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..
% ?3 N+ E8 C) @3 O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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声明部分
% N# R3 t& _" B$ k//二种声方法均可
; D( w0 l" D. D* |//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’; , w- h" ^! W) r/ x1 n: e
const gamestr : array [0..127] of byte = " `* m1 z+ p/ g1 D* E. F, g
($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); ' e# g0 I, H7 Q9 {
G5 S5 s- r% U代码部分
+ o" ^- k; a( _function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
+ L" \* j9 b& b6 `4 L3 evar
7 E6 ^5 h7 q2 R- F5 Q a1, d1: byte;
$ }1 |- z4 Z& H: j! { j, count, di, si :integer; 8 ^9 \. _. {5 c* y$ [2 ^+ H
begin ( u: v& b. j! }! E. h. u/ u
decode := len div 4 * 3 ; //返回解密后数据长度 ( h2 `0 q. m% o; H( K
j := 0;
3 u1 f$ T. i; t( G7 }7 W8 a while i < len do
, M( {7 s$ j2 r1 o4 Q0 g+ h7 z begin 1 M* R5 h; \. E) W; W! B: V1 S5 p
d1 := byte (inchar[j] ); ; T: R% `! X* K" j$ r6 o4 b# z
if ( d1 = $3B ) or (d1 = $7A) then + T4 b4 h/ [, O8 H3 f/ e6 F1 \7 u: ?1 \6 V
begin n& a# Z! Q! }% |( C
end; ' d: e) M9 h- E. T- X3 S x" u
d1 := d1 and {&content}FF;
8 {5 M* j, v4 H3 b* @! n d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
+ e& `8 A* N" S: A byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 6 G9 W+ m4 W0 Y8 \; \, M6 D5 Y3 Y- R
inc (i);
" u! o5 B+ ] @* u end;
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di := 0 ; 1 x) \& l$ \( L1 n
si := 0 ;
" O& {/ L) \+ T; F1 d; W0 m) e8 K count := len div 4 ; //循环次数 - q. |' w9 e( b1 n
for j := 1 to count do
2 j* v5 m3 X) ^$ y# |4 g! a0 s begin
, X2 V, d9 Q k Z; G' t U a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) 7 N9 F1 a S! F9 e0 o
a1 := a1 shl 2 ;
5 B5 X8 p9 [: U: C9 p* o7 q- h4 G d1 := ord ( inchar [di + 1]);
! t7 W2 i5 B! [; T5 j3 Y- p' s d1 := d1 shr 4 ; 9 V! u& b: C/ G5 H. w) n5 H1 C
a1 := a1 or d1 ; $ [" m+ B- R( l+ o M2 R4 w! W
outchar[si] := chr (a1); + k$ E# B3 l8 d5 f" V& b
. q' `7 D9 }' n* z, C x: S) I" i+ m2 @$ O a1 := ord ( inchar [di + 1]);
5 M$ S* ~3 k. P a1 := a1 shl 4 ; ; `, N) f9 [# U& V& d ~9 I
d1 := ord ( inchar [di + 2]); 0 h! J: R2 M& ~! E5 ?' Q$ s. S0 ^
d1 := d1 shr 2 ; 8 q1 X* \: Y6 f& i3 O2 d& t( ]
a1 := a1 or d1 ;
$ R0 h6 a2 I6 Z' o0 k outchar[si + 1] := chr (a1); 8 O, ^3 \/ F& i7 u! s
+ Q5 U' H0 T* H' q% }8 [ a1 := ord ( inchar [di + 2]);
7 d% ^3 l1 `6 P; E a1 := a1 shl 6 ; / V0 y' \ A# M/ a$ H, z& c! H
d1 := ord ( inchar [di + 3]); / J/ S( M4 x, X. \0 T
a1 := a1 or d1 ;
+ O" n) V# L2 r$ d/ E" q outchar[si + 2] := chr (a1); / j- r" O7 h3 S- x+ K( ~
) Q1 Y& R+ y; j/ A5 e, ?0 u8 i" h9 }
di := di + 4 ;
9 }# D4 j% I! p) k si := si + 3 ;
/ K# Y! \% d! H6 i end;
9 X& d$ x8 V+ y. d7 ~end; & \* ~* P" i9 z, o3 t) T
