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我在旧网站上发现了这个代码（无法访问所有者），它计算用户输入数字的阶乘（最高255），并且它工作正常。我的问题是，我可以弄清算法是如何工作的，但我不明白。如果有人会告诉我它如何在人类语言中起作用，我将不胜感激。这段代码的算法是如何工作的？
data segment
b1 db 512 dup(0)
b2 db 512 dup(0)
msg db 0dh,0ah,"Number:$"
fac db 0dh,0ah,"Factorial",0dh,0ah,"$"
data ends
bignum_get macro var
local exit,get,loo,skip
xor cx,cx
lea si,var
mov di,si
mov ah,01h
get:
int 21h
cmp al,0dh
je exit
sub al,30h
mov [si],al
inc si
inc cx
jmp get
exit:
shr cx,1
jz skip
dec si
loo:
mov ah,[si]
mov al,[di]
mov [si],al
mov [di],ah
dec si
inc di
loop loo
skip:
endm
geti macro
local exit,get
xor bx,bx
mov dx,bx
mov cx,00ah
get:
push bx
;Read character
mov ah,01h
int 21h
xor ah,ah
pop bx
;If enter stop
cmp al,0dh
jz exit
sub al,30h
;Multply By 10
push ax
mov ax,bx
mul cx
mov bx,ax
pop ax
;add
add bx,ax
;redo
jmp get
exit:
mov ax,bx
endm
bignum_put macro var
local put,en,nxt,exit
lea si,var
mov di,si
mov cx,0200h
add di,cx
dec di
en:
cmp BYTE PTR [di],00h
jne nxt
dec di
jmp en
nxt:
mov ah,02h
put:
mov dl,[di]
add dl,30h
int 21h
cmp si,di
je exit
dec di
loop put
exit:
endm
bignum_add macro b1,b2
local ader,exit
lea si,b1
lea di,b2
mov cx,0200h
mov bl,10
ader:
mov al,[di]
mov ah,[si]
add al,ah
jz exit
xor ah,ah
div bl
mov [si],ah
inc si
add [si],al
inc di
loop ader
exit:
endm
bignum_mul macro b1
local loo,exit
cmp dl,01h
jz exit
lea si,b1
mov dh,10
xor bx,bx
mov cx,0200h
loo:
mov al,[si]
xor ah,ah
mul dl
add ax,bx
div dh
mov [si],ah
inc si
mov bl,al
loop loo
exit:
endm
puts macro msg
push ax
push dx
mov dx,offset msg
mov ah,09h
int 21h
pop dx
pop ax
endm
assume cs:code,ds:data
code segment
start:
mov ax,data
mov ds,ax
lea si,b1
mov BYTE PTR [si],01h
puts msg
geti
mov dl,al
loo:
push dx
bignum_mul b1
pop dx
dec dl
jnz loo
puts fac
bignum_put b1
mov ah,4ch
int 21h
code ends
end start