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.. ; g! `7 n3 ^+ B, i" \; E; K3 K7 h M
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) + F; q# v% A& R% P ^
i$ Q2 w# e% Z6 w! w) J2 }声明部分
' H8 [; k0 {1 w8 E: \9 w//二种声方法均可 : d! y2 [# C3 G
//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’;
k2 z/ f3 h' a2 @9 B* w+ |const gamestr : array [0..127] of byte = O. ~& B' d& g& V7 a% Z' n
($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, $1A);
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5 u3 [& k/ {, f9 A代码部分
! _; H$ u; ~# H3 H* n) T- T- r. Ffunction decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
* [1 J" L3 w' G. `7 v' ^+ S0 h8 Gvar - W1 t/ x0 [3 W8 D
a1, d1: byte;
4 e$ \% x8 y$ W2 G' }6 J j, count, di, si :integer;
1 i# H9 w* A" ?; M, x+ L4 Abegin
+ K/ l) T) j, O9 Y0 C decode := len div 4 * 3 ; //返回解密后数据长度 5 t, [/ O/ v; R* S5 n8 c
j := 0; ! t P, w$ K- g% t* o' P6 b
while i < len do $ x; e+ g, Y7 }% M N" s
begin
) k8 q( t# ^, z$ b0 R' T d1 := byte (inchar[j] );
/ Y. x7 F3 r \# }% H7 p p if ( d1 = $3B ) or (d1 = $7A) then 3 n$ ]3 N) d, h2 ^
begin 7 X+ E/ K' K: O0 }
end;
4 ~: o/ `2 K+ n: J d1 := d1 and {&content}FF; ' u2 n2 y, S- H; @1 P
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
$ B5 H7 A7 t+ H$ U+ I' ]& E# }) ` byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
: R# O, b4 V/ N ] inc (i);
! {( y5 v: ]( H ~ end;
5 s6 f8 i7 Z$ o# x) t" [( ~ Y. ^0 @1 _
di := 0 ;
. |& k7 h6 a/ S0 ]' ~! W9 a si := 0 ;
5 H' B% O3 d8 A* j' m count := len div 4 ; //循环次数
$ V3 [* z: ]& z' C, Q7 o for j := 1 to count do
( T: \: K5 r6 C( k1 V, ?( _ begin 5 I7 g/ M f0 ?) P5 k
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个)
- T+ w+ y# X2 r7 q- } a1 := a1 shl 2 ;
7 v# L# r; P5 z' b6 C' F# [1 r1 U0 t; D d1 := ord ( inchar [di + 1]); ' b, u0 c. S- ]
d1 := d1 shr 4 ; 3 p. B9 y5 B+ [1 h7 F' Z- X6 _# B0 {
a1 := a1 or d1 ; 8 X6 P* Q9 Q6 ?* J% ?+ K9 C
outchar[si] := chr (a1);
" U2 {8 @+ i2 u; N j; o* L" d0 i2 i" a
a1 := ord ( inchar [di + 1]); ) O3 M* V1 Z7 o$ J7 q% x
a1 := a1 shl 4 ; ( e* p$ E- v, k( S
d1 := ord ( inchar [di + 2]);
* N# a# o: U/ @2 q! r F9 B& I2 Q d1 := d1 shr 2 ;
5 P; N- A* t$ q9 \2 O( n& @ a1 := a1 or d1 ; 1 k) ^9 {9 X( t1 N
outchar[si + 1] := chr (a1); ! R+ b. x' {6 S5 n+ A8 M( _
# t1 S- `4 V( S8 O4 E a1 := ord ( inchar [di + 2]);
5 B7 Q: u: E' _! W a1 := a1 shl 6 ;
1 s) L6 p* M" i+ }0 v7 B8 V! }! a. h d1 := ord ( inchar [di + 3]);
) {" H: l0 E+ `4 Z: Y9 \ a1 := a1 or d1 ; 7 j/ `* O9 S0 o v, L6 p8 K
outchar[si + 2] := chr (a1); % ^( ?# f, [ H2 ^
( y& N7 Z, J4 {2 t; | di := di + 4 ; 0 x0 ]+ J0 |! @* |( P# G/ I
si := si + 3 ; % X' g2 M7 t. b I
end;
' E/ i# }- G! ], j* G8 M( z( T8 ^end;
