計算機等級考試三級PC技術上機（5）

1、2013年計算機等級考試三級pc技術上機（5）請編制程序，其功能是：內存中從source開始連續(xù)存放著21個八位有符號數(shù)(補碼)，其相鄰兩數(shù)之間差值不超過-8至7。對這種變化緩慢的數(shù)據(jù)可采用差分方法進行壓縮。即第一個數(shù)據(jù)不變，其后的數(shù)據(jù)取與前一數(shù)據(jù)的差值并用四位二進制補碼表示，兩個差值拼成一個字節(jié)，前一個差值放在高四位，后一個差值放在低四位。例如：原數(shù)據(jù)(xn)：23h,27h,2ah,29h,22h壓縮后(yn)：23h, 43h, f9h編程按上述方法進行壓縮，結果保存在result開始的內存單元中。部分程序已給出, 其中原始數(shù)據(jù)由過程load從文件input1.dat中讀入source開

2、始的內存單元中。運算結果要求從result開始存放,由過程save保存到文件output1.dat中。填空begin和end之間已給出的一段源程序使其完整(空白已用橫線標出,每行空白一般只需一條指令, 但采用功能相當?shù)亩鄺l指令亦可),或刪除begin和end之間原有的代碼并自行編程來完成要求的功能。對程序必須進行匯編,并與io.obj鏈接產生prog1.exe執(zhí)行文件,最終運行程序產生結果(無結果或結果不正確者均不得分)。調試中若發(fā)現(xiàn)整個程序中存在錯誤之處, 請加以修改。試題程序：extrn load:far,save:farn equ 10stac segment stackdb 128 d

3、up(?)stac endsdata segmentsource db 2*n+1 dup(?)result db n+1 dup(0)name0 db input1.dat,0name1 db output1.dat,0ndata db n dup(0)pdata db n dup(0)data endscode segmentassume cs:code,ds:data,ss:stacstart proc farpush dsxor ax,axpush axmov ax,datamov ds,axmov es,ax ;置附加段寄存器lea dx,source ;數(shù)據(jù)區(qū)起始地址lea si,

4、name0 ;原始數(shù)據(jù)文件名mov cx,2*n+1 ;字節(jié)數(shù)call load ;從input1.dat中讀取數(shù)據(jù);*begin*lea si,sourcelea di,resultcldmovsb ;y0=x0xor bx,bx ;flag=0mov dx,n*2 ;countercompress:lodsb ;xn-xn-1sub _ ;flag=not flag_j_ low_hexmov cl,4_ al,clmov ah,aljmp nextlow_hex:_or al,ahstosbnext: _jne compress;*end*lea dx,result ;結果數(shù)據(jù)區(qū)首址le

5、a si,name1 ;結果文件名mov cx,n+1 ;字節(jié)數(shù)call save ;保存結果到文件retstart endpcode endsend startc29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795

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