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..
$ Q. @0 [$ Y# _% J( }* L0 Q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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4 A& {& ]/ Z/ X) h$ ~0 ]9 y声明部分
2 I5 k! d/ l! h. H* {//二种声方法均可 3 g1 _, n) }1 H2 G5 f& M1 @' k
//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’;
$ `& {% q& c" T# Qconst gamestr : array [0..127] of byte = 4 x9 x6 y& r8 u4 c6 j! {& `+ u1 F
($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, $1A); 8 i, y+ e! V6 O# Q6 s0 r2 ^
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代码部分
1 ]# P" B$ _+ e& Hfunction decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
5 O, j: p9 F' M7 Dvar
/ s% k5 W9 v q+ k- L a1, d1: byte;
. I2 ?& Z( _) k) n j, count, di, si :integer; ! C& b* L% y* R7 s8 @" e9 V7 k
begin ( [8 i s3 D/ b* Z( s" e
decode := len div 4 * 3 ; //返回解密后数据长度 * d6 Z, Q) i5 ~5 E5 k% J$ P
j := 0; 6 \$ O+ j7 i; M+ T! X
while i < len do
. |& y2 K) j: \% L" n- c1 m a% q begin
' A, L0 }5 \( ~ d1 := byte (inchar[j] );
2 X( H2 l3 n+ k5 V& Y& j) L if ( d1 = $3B ) or (d1 = $7A) then 8 F, H, N9 C6 G; `, I! C
begin
& A& B( _) s: v end; 4 o# R& @ Q6 K g
d1 := d1 and {&content}FF; 9 y6 W3 W) z4 x9 j5 y X) O
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
: Z6 V% U1 \2 [( l3 P/ d byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
& p! R" p0 L. m' a% h inc (i); * ^' J/ P& s. S( R
end;
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5 [* E1 j D4 L/ P di := 0 ;
$ v, ?+ Y" s- b5 O% W( [2 O si := 0 ;
8 s5 v1 @' ?( c count := len div 4 ; //循环次数 % O; K% i/ H) h, F
for j := 1 to count do
( o! F3 P% X) ?5 x+ \; u2 D# } begin ) w) [% z' a W: o9 N. t/ i/ A9 z
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个)
2 _* ?: M: E# C/ q& G% c a1 := a1 shl 2 ;
7 n4 g8 y$ q1 e/ K* W! M t- ^ d1 := ord ( inchar [di + 1]);
5 f* n" J( ~' Y: ?5 s7 T: u d1 := d1 shr 4 ; 8 I) {; r9 M5 F1 _! q$ z5 }
a1 := a1 or d1 ;
5 w$ _5 s* `$ T$ I1 L0 U6 `9 G outchar[si] := chr (a1);
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- k% I& X3 U$ E, v a1 := ord ( inchar [di + 1]);
$ J6 S. p y G7 O9 m2 J a1 := a1 shl 4 ;
% ^4 k! ^. T/ i& R# `( l d1 := ord ( inchar [di + 2]);
. G& {. T. c9 U9 z1 C d1 := d1 shr 2 ;
% W D" H) e/ ?& a) B5 k" ?# h6 u a1 := a1 or d1 ;
" e ]# j, |2 h9 y. W0 G outchar[si + 1] := chr (a1); 0 [. a _( b* W9 @# K1 q1 K
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a1 := ord ( inchar [di + 2]); " \; r: @! R4 a& ~) K$ A. i+ d
a1 := a1 shl 6 ;
/ x: E/ m- B5 s6 y: Q, c/ x( f d1 := ord ( inchar [di + 3]); $ L* u2 p+ `' v* Z x
a1 := a1 or d1 ; ! s4 i5 o3 A3 D/ Y
outchar[si + 2] := chr (a1); \# F" ?) X% U2 C8 l
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di := di + 4 ;
' h: }6 P7 `0 R; S, \3 m4 \$ [! X si := si + 3 ; 4 C* I9 ~$ d+ y* E
end; ) F6 O% d! B; s4 L2 d