7 ?9 T8 a" B, |% @function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
4 m: \: W$ h/ F3 M( P2 `1 _var 3 {8 R4 e. [* A" C6 t$ L
a1, d1 :byte;
: r: H2 U2 x `7 z5 Q D j,count,di,si :integer; # V$ t. j3 `6 `$ l; e7 c& ?7 d
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begin 5 q# p) l1 a) s9 e# A ]9 |# i
encode := len div 3 * 4 ; //返回加密后数据长度 8 |4 G( f* `8 R
di := 0 ;
" U& R3 ^8 V8 J( A) J; j si := 0 ;
: v4 ^9 N7 q* C c$ O; f& c count := len div 3; //定义循环次数 % s; V4 x/ a- b5 [/ A
for j := 1 to count do * u( p4 Q' e1 u
begin / O8 `' o5 r9 J7 ~) p0 z( J
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
4 i- N7 C0 [" }# b/ W. i a1 := a1 shr 2 ;
( N1 U6 P( n1 ?3 d6 _ outchar [si] := chr (a1);
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2 q; h F" j5 ^! x c- W a1 := ord (inchar[di]);
. ~7 a5 ~7 ?. {8 p+ Q a1 := a1 and 3 ; v, M9 g$ X1 E3 b
a1 := a1 shl 4 ;
& g* W% m# m Q, D: g d1 := ord (inchar[di +1]);
2 J4 K9 W1 x1 Z% _: A. x! t d1 := d1 shr 4 ;
Q) ?, @: Q& X) ? a1 := a1 or d1 ; . Y, Z5 J$ x/ z+ S0 _1 C, }
outchar[si +1] := chr (a1) ;
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a1 := ord ( inchar[di + 1]) ;
" E, P( X! O2 K( [3 K: { a1 := a1 and {&content}F ;
$ F- ]& x* h3 s! z a1 := a1 shl 2 ; + G9 O. h/ D$ z, {' X0 v0 F
d1 := ord ( inchar[di + 2]) ;
, e0 w; d! K5 Q* T5 z2 O d1 := d1 shr 6 ;
* j" R) W0 Q1 T' a U- ~ a1 := a1 or d1 ;
6 n/ H. p9 {, t7 G9 x" }- Q outchar[si + 2] := chr (a1);
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: W7 n# o- m% n+ B* v a1 := ord ( inchar[di + 2]) ;
! A1 Y: ?6 u! e a1 := a1 and $3F;
9 k( B# j2 D( X: @! M6 \ outchar[si + 3] := chr (a1) ; ; |+ Y9 _% I" Q5 Q& l8 w
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//第二次转换
% v" }! l& \! a$ V( W d1 := ord ( outchar [si]); //第 1 个 3 ]9 P$ U, z0 [+ O" J0 W" B: q7 N
d1 := gamestr [d1];
8 G/ Q5 O. T9 i& f8 x outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个 : R( T# P+ x$ i# R3 _4 j8 ]( @
d1 := gamestr [d1]; - n4 S5 w( L8 \) T. h, w- [; b/ H
outchar [si + 1] := chr (d1);
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; } m8 o/ s9 n; _8 f0 @+ N d1 := ord ( outchar [si + 2 ]); //第 3 个 F4 z/ ?% b. E/ @% u
d1 := gamestr [d1];
5 B# X- \* Y& P* { outchar [si + 2] := chr (d1); / h" J9 A) Q- e+ |$ m- ]* R4 u
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d1 := ord ( outchar [si + 3]); //第 4 个
* j h& p7 M* b d1 := gamestr [d1]; $ W; a" G8 w6 s) z7 Z; t7 Z* N d
outchar [si + 3] := chr (d1);
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di := di + 3 ; , M: o* f" n U5 }1 O3 N8 k W
si := si + 4 ;
+ D9 Z4 v. O) ^$ s1 C2 N: ` end; * F6 g4 D& c6 J( [1 ?+ @
end;
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