千年加密解密代码....虽然看起来挺复杂的.我是对应汇编的.编译后几乎和游戏客户端里的一样...保证了速度.. 0 F& L+ I* ~& h: G
二次解密代码我还没去搞.不过一次解密足够了.可以制出很多功能外挂.如自动拾取.吃药.等等...二次解密加密主要用于怪物坐标用的.(自动练功需要解析怪物坐标要用)
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0 Y6 a: y) u2 ^: k8 e3 F( }声明部分
) H- A8 X: A" m& g/ b" K//二种声方法均可 ! M- l( s4 r9 `; G! H" T* t' M
//const gamestr:string[255] = ’N>TSVUJlwdcBMFjnAKb?qxvyeGzfLP=_ER@Z\am]ChgoWD;QuX[<tkpr^`iIHOYs.3" (- ’+chr(13)+’<;’+chr(12)+chr(00)+’=/!,1>#2$’+chr(39)+’89%’+chr(10)+chr(9)+’*):5&+67?40’; 7 N3 `0 K* E6 V* N: S; `1 L& G
const gamestr : array [0..127] of byte = 2 \" u4 S8 Z; ]8 Z1 K1 P6 p: ?/ p
($4E, $3E, $54, $53, $56, $55, $4A, $6C, $77, $64, $63, $42, $4D, $46, $6A, $6E, $41, $4B, $62, $3F, $71, $78, $76, $79, $65, $47, $7A, $66, $4C, $50, $3D, $5F, $45, $52, $40, $5A, $5C, $61, $6D, $5D, $43, $68, $67, $6F, $57, $44, $3B, $51, $75, $58, $5B, $3C, $74, $6B, $70, $72, $5E, $60, $69, $49, $48, $4F, $59, $73, $2E, $33, $1E, $01, $13, $22, $10, {&content}B, $28, $2D, $20, {&content}D, $19, $3C, $3B, $06, $11, $1C, {&content}C, {&content}, $3D, $1D, $2F, $21, $03, $02, $05, $04, $2C, $31, $3E, $23, $32, $24, $27, $38, $1F, $39, $25, $12, {&content}A, $09, $18, $1B, $2A, $29, $3A, {&content}E, $35, $07, $26, {&content}F, $2B, $36, $14, $37, $3F, $34, $30, $16, $08, $15, $17, $1A);
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代码部分 ]4 i' k- S& l8 V( ]* `' K
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 # F* x, @ u9 E7 e- ]# [; O; ~
var $ a0 [+ w# R2 O1 w3 N
a1, d1: byte;
0 Y2 \) j0 \4 ^0 j' n j, count, di, si :integer;
# k: L. ^/ S, C, B. h! V Hbegin ) G& C* q3 }# k0 ~7 g0 K* a
decode := len div 4 * 3 ; //返回解密后数据长度 ; N& [3 U& e9 j! v
j := 0; 5 `; ^0 X5 P/ o! o
while i < len do
8 ~: Z) J K8 t# J; G begin
3 }! @1 t, U* A( Y d1 := byte (inchar[j] );
" s. k/ j; A, C' Z' w) x if ( d1 = $3B ) or (d1 = $7A) then % Q# w" o; H4 n+ s0 |! t# i/ u
begin
- v! g; o! }5 T: E8 l. k end;
1 x% ~: \. ^3 T d1 := d1 and {&content}FF;
1 l$ ~ }! g. ?% g d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] ); " w# n4 e1 `, N7 M
byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
4 ]" [7 o( t0 |5 `* V inc (i); 6 i( W8 H2 d( t2 Q1 w+ V
end; * G* T1 T0 h: S
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di := 0 ;
! Z; S- s8 o i1 ^& C si := 0 ; % r* z5 X3 F: U, K5 g/ z
count := len div 4 ; //循环次数 . E3 s0 T+ q5 l0 H5 _. q6 w
for j := 1 to count do % g" U0 y1 i3 a4 _2 m _' V
begin
# M- `/ ~) @% Y2 n* b a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) . W. l/ W, p4 \1 u
a1 := a1 shl 2 ; 2 L2 \3 g) j- b4 J7 D
d1 := ord ( inchar [di + 1]); ! }8 k- j# Q) R$ H& W: e
d1 := d1 shr 4 ; 0 A0 U+ B$ U2 h" Q* G
a1 := a1 or d1 ;
1 b4 N6 Z' a, P outchar[si] := chr (a1);
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a1 := ord ( inchar [di + 1]);
0 V: v; z& u" ^8 }9 O( I a1 := a1 shl 4 ;
2 `& O) F9 o; h; N! k a% G d1 := ord ( inchar [di + 2]); ) v" V$ M6 ^7 T: W8 ], F/ A {
d1 := d1 shr 2 ;
2 N. p: U1 J2 @$ l a1 := a1 or d1 ; 8 t: V3 k4 O+ y P @! Z7 [: S
outchar[si + 1] := chr (a1); 6 S) w, G2 Y- k; q0 L
