不懂，求大佬解答

Answer 1

;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