实验7

野狼/mg

assume cs:codesg
data segment
db '1975','1976','1977','1978','1979','1980','1981','1982','1983'
db '1984','1985','1986','1987','1988','1989','1990','1991','1992'
db '1993','1994','1995'

dd 16,22,382,1356,2390,8000,16000,24486,50065,97479,140417,197514
dd 345980,590827,803530,1183000,1843000,2759000,3753000,4649000,5937000

dw 3,7,9,13,28,38,130,220,476,778,1001,1442,2258,2793,4037,5635,8226
dw 11542,14430,15257,17800
data ends

table segment
db 21 dup ('year summ ne ?? ')
table ends

codesg segment
start:mov ax,data
mov ds,ax
mov ax,table
mov es,ax

mov bx,0
mov si,0
mov di,0
mov cx,21

s0: mov al,[bx]
mov es:[di],al
mov ax,[bx+1]
mov es:[di+1],al
mov al,[bx+2]
mov es:[di+2],al
mov al,[bx+3]
mov es:[di+3],al

mov ax,54h[bx]
mov dx,56h[bx]
mov es:5h[di],ax
mov es:7h[di],dx

mov ax,0a8h[si]
mov es:0ah[di],ax

mov ax,54h[bx]
mov dx,56h[bx]
div word ptr ds:0a8h[si]
mov es:0dh[di],ax

add bx,4
add si,2
add di,16
loop s0

mov ax,4c00h
int 21h
codesg ends

end start

按照鱼老师的思路写的，想不出怎么简化了

莫名其妙

29.s0: mov al,[bx]

30.mov es:[di],al

31.mov ax,[bx+1]

32.mov es:[di+1],al

33.mov al,[bx+2]

34.mov es:[di+2],al

35.mov al,[bx+3]

36.mov es:[di+3],al

可以用mov ax,[bx]
mov es:[di],ax  
mov ax,[bx+2]
mov es:[di+2],ax  
  一次保存两个数据
mov ax,54h[bx]
mov dx,56h[bx]

mov es:5h[di],ax

mov es:7h[di],dx
mov ax,0a8h[si]

mov es:0ah[di],ax
mov ax,54h[bx]

mov dx,56h[bx]

div word ptr ds:0a8h[si]

mov es:0dh[di],ax

1
mov ax,0a8h[si]

mov es:0ah[di],ax
2
mov ax,54h[bx]
mov dx,56h[bx]
mov es:5h[di],ax

mov es:7h[di],dx
3
div word ptr ds:0a8h[si]
mov es:0dh[di],ax
因为已经确定了每个数据存放的位置 可以提前吧人数存放进相应的单元 然后 在把总钱数 存进相应的ax dx中 然后直接进行除法
PS: 楼下有个 写实验7的 可以参考下!~