刚开始觉得这题目超难...

做了一个多小时没什么头绪.

参考了下论坛上牛人的代码.再自己分析了下.弄懂了.贴出我的代码.我用了2种方法.一种常规的.就是一行一行的覆盖table表.另外一种是用到了堆栈.感觉第二种方法比较好理解.贴出来备忘.

assume cs:code,ss:stack

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
        ;db '1975' = db '1','9','7','5'.每个数字占一个字节.所以一共为21*4=84个字节.
        
        dd 16,22,382,1356,2390,8000,16000,2446,50065,97479,140417,197514
        dd 345980,590827,83530,1183000,1843000,2759000,3753000,4649000,5937000
        ;以上是表示21年公司总收入的21个dword型数据.dd是双字=4个字节.所以一共占21*4=84个字节.
        
        dw 3,7,9,13,28,130,220,476,778,1001,1442,2258,2793,4037,5636,8226
        dw 11542,14430,15257,17800
        ;以上表示21年公司雇员总数的21个word型数据,dw是字=2个字节.所以一共占21*2=42个字节.
        
data ends

table segment

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

stack segment
        db 16 dup (0)

stack ends

code segment

start:
        
                mov ax,stack
                mov ss,ax
                mov sp,10H
                mov ax,data
                mov ds,ax
                mov ax,table
                mov es,ax
                
                mov si,0
                mov di,0
                mov bx,0
                mov cx,21
                
        s: push [si+0]    ;年份前2位入栈
                push [si+2]    ;年份后2位入栈
                push [si+84]   ;收入前2位入栈
                push [si+84+2] ;收入后2位入栈
                push [168+bx]  ;雇员数量入栈
                
                pop es:[di+10] ;雇员数量出栈
                pop es:[di+7]  ;收入后2位入栈
                pop es:[di+5]  ;收入前2位入栈
                pop es:[di+2]  ;年份后2位入栈
                pop es:[di+0]  ;年份前2位入栈
                
                mov ax,es:[di+5] ;高8位放到ax
                mov dx,es:[di+7] ;低8位放到dx
                div WORD ptr es:[di+10] ;除以雇员数量
                mov es:[di+13],ax ;ax是所得的商
                
                mov es:[di+4],20H
                mov es:[di+9],20H
                mov es:[di+12],20H
                mov es:[di+15],20H
                
                add si,4
                add bx,2
                add di,16
                
                loop s
                
                
                
                mov ax,4c00h
                
                int 21h
                
                code ends
                
                
                end start
ps:网上牛人真多...哎.我怎么就没想到用栈呢...