12 2007 档案
摘要：开超高画质的方式? 它的差异在哪?在「我的文件夹」裡有一个 Battlefield2 目录﹙正式版﹚或 Battlefield2 Demo﹙试玩版﹚找到 Profiles/xxxxxx 目录裡的 video.con用笔记本开启它﹙记得要备份，以免发生问题﹚(xxxx 要看你的帐号数量来决定, 离线是 default, 第一个是 001, 依此类推)然后依照下列数值修改︰VideoSettings.setTerrainQuality 4VideoSettings.setGeometryQuality 4VideoSettings.setLightingQuality 3VideoSettings
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摘要：将bcd码转化为二进制数data segmentbcd db 1,2,3,4,5 ;bcd码a dw ? ; 用于存放bcd码转化后的十六进制数data endsstack segment para stack'stack'stapn dw 50 dup(?)stack endscode segmentmain proc farassume ds:data,cs:code,ss:stackstart:;set up stack to retuan ;set DS register to current segment mov ax,data mov ds,ax mov ax,s
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摘要：输入1-12 ,显示月份缩写title home datasg segmentthree db 3monin db 3,4 dup(?)alfmon db '???','$'montab db 'JNA','FEB','MAR','APR','MAY','JUN' db 'JUL','AUG','SEP','OCT','NOV','DEC'datasg endscodesg
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摘要：;程序实现的功能如下：;100字数组的地址为mem，编写的程序删除数组中所有为零的项，;并将后续向前压缩，最后将数组的剩余部分补零。data segmentmem dw 20 dup(0,1, 2, 3, 4) n equ $-datadata endsstack segmentstapn dw 50 dup(?)stack endscode segmentmain proc farassume ds:data,cs:code,ss:stackstart:;set up stack to retuan push ds sub ax,ax push ax ;set DS register to
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摘要：找出A,B中相同的整数,存入C中datarea segmenta dw 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15b dw 2,4,6,8,10,20,21,22,23,24,25,26,27,28,29,1,3,5,7,9c1 dw 15 dup(0)datarea endsprognam segmentmain proc farassume ds:datarea, cs:prognamstart:;set up stack to retuan push ds sub ax,ax push ax;set DS register to current segment
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摘要：输入3-9之间的一个数 生成一个口字:如 4XXXXX XX XXXXXdata segmentN dw ?data endscode segmentmain proc farassume ds:data,cs:codestart:push dsxor ax,axpush axmov ax,datamov ds,axmov ah,1int 21hcmp al,'3'jl exitcmp al,'9'jg exit sub al,30hand ax,0fhmov [N],ax mov ah,2 ;linefeedmov dl,0ah int 21h mov dl,
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摘要：接受键盘命令 T,N T显示生日,N显示名字data segmentbirth db '2926/16/37','$'nm db 'Bill Gates,'$'cmd db 'Please enter the command:','$'linefeed db 13d,10d,'$'data endscode segmentmain proc farassume ds:data,cs:codestart:push dsmov ax,0push axmov ax,datamov ds,axmov
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摘要：人机对话,'What is your name?',将输入存入内存中data segmentpars db 100,?,100 dup(?)mesg db 'What is your name?','$'data endsstack segment para stack 'stack'db 100 dup(?)stack endscode segmentassume cs:code,ds:data,ss:stacksart proc farpush dsmov ax,0push axmov ax,datamov ds,axmov
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摘要：;实训汇编语言以格式“SUNXXXX”显示一段英文中SUN出现的次数。data segmentSTO db 100 dup(?)Tip db 'Please enter a sentence include SUN:',13,10,'$'SUN db 13,10,'SUN'NUM db 4 dup(?),'$'data endsstack segment stack str dw 256 dup(?)stack endscode segmentmain proc farassume ds:data,cs:code,ss:stack
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摘要：比较两个字符串,相同显示Y否则显示Ndata segmentstring1 db 'gaoyufeng'l1 equ $-string1string2 db 'gaoyubing'l2 equ $-string2data endsstack segment para stack'stack' stapn dw 50 dup(?)stack endscode segmentmain proc farassume ds:data,cs:code,ss:stackstart: push dx sub ax,ax push ax mov ax,data
