发布时间 : 星期六 文章《微型计算机原理与接口技术》第二版问题详解张荣标更新完毕开始阅读
文档
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