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 7 w8 z7 Y. v. A3 ^2 r$ K
var 5 V/ ^" K. X+ c( k5 T" r
a1, d1 :byte; ) h0 A) d+ K7 Z! t v9 t# x
j,count,di,si :integer; : H. s( ]/ E% d! J# \% x3 ?9 c
) Y5 v- c. X* t3 L5 |+ n! n( B' lbegin 7 x0 D7 p' G4 I$ { B
encode := len div 3 * 4 ; //返回加密后数据长度 / l# e, `& ~& {
di := 0 ; 0 n9 p0 m' T) s* F; [
si := 0 ;
; X$ j; D9 o; s: q count := len div 3; //定义循环次数 & H% ^; ?, P1 f. z
for j := 1 to count do
/ B$ `1 W% X8 y$ e# A: U0 B1 ^/ d begin
& X( b% A( I" K! G+ | a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
% K* m) p5 N- T& T, c a1 := a1 shr 2 ; ) ^ m* A$ ^5 I- C2 D+ P
outchar [si] := chr (a1); / f' ?0 d$ J/ z* z8 C5 x
2 E" Z' N, D) ] a1 := ord (inchar[di]);
8 F8 C! O+ j/ { a1 := a1 and 3 ;
7 m2 [: s, P5 |: K+ R$ T9 V a1 := a1 shl 4 ; 4 I3 Q7 p+ f( Z2 M% i& W
d1 := ord (inchar[di +1]);
3 L6 N; b# X- m7 I [4 K- q d1 := d1 shr 4 ; + w& K7 M F) u. H1 K" t
a1 := a1 or d1 ;
4 v; ^' y+ {, c3 V' e! ^+ I outchar[si +1] := chr (a1) ;
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6 J* Q8 m9 J+ m( E a1 := ord ( inchar[di + 1]) ; + ?. j C' g+ H6 h* t5 O5 [5 L
a1 := a1 and {&content}F ; ! `$ f4 H8 `1 b; ~( I8 R& O
a1 := a1 shl 2 ; - N [; T; j/ Y
d1 := ord ( inchar[di + 2]) ;
) K* A1 H0 Q' H3 `) ]% X d1 := d1 shr 6 ;
P& l; t8 o3 a0 { a1 := a1 or d1 ; # @. l7 s- S" V/ i8 a; ]# N
outchar[si + 2] := chr (a1);
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' @6 D& a/ e1 P0 z5 [# z a1 := ord ( inchar[di + 2]) ; ' u" q$ o. V7 I6 h7 p! r/ M5 O w
a1 := a1 and $3F;
) @3 Q& z" G" b0 x: T outchar[si + 3] := chr (a1) ;
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5 ]( ^& C) d( u) y$ g //第二次转换
( `( |* |& }3 s" N5 S4 v. r- B3 @6 M6 B d1 := ord ( outchar [si]); //第 1 个 $ v' H0 a4 e# h' y, d8 N2 s
d1 := gamestr [d1];
& D( f+ W4 E* X" M' L) _ I outchar [si] := chr (d1);
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7 w* q8 P" f: V2 k, p d1 := ord ( outchar [si + 1]); //第 2 个 , o, W5 Z9 G3 d5 o
d1 := gamestr [d1];
1 S# r/ Q7 G* n1 r2 }, `* ~ outchar [si + 1] := chr (d1);
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d1 := ord ( outchar [si + 2 ]); //第 3 个 ; ?" i# _3 Z3 v1 L4 D8 L% h
d1 := gamestr [d1];
. S' h( j5 g" O! ?! v4 L9 D outchar [si + 2] := chr (d1);
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% b' ]( p7 Z6 W; D0 q& M d1 := ord ( outchar [si + 3]); //第 4 个 ; J1 J0 M! W1 t! S
d1 := gamestr [d1]; - v* Y. L2 S& H6 ~
outchar [si + 3] := chr (d1); * l7 C. A [- n6 Y; S% w# N8 G, v
1 o$ d6 U" `! v3 u. q% m& x di := di + 3 ; 5 M" _0 h; c& K3 Q8 J% T% e& \5 s
si := si + 4 ; # Q$ j; g2 J' \ v' p, \; P
end;
9 j; \- W" t3 J( @/ p: G- W0 Oend; , F% t% r. a! f( I, x# F, }
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