《微型计算机原理与接口技术》第二版问题详解张荣标

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DATA SEGMENT

DAT 0,1,8,27,64,-9,-5,6,77,-5,55,88,99,33,55,1,3,5,6,87

SUM DW ?

DATA ENDS

CODE SEGMENT

MAIN PROC FAR

ASSUME CS:CODE,DS:DATA

START:

MOV AX,DATA

MOV DS,AX

MOV AX,0

MOV SUM,AX

MOV SI,OFFSET DAT

MOV CX,20

LP1: ADD AX,[SI]

INC SI

INC SI

LOOP LP1

MOV SUM,AX

MOV AH,4CH

INT 21H

DW

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MAIN ENDP

CODE ENDS

END START

14 编一字程序，计算f（t）=at3+bt2+ct+d。设a，b，c，d，t均为一位十进制数，结果存入RESULT单元。

DATA SEGMENT

AA1 DB 8

BBB DB 5

CCC DB 5

DDD DB 5

TTT DB 2

QQQ DW 3 DUP(?)

RESULT DW ?

DATA ENDS

CODE SEGMENT

MAIN PROC FAR

ASSUME CS:CODE,DS:DATA

START:

MOV AX,DATA

MOV DS,AX

MOV SI, OFFSET QQQ

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MOV BL,TTT

MOV BH,0

MOV AX,BX

MUL BX

MOV [SI],AX ;存t2

MUL BX

MOV [SI+2],AX ; 存t3

MOV AL,AA1

MOV AH,0

MUL WORD PTR[SI+2]

MOV RESULT,AX

MOV AL,BBB

MOV AH,0

MUL WORD PTR[SI]

ADD RESULT,AX

MOV AL,CCC

MUL TTT

ADD RESULT,AX

MOV AL,DDD

MOV AH,0

ADD RESULT,AX

MOV CX,RESULT

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MOV AH,4CH

INT 21H

MAIN ENDP

CODE ENDS

END START

15 求出前20个Fibonacci数，存入数据变量FN开始的区域中。

Fibonacc数的定义为：f0=0，f1=1，fn+2=fn+fn+1

DATA SEGMENT

F1 DW 0

F2 DW 1

FN DW 20 DUP(?)

RESULT DW ?

DATA ENDS

CODE SEGMENT

MAIN PROC FAR

ASSUME CS:CODE,DS:DATA

START:

MOV AX,DATA

MOV DS,AX

MOV CX,18