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a1 := ord ( inchar [di + 2]); ; w r8 _' F c
a1 := a1 shl 6 ; " F- J. W' J5 V0 A$ z5 j+ p
d1 := ord ( inchar [di + 3]);
' t- d5 d' w- A# L) X$ u a1 := a1 or d1 ;
' e; q: ?0 g' h5 H$ } outchar[si + 2] := chr (a1);
( Z; V! z! A4 d2 {
& Q( x) g3 a! d1 ?( U di := di + 4 ; # `8 G% f$ k6 {0 `' _6 m" c0 M
si := si + 3 ;
& y$ E' b/ W. q$ Y0 ^1 l6 | end; $ i; P' r, p- ?! ~
end;
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" w; [3 X, X1 M% q2 [8 i5 Y7 g1 ifunction encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
@+ v2 T8 ?( I; ?var : }) w' b* Y- U1 v
a1, d1 :byte; * P; F( V3 D: u
j,count,di,si :integer;
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' Z0 Y$ V+ l# g. P& \- c* \begin : r# k$ [; }# C q0 {, \
encode := len div 3 * 4 ; //返回加密后数据长度
- L8 I4 m+ b* w" i/ P3 P9 `& X di := 0 ;
0 Y' `$ ^; S2 I; w/ @' i si := 0 ;
4 n2 F# S. z n* Z count := len div 3; //定义循环次数 9 ^- C+ u, A" y) @! b& n
for j := 1 to count do
+ a5 q" z+ B' u, P4 o begin
; ^, t3 o& T# {; _/ L a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
) o9 q& z! Z- O a1 := a1 shr 2 ;
9 M% p/ |7 `/ [6 Q. M. U outchar [si] := chr (a1); & Z3 X$ `2 M; v) f+ b
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a1 := ord (inchar[di]); " z: A+ n6 a( \4 F3 @) \
a1 := a1 and 3 ; 0 a/ j# H7 B9 D6 V
a1 := a1 shl 4 ;
$ O& x6 u+ l. Z* v! |, G8 Z d1 := ord (inchar[di +1]); $ J1 }" I' D9 v/ O1 B/ j8 E% p
d1 := d1 shr 4 ;
+ I$ l" }. n# D3 K' { a1 := a1 or d1 ; ) C4 i+ |( }& z
outchar[si +1] := chr (a1) ;
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( G; s6 n" b- Z6 P. r G a1 := ord ( inchar[di + 1]) ;
+ m x% n: P, W a1 := a1 and {&content}F ; # f8 g( w$ e6 z. @8 h% }/ J; b* h* U# d
a1 := a1 shl 2 ;
3 z9 x% p8 J+ B) S* P d1 := ord ( inchar[di + 2]) ; * N) g% @0 q7 n' _2 S
d1 := d1 shr 6 ;
6 p$ c) S+ }" {& c a1 := a1 or d1 ; 8 A- ~ N: I/ U5 C% i; B
outchar[si + 2] := chr (a1);
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! F7 V, q0 y; u5 J$ { a1 := ord ( inchar[di + 2]) ;
5 ]1 A; i ]9 ~) U a1 := a1 and $3F; ( F4 s6 u( s) R$ v6 N2 G
outchar[si + 3] := chr (a1) ;
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//第二次转换 ( I. f5 A. D5 ^4 `' ^# e
d1 := ord ( outchar [si]); //第 1 个 ) h. C7 ^# A; E! D3 a
d1 := gamestr [d1]; 9 ]* y6 M# B0 q9 T4 |( v2 x* o, E8 q- o
outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个 : Y7 d/ U% A |5 H' ?4 L
d1 := gamestr [d1];
, a1 l0 w# s. v4 ` outchar [si + 1] := chr (d1); 6 f3 ~) m" d9 r6 a8 \8 E5 Y2 v
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d1 := ord ( outchar [si + 2 ]); //第 3 个
8 z f* }% j4 g, ] d1 := gamestr [d1];
3 x$ N' d, }$ T d/ p6 g+ B9 d/ v outchar [si + 2] := chr (d1); 9 x- V8 m8 I6 g) a7 v7 f
) @' s+ E4 [, C% | F d1 := ord ( outchar [si + 3]); //第 4 个
% l. z: }* W% _- `7 S d1 := gamestr [d1]; 2 F q; { G* @0 ]9 |$ k
outchar [si + 3] := chr (d1); 5 F( y) j: ?& o
8 ?7 J8 Z$ s; \1 Q2 Y' S0 Q) b di := di + 3 ; 1 y7 `8 t1 z& N& L7 d% y( V
si := si + 4 ; 2 Q7 O2 Y) h$ I4 {# c7 b- ]& I
end;
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