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发表于 2003-4-27 10:16:09
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becker be1431 的解碼方法
becker be1431的密碼IC 是85c82,3 f |6 r# c# D$ Q1 F7 [7 e! n( t3 b
$26=8a (wait) k- G9 `/ H8 i. g6 S+ R1 f5 E* k8 ]
$1d~$29 是重複三組的密碼, 密碼不能直接讀出.: ^/ c, e, E6 d o; |; x: X) X
IC 內容與密碼的對照如下:( Q, k+ H1 J. X: S
IC----------CODE @1 r1 [% D5 d/ k" z7 N( O
F-----------55 y& k7 s4 _ O: s
E-----------4
8 q Q/ M1 R+ a$ h9 e) R4 VD-----------7$ g9 @0 J& x Q* Z9 K/ a; }# T
C-----------6( I; u9 l1 Z) L8 d& b
B-----------1; F& O, I- m/ Q0 ^: {
A-----------0
9 r! A3 Y9 d3 X6 }8 {8 T a9-----------3
- Y7 r. z& O5 l8-----------24 W$ H3 _& }* V6 I
3-----------9
4 d; E% i9 t" {, Y5 _6 h! i# \# d( o2-----------8
8 Y% ^1 Z |3 H+ c8 j& {將$1d~$29密碼內容改為AA可以變成沒有密碼.
8 ~9 W& }9 Z0 L) ?7 `$ n5 ?# t2 _& h. s5 q5 q! d( ?0 X: ~, N: C( e
實際內容:4 M: K( X* R% n0 ~7 o( Y
becker be1431 85c82 code=8710
1 ]. ?# N2 ` E* [0000: 00 00 00 00 00 00 00 00
2 ]4 c) j. G6 d: G+ o# s" U) m0008: 00 00 00 00 00 00 00 00
9 ^; h: F3 w' {4 ~0010: 00 00 00 00 00 00 00 00
5 ]1 P. g. G7 w2 ]- G5 S0018: 00 00 00 00 00 A2 AD AB s! n2 u0 s! } y; n: j
0020: AA A2 AD AB AA 00 99 A2
- f% J; Q& z# Z% K0028: AD AB AA 00 AA 00 00 00
7 T3 X0 `0 t% Z6 `0030: 84 8A 86 96 8A A4 70 6E% a) Q/ M' t' Q4 Z
0038: 6C 40 86 88 40 96 90 B4 v5 [; K. v/ z& W) Y( m& ~! v3 e
0040: 00 00 00 00 00 00 00 00
3 {* b/ b$ n9 c4 O0048: 00 00 00 00 00 00 00 00
: F# T+ F6 r S5 I$ d0050: 00 00 00 00 00 00 00 00
0 ?0 [8 B7 N. u" Q0058: 00 00 00 00 00 00 00 00
$ g: I. q& f9 |& X& m0060: 00 00 00 00 00 00 00 00
6 u, M' }6 O: v b0068: 00 00 00 00 00 00 00 000 ^( c, \+ k. O6 E: W
0070: 02 08 02 05 02 05 00 001 d0 }1 W* }8 @% q8 S4 c+ O/ b
0078: 00 00 00 00 00 00 00 67! Q9 ]1 A- N( E) o) M
0080: AA AA AA AA AA AA AA AA
7 F% f4 Q* P% ?4 E- S6 ?0088: AA AA AA AA AA AA AA AA: y3 ^- g3 C+ r y M& k! {
0090: 55 55 55 55 55 55 55 552 H" J8 b4 Q; n3 a w3 p1 D" P
0098: 55 55 55 55 55 55 55 557 n6 m8 L9 ]- p2 o* p2 {
00A0: AA AA AA AA AA AA AA AA3 F$ ?4 ?' v/ h6 {% w
00A8: AA AA AA AA AA AA AA AA
6 x4 u5 [& ?( Q( ~2 }8 B00B0: 55 55 55 55 55 55 55 55
# v1 t# n5 u+ a- u2 [9 W00B8: 55 55 55 55 55 55 55 55
5 h' z1 q& X. C- m' g00C0: AA AA AA AA AA AA AA AA
! D: X+ [$ J& ^6 q/ L I00C8: AA AA AA AA AA AA AA AA7 z7 Q, O. R8 c3 T; k Z
00D0: 55 55 55 55 55 55 55 559 }" ?* ?3 M5 x/ u: d$ M1 w
00D8: 55 55 55 55 55 55 55 554 I' ~) L2 @1 ~ M$ f3 P7 f
00E0: AA AA AA AA AA AA AA AA
& Z l- C2 ~ F& ^8 G00E8: AA AA AA AA AA AA AA AA- e* D; _; i7 l" z& ?$ ]
00F0: 55 55 55 55 55 55 55 55
4 o$ r0 A7 s, y00F8: 55 55 55 55 55 55 55 55 |
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