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注册日期：2019-05-08 16:38 |
assume cs:codesg,ds:data,ss:stack
data segment
dw 1975,1976,1977,1978,1979,1980,1981,1982,1983,1984,1985,1986,1987,1988,1989,1990,1991,1992,1993,1994,1995,0
dd 16,22,382,1356,2390,8000,16000,24486,50065,97479,140417,197514
dd 345980,590827,803530,1183000,1843000,2759000,3753000,4649000,5937000,0
dw 3,7,9,13,28,38,130,220,476,778,1001,1442,2258,2793,4037,5635,8226,11542,14430,15257,17800,0;以上是表示21年公司雇员人数的21个word型数据
dw 5,3,42,104,85,210,123,111,105,125,140,136,153,211,199,209,224,239,260,304,333,0
db 200 dup(0)
data ends
stack segment
db 96 dup(0)
stack ends
codesg segment
start:
mov bx,stack
mov ss,bx
mov sp,60h
mov bx,data
mov ds,bx
mov si,0
mov bx,0
mov dx,0
mov cx,0
mov ax,0
call dtoc
mov bp,0
mov si,0
mov dh,1 ;行号
s1: mov dl,1 ;列号
mov cl,2 ;颜色
call show_str
mov cx,21 ;排列21个数
inc bp
sub cx,bp
jcxz s2
inc dh
jmp s1
s2:
mov bx,44
call dtocm
mov bp,0
mov si,0
mov dh,1 ;行号
s3: mov dl,11 ;列号
mov cl,2 ;颜色
call show_str
mov cx,21 ;排列21个数
inc bp
sub cx,bp
jcxz s4
inc dh
jmp s3
s4:
mov bx,132
call dtoc
mov bp,0
mov si,0
mov dh,1 ;行号
s5: mov dl,21 ;列号
mov cl,2 ;颜色
call show_str
mov cx,21 ;排列21个数
inc bp
sub cx,bp
jcxz s6
inc dh
jmp s5
s6:
mov bx,176
call dtoc
mov bp,0
mov si,0
mov dh,1 ;行号
s7: mov dl,31 ;列号
mov cl,2 ;颜色
call show_str
mov cx,21 ;排列21个数
inc bp
sub cx,bp
jcxz s8
inc dh
jmp s7
s8:
mov ax,4c00h
int 21h
dtocm :
push si
push cx
mov si,0
mov dx,0
mov cx,0
mov ax,0
mov bp,0 ;初始化
dtoc2m :
mov ax,ds:[bx] ;把第一个数低位付给ax
mov dx,ds:[bx+2] ;把第一个数高位位付给dx
dtoc3m : mov cx,10 ;除数为10
call divdw
add cl,30h ;余数低位+30h输出数字
push cx ;存入数字
mov cx,ax
jcxz dt1m ;把商的低位给cx,为0则跳dt1,否则继续
inc bp ;bp记录有多少个数字
jmp dtoc3m ;跳dtoc3不断转换数字
dt1m:
mov cx,dx ;把商的高位给dx，为0则跳OK2，否则继续
jcxz ok2m
inc bp ;bp记录有多少个数字
jmp dtoc3m ;跳dtoc3不断转换数字
ok2m: inc bp ;bp+1才是转换的数字的字节
ok3m:
pop cx ;取出存入的cx值，倒着读余数
mov byte ptr [si+140h],cl ;cx低字节，余数存入ds:10h地址
sub bp,1 ;bp-1少了一个转换字节
mov cx,bp ;bp赋给cx
jcxz ok4m ;转换完则跳ok4，否则继续执行
inc si ;下个地址
jmp ok3m
ok4m:
inc si
mov byte ptr [si+140h],20h ;空一格
inc bx
inc bx
inc bx
inc bx
mov cx,[bx] ;第二个dd数低位看是否为零，
jcxz ok5m ;等于零则跳ok5,不等则继续读下个数
inc si
mov bp,0 ;bp清零，si+1存入下个字节
jmp dtoc2m
ok5m:
mov cx,[bx+2] ;第二个dd数高位看是否为零，
jcxz ok6m ;等于零则跳ok6,不等则继续读下个数
inc si
mov bp,0 ;bp清零，si+1存入下个字节
jmp dtoc2m
ok6m: pop cx
pop si
ret
dtoc :
push si
push cx
mov si,0
mov dx,0
mov cx,0
mov ax,0
mov bp,0
dtoc2 :
mov ax,ds:[bx]
mov dx,0
dtoc3 : mov cx,10
call divdw
add cl,30h
push cx
mov cx,ax
jcxz ok2
inc bp
jmp dtoc3
ok2: inc bp
ok3:
pop cx
mov byte ptr [si+140h],cl
sub bp,1
mov cx,bp
jcxz ok4
inc si
jmp ok3
ok4:
inc si
mov byte ptr [si+140h],20h
inc bx
inc bx
mov cx,[bx]
jcxz ok5
inc si
mov bp,0
jmp dtoc2
ok5: pop cx
pop si
ret
divdw: push bx
push ax
mov ax,dx
mov dx,0
div cx
mov bx,ax
pop ax
div cx
mov cx,dx
mov dx,bx
pop bx
ret
show_str:
push cx ;保存cx值
mov bx,0
mov ax,0B800H
mov es,ax
mov al,dh
sub al,1 ;al=(dh-1)
mov bl,0A0H
mul bl ;ax=(dh-1)*A0
add bx,ax ;BX=(dh-1)*A0
push bx ;存bx值
mov al,dl
sub al,1 ;al=(dl-1)
mov bl,2
mul bl ;ax=(dl-1)*2
pop bx ;取出第一个bx值
add bx,ax ;bx=(dh-1)*A0+(dl-1)*2
show_str2:
mov cl,[si+140h] ;把ds:[si+60h]赋给cl
sub cl,20h
mov ch,0
jcxz ok ;cx=0则结束转换，否则继续执行
mov al,[si+140h] ;把ds:[si+60h]赋给al
mov ah,0
mov es:[bx],al ;把al的值存到显存中
pop cx ;取出cx值
mov ch,0
mov byte ptr es:[bx+1],cl ;cl值赋予颜色
push cx ;保存cx值
inc bx
inc bx ;指向显存下个地址
inc si ;下个要转换的字节
jmp show_str2
ok:
inc si ;跳过空格
pop cx
ret
codesg ends
end start | | |
