王爽汇编第十章课程设计精简设计~（整理）

 assume cs:code
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 11542,11430,15257,17800
data ends
stack segment
dw 16 dup (0)
stack ends
code segment
              start:mov ax,data
                  mov ds,ax
                   mov ax,stack
                   mov ss,ax
                   mov sp,32
                   mov ax,0b800h
                   mov es,ax
                   mov bx,0
                   mov bp,0
                   mov di,0
                   mov cx,21
                s: mov si,10      ;21次loop    注意SI的值为列数一定要清10后以bp为变量每次加160
                push cx     ;保存外层CX值
                   mov cx,4       ;内层年份4次
                   push bx    ;因为loop后bx会变，所以这里先保存
                n1:push [bx]        ;年份  从第10列开始显示    
                   pop es:[bp+si]
                   mov es:1[bp+si],02  ;加色
                   inc bx
                   add si,2
                   loop n1
                   pop bx
                   mov si,40     ;收入从40列开始
                   mov ax,84[bx]
                   mov dx,86[bx]
                   call dtoc
                   mov si,80    ;人数从80列开始
                   mov ax,168[di]
                   mov dx,0
                   call dtoc
                   mov si,120   ;人均收入从120列开始
                   mov ax,84[bx]
                   mov dx,86[bx]
                   mov cx,168[di]
                   call divdw
                   call dtoc
                   add bx,4     ;年份和收入为4字节，每次加4
                   add di,2     ;人数为2字节每次加2
                   add bp,160   ;bp每次加160用来换行
                   pop cx
                   loop s
                   mov ax,4c00h
                   int 21h
               
          divdw: push bx            ;这是大除法子程序dx+ax除以cx后商高为DX低位为AX余数为CX
              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
          dtoc:push bx     ;这是把高16位=dx低16位=ax的数转化为ascll码显示在屏幕上的子程序
                push cx
                  mov bx,0   ;一定要重置为0
            b1:  mov cx,10      ;注意用商来判断一定要先存余数后再判断
               call divdw
                 mov ch,02h ;加上显示的颜色
                 add cl,30h ;加上30h用ASCLL码显示
                 push cx   ;保存加色后的余数到载
                 inc bx  ;用来记录push了几次
                  mov cx,ax;下面是判断商是否为0
                  jcxz jdx
                jmp b1
          jdx:   mov cx,dx
               jcxz ok 
                 jmp b1
        ok:     mov cx,bx ;push几次直接pop 几次
           a1: pop es:[bp+si]   ;这里的bp和si直接来至主程序
                add si,2
                 loop a1
                 pop cx
                 pop bx
                 ret 
code ends
end start