《汇编语言》课程设计1答案

assume cs:code,ds:data,ss:stack,es:table
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'
;以上是表示21年的21个字符串
dd 16,22,382,1356,2390,8000,16000,24486,50065,97479,140417,197514
dd 345980,590827,803530,1183000,1843000,2759000,3753000,4649000,5937000
;以上是表示21年公司总收入的21个dword型数据
dw 3,7,9,13,28,38,130,220,476,778,1001,1442,2258,2793,4037,5635,8226
dw 11542,14430,15257,17800
;以上是表示21年公司雇员人数的21个word型数据
data ends
table segment
db 420 dup(0)
table ends
stack segment
db 160 dup (0)
stack ends
code segment
start:
;--------------先把平均工资，雇员人数，总收入放入table中
mov ax,data
mov ds,ax
mov ax,table
mov es,ax
mov ax,stack
mov ss,ax
mov sp,160
sub bx,bx   ;bx指向data段
sub si,si     ;si指向data段
sub di,di    ;di指向table段
push bx
push si
push di
mov cx,21
s1:
push cx
mov ax,[bx+84] ;ax=总收入低位
mov dx,[bx+86] ;dx=总收入高位
div word ptr [si+168]   ;求平均工资,此时ax为商，dx为余数  此时ax为平均工资（16进制）
mov cx,10
call dtoc_word_divdw     ;执行后，es:di~es:di+5为平均工资十进制
add bx,4
add si,2
add di,20
pop cx
loop s1
pop di
pop si
pop bx
;------------------------------接下来把总收入十进制放进table中
push bx
push di
mov cx,21
s2:
push cx
mov ax,[bx+84]   ;ax=总收入低位
mov dx,[bx+86]   ;dx=总收入高位
mov cx,10
call dtoc_dword_divdw
add bx,4
add di,20
pop cx
loop s2
pop di
pop bx
;------------------------------------接下来把人数十进制放进table
push bx
push di
mov cx,21
s4:
push cx
mov ax,[bx+168]
mov cx,10
call dtoc_word_divdw1
add bx,2
add di,20
pop cx
loop s4
pop di
pop bx
;--------------------------------------接下来把table中的数据放到显示屏中
mov ax,0b800h
mov es,ax
mov cx,21
push bx
push di
s6:
push cx
push di
mov cx,4
s7:
mov al,[bx]
mov ah,2
mov es:[di+160*3],ax
inc bx
add di,2
loop s7
pop di
add di,160
pop cx
loop s6
pop di
pop bx
;-------------------------------------------------------
mov ax,es
mov ds,ax
mov ax,table
mov es,ax
mov cx,21
s8:
push cx
push di
push bx
s9:
mov al,es:[di]
mov ah,2
sub cx,cx
mov cl,al
jcxz ok1
mov [bx+160*3+120],ax
add bx,2
inc di
jmp s9
ok1:
pop bx
pop di
push bx
push di
s10:
mov al,es:[di+6]
mov ah,2
sub cx,cx
mov cl,al
jcxz ok2
mov [bx+160*3+40],ax
add bx,2
inc di
jmp s10
ok2:
pop di
pop bx
push bx
push di
s11:
mov al,es:[di+14]
mov ah,2
sub cx,cx
mov cl,al
jcxz ok3
mov [bx+160*3+80],ax
add bx,2
inc di
jmp s11
ok3:
pop di
pop bx
add bx,160
add di,20
pop cx
loop s8
mov ax,4c00h
int 21h
;--------------------------------------------------------------
dtoc_word_divdw1:
push bx         ;bx当做计数器
push di
sub bx,bx
change2:
sub dx,dx
div cx           ;ax=商，dx=余数
add dx,30h
push dx
inc bx
mov cx,ax
jcxz next
mov cx,10
jmp change2
next:
add cx,bx
s5:
pop ax
mov es:[di+14],ax
inc di
loop s5
pop di
pop bx
ret
;-------------------------------------------------------------
dtoc_dword_divdw:
push si     ;si为计数器
push bx
push di
sub si,si
change1:
push ax      ;存数数据低位
mov ax,dx
sub dx,dx
div cx     ;ax=(h/n)的商，dx=(h/n)的余数
mov bx,ax   ;bx=(h/n)的商
pop ax     ;L=数据低位
div cx    ;ax=[rem(h/n)*65536+L]/n的商，dx=[rem(h/n)*65536+L]/n的余数
add dx,30h
push dx
inc si
mov dx,bx   ;dx:ax=int(h/n)*65536+[rem(h/n)*65536+L]/n
mov cx,dx
jcxz dxis0
mov cx,10
jmp change1
dxis0:
mov cx,ax
jcxz axis0
mov cx,10
jmp change1
axis0:
mov cx,si
s3:
pop ax
mov es:[di+6],ax
inc di
loop s3
pop di
pop bx
pop si
ret
;-------------------------------------------------------------
dtoc_word_divdw:
push bx
push ax
push dx
push di
sub bx,bx    ;bx为计数器
change:
sub dx,dx
div cx       ;ax为商，dx为余数
add dx,30h
push dx
inc bx
mov cx,ax
jcxz ok
mov cx,10
jmp change
ok:
add cx,bx
s:
pop ax
mov es:[di],ax
inc di
loop s
add cx,10
pop di
pop dx
pop ax
pop bx
ret
;-----------------------------------------------------------------
code ends
end start