//ls_pass:659874532156987116155987532157 //密钥
//as_id:898016005418048405705040871239 //密文
//li_cardlen:30 //密文长度
if len(as_id) <> li_cardlen then
return as_id
end if
li_len = integer(mid(as_id,13,2))
li_total = integer(mid(as_id,11,2))
li_cardlen = li_cardlen - 4
ls_newid = left(as_id,10) + mid(as_id,15)
for li_ifor = li_total to li_cardlen
ls_newpass = ls_newpass + mid(ls_pass,li_ifor,1)
next
li_num = li_total - 1
for li_ifor = 1 to li_num
ls_newpass = ls_newpass + mid(ls_pass,li_ifor,1)
next
for li_ifor = 1 to li_cardlen
ls_temptxt = ls_temptxt + string(mod(integer(mid(ls_newid,li_ifor,1)),10))
next
ls_newid = ""
for li_ifor = 1 to li_cardlen
li_tmp = integer(mid(ls_temptxt,li_ifor,1)) - integer(mid(ls_newpass,li_ifor,1))
if li_tmp < 0 then
li_tmp = 10 + li_tmp
end if
ls_newid = ls_newid + string(li_tmp)
next
ls_result = left(ls_newid,li_len)
return ls_result
这是一段解密算法，如何才能做出一个程序，用来解密手中掌握的密文？密钥是固定的。
目测是basic。