本人初学反编译，以下程序段如何解释？
ORG0000H
L0:ORLA,R5
ANLA,R2
MOVDPTR,#0003H
INCA

L2:
MOVR7,A
MOVR7,A
CJNER0,#00H,L3
L3:

L9:JCL7
L7:
L4:
L5:
L6:MOVXA,@DPTR
INCR6
DECR7
CJNER2,#0EH,L8
L8:CJNEA,#09H,L9
AJMPL10
;============================
DB01H,4CH,CDH,21H,54H,68H,69H,73H
DB20H,70H,72H,6FH,67H,72H,61H,6DH
DB20H,63H,61H,6EH,6EH,6FH,74H,20H
DB62H,65H,20H,72H,75H,6EH,20H,69H
DB6EH,20H,44H,4FH,53H,20H,6DH,6FH
DB64H,65H,2EH,0DH,0DH,0AH,24H,00H
DB00H,00H,00H,00H,00H,00H,70H,D3H
DB70H,47H,34H,B2H,1EH,14H,34H,B2H
DB1EH,14H,34H,B2H,1EH,14H,4FH,AEH
DB12H,14H,35H,B2H,1EH,14H,B7H,AEH
DB10H,14H,3AH,B2H,1EH,14H,DCH,ADH
DB14H,14H,0AH,B2H,1EH,14H,BCH,AEH
DB0EH,14H,31H,B2H,1EH,14H,56H,ADH
DB0DH,14H,31H,B2H,1EH,14H,34H,B2H
DB1FH,14H,02H,B2H,1EH,14H,DCH,ADH
DB15H,14H,36H,B2H,1EH,14H,52H,69H
DB63H,68H,34H,B2H,1EH,14H,00H,00H
DB00H,00H,00H,00H,00H,00H,00H,00H
DB00H,00H,00H,00H,00H,00H,50H,45H
DB00H,00H,4CH,01H,03H,00H,62H,01H
DBBFH,41H,00H,00H,00H,00H,00H,00H
DB00H,00H,E0H,00H,0FH,01H,0BH,01H
DB06H,00H,00H,70H,00H,00H
;============================
最后DB那一段基本就是EXE文件的头，而程序一开始我就不太明白，累加器A和寄存器逻辑或，2者初始值是多少，哪位能详细解释这段啊，多谢了。。。

[解决办法]
0 是 PE 文件么？

1 侦过壳么？
[解决办法]
PE 是Win32可执行文件的标准格式，任何Win32应用程序(.exe .dll .cpl .ocx ...)都必须符合PE的格式定义。
壳 是现在常用的软件保护手段，先进的加密壳通常具有花指令，多态变形等反反汇编技术，静态反汇编加过壳的程序极难得到正确结果