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摘要：;本程序完成功能如下：;设有十个学生的成绩分别为56 69 84 82 73;88 99 63 100 80。编制程序分别统计低于60分，60-90;67-97，80-89，90-99和100分的人数，并存放于s5,s6,s7,s8,s9,s10中。data segmentscore dw 59,69,84,82,73,88,99,63,100,80 s5 dw 0 s6 dw 0 s7 dw 0 s8 dw 0 s9 dw 0 s10 dw 0data endscode segmentmain proc farassume ds:data,cs:codestart: ;set stack f
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摘要：将内存first区多字节10进制数与second区相同10进制数相加,结果保存到dest区DATA SEGMENTFIRST DB 11H, 22H, 33H, 44H, 55H, 66H, 77H, 88H, 99H, 00H,SECOND DB 88H, 77H, 66H, 55H, 44H, 33H, 22H, 11H, 00H, 99H,DEST DB 10 DUP (?)DATA ENDSSTACK SEGMENT PARA STACK 'STACK'STAPN DW 50 DUP(?)TOP EQU LENGTH STAPNSTACK ENDSCOSEG SEGME
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摘要：将area1存储区的16个数据传输到area2NAME LOADDATA SEGMENTAREA1 DB 00H, 01H, 02H, 03H, 04H, 05H, 06H, 07H, DB 08H, 09H, 0AH, 0BH, 0CH, 0DH, 0EH, 0FHAREA2 DB 16 DUP(?)DATA ENDSSTACK SEGMENT PARA STACK 'STACK'STAPN DW 50 DUP (?)TOP EQU LENGTH STAPNSTACK ENDSCODE SEGMENT ASSUME CS: CODE, DS: DATA, ES: DATA,
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摘要：重装系统游戏启动错误解决方法，我知道重装游戏可以，但是这个方法不是我们懒人干的Delta Force Xtreme重装系统以后启动会提示出错，就是常见的非法操作内存错误用记事本写入以下内容，以*.reg后缀名保存，双击保存文件，选择"是"Windows Registry Editor Version 5.00[HKEY_LOCAL_MACHINE/SOFTWARE/NovaLogic/Delta Force: Xtreme/1.00.0000]"CDKIID"="IKJLMNEPCANMLBAAJHIIBBFEJPMDOBJHJHIPIBNLO
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摘要：一个经典算法result love(boy, girl){ if( boy.有房() and boy.有车() ) { boy.set(nothing); return girl.嫁给(boy); } if( girl.愿意等() ) { while(!(boy.赚钱 > 100,000 and girl.感情 > 8 ) ) { for( day=1; day <=365; day++) { if( day == 情人节 ) if( boy.givegirl(玫瑰) ) girl.感情++; else girl.感情--; if( day == girl.生日 ) if(
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摘要：练习实训接受一个小写字母输出对应大写字母data segmenta db '输入一个字符',0dh,0ah,'$'c db '下一个字符',0dh,0ah,'$'data endscseg segmentassume cs:cseg,ds:datastart:mov ax,datamov ds,axmov dx,offset a mov ah,9h int 21hmov ah,1int 21hmov dl,alsub dl,20hmov ah,2int 21hexit:mov ah,4chint 21hcseg endsend s
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摘要：练习实训 接受一个小写字母，找出前个字符和后续字符：data segmenta db '输入一个字符',0dh,0ah,'$'c db '下一个字符',0dh,0ah,'$'data endscseg segmentassume cs:cseg,ds:datastart:mov ax,datamov ds,axmov dx,offset a mov ah,9h int 21hmov ah,1int 21hmov dl,aldec dlmov cx,3L1:mov ah,2int 21hinc dlloop L1exit:mov a
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摘要：练习实训：AX中16位数分为4组（1234分为1,2,3,4），每组4位分别放在AL,BL,CL,DL答:data segmentdata endscseg segmentassume cs:cseg,ds:datastart:mov ax,1234hmov cl,4mov dx,axand dx,0fhmov bx,axshr bx,cland bx,0fhmov ch,blmov bx,axshr bx,clshr bx,cland bx,0fhshr ax,clshr ax,clshr ax,cland ax,0fhmov cl,chand cx,0fhmov bh,blmov dh,d
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摘要：实训内容、程序及操作要领 排序： datasegment memdb10,9,8,7,6,5,4,3,2,1 lendw$-mem dataends codesegment mainprocfar assumecs:code,ds:data start:pushds subax,ax ...
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