end; : W" E; Y+ G, i" a! G5 L
$ B- m* V" k3 w, ifunction encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
8 n& B! \; M6 Lvar
+ G H2 X' r! C' a2 w a1, d1 :byte; . Q& K6 b2 J4 M& Z L, V
j,count,di,si :integer; ) h' A X( J5 L2 J- D
8 D/ e0 r' G. u9 ]# L( h; s/ c, h6 A) ]begin " L! ^( x0 ^( r" x- q
encode := len div 3 * 4 ; //返回加密后数据长度 0 w1 m' N( j, o; `9 R% \6 N) f1 M
di := 0 ;
4 @" Q# b# o! v7 H9 j( R G A2 x$ E si := 0 ;
. H+ R3 z5 x' t# L count := len div 3; //定义循环次数 $ Z4 f8 \7 y# D' I" H) U" p. Y
for j := 1 to count do 7 C2 `! C; r5 Z( V) u
begin
8 Z" E! A( @. J6 z- \4 n; h a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
' w; n6 O% F" E9 D a1 := a1 shr 2 ;
) H( U) }* {: D* B, p outchar [si] := chr (a1); 2 P6 m! j7 E; t3 U- q7 N
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a1 := ord (inchar[di]); + Z( u7 M9 k1 l2 w ~0 J
a1 := a1 and 3 ; ! g: N3 J# v Q( N
a1 := a1 shl 4 ;
7 P @# J' A$ n* i* ]5 E, c) N# z0 ` T d1 := ord (inchar[di +1]); * {$ S2 [+ Y- B2 E' g |0 ?2 e# a% s
d1 := d1 shr 4 ; 5 `$ @$ l, z4 o/ p; J
a1 := a1 or d1 ; 2 t! c. F' [& h1 T# U* P* w3 N: E
outchar[si +1] := chr (a1) ;
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0 q- P- [5 _2 d n; ^& J a1 := ord ( inchar[di + 1]) ;
6 w& A2 g5 Z. m/ f. w; \% Z9 ^- z a1 := a1 and {&content}F ; 3 X8 r0 N8 m: {% ]* {; s
a1 := a1 shl 2 ; 5 t4 l" Q6 o/ y0 w9 a" Y% K
d1 := ord ( inchar[di + 2]) ;
; v* H2 F4 F& G* Q+ D% @* x9 i d1 := d1 shr 6 ;
* s1 S* }$ u" f# {3 ~. Q+ D a1 := a1 or d1 ; 9 y0 p/ Z; p$ K! e' G' c
outchar[si + 2] := chr (a1);
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a1 := ord ( inchar[di + 2]) ;
! \9 d6 H2 H5 u1 n! `) X+ y a1 := a1 and $3F;
7 i4 t) i$ @& t1 A0 ?7 g; i. j outchar[si + 3] := chr (a1) ; ; s- v$ B1 P* s. {$ b
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//第二次转换
$ \3 n, l* m+ H d1 := ord ( outchar [si]); //第 1 个
8 r9 ^& c. R2 Q* G$ o d1 := gamestr [d1]; $ X7 ~+ R% h4 L c
outchar [si] := chr (d1); $ P6 M# R3 G' P1 |+ L
/ ?; f2 v* F5 g- x* ?& F# c2 h d1 := ord ( outchar [si + 1]); //第 2 个
: ^7 \ P! C1 L/ ~( _ d1 := gamestr [d1];
9 K1 X9 t/ v- \. p! N0 G8 Q; e outchar [si + 1] := chr (d1);
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d1 := ord ( outchar [si + 2 ]); //第 3 个
2 }0 P( [' {. t d1 := gamestr [d1];
9 q' i4 a5 g) V( m: Z4 N$ O: _ outchar [si + 2] := chr (d1);
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2 {2 e; ?$ q1 y7 R' X d1 := ord ( outchar [si + 3]); //第 4 个
; M4 ^% O3 b# H2 ?6 r, m2 y d1 := gamestr [d1]; ! q: |/ `5 u9 s: K! [1 ^! i
outchar [si + 3] := chr (d1);
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: _: ]; U. s: W/ T$ f: N di := di + 3 ; " c1 Y" G9 ^. j1 W0 I
si := si + 4 ; : [% R$ A" L4 x+ y2 y
end;
* U, U5 S1 X3 F+ mend;
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