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.. . J% B- k/ s$ ?
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) 1 G! J! B9 [1 `/ s: C4 T' |
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声明部分 5 g# [% U# U3 e, b% A
//二种声方法均可 * m, r% O5 A3 o8 q
//const gamestr:string[255] = ’N>TSVUJlwdcBMFjnAKb?qxvyeGzfLP=_ER@Z\am]ChgoWD;QuX[<tkpr^`iIHOYs.3" (- ’+chr(13)+’<;’+chr(12)+chr(00)+’=/!,1>#2$’+chr(39)+’89%’+chr(10)+chr(9)+’*):5&+67?40’;
1 V. S5 y$ ?+ I$ W3 F. s9 tconst gamestr : array [0..127] of byte =
' I; k3 t/ Y9 {. ^8 _($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, $1A);
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' y" o9 @/ E# t* v代码部分 * U6 A, ^. C6 P: Y U9 F- u, C& T
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
8 }* }! N$ R9 L Uvar
; S$ p) G( `4 b% H7 b1 D a1, d1: byte;
; r: S$ y0 L( P5 p/ L j, count, di, si :integer;
. {% N6 P2 V# }& d5 D2 ^/ N7 tbegin 4 ] {* v6 ?$ X0 t" f8 z5 E
decode := len div 4 * 3 ; //返回解密后数据长度
. A* W& K9 q) R. P: D9 k j := 0;
* q9 V9 } J. C# ^2 R5 F/ v while i < len do
2 [( r4 N' m+ i' W* t, `5 k begin 2 L# r/ r3 i8 b2 v
d1 := byte (inchar[j] );
/ C) a O8 @: L8 p' Q& j if ( d1 = $3B ) or (d1 = $7A) then + W$ o) G. [6 J# w
begin # G) n# W2 @1 u* _# I; @% m. C: i
end; 1 P5 K- D: F1 C0 `3 Y4 P: m5 [
d1 := d1 and {&content}FF; ' Z; @. M' k5 M$ b" {) q8 O8 |
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
+ b3 I8 o# T$ Y byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 + l/ s1 E+ u" s1 C6 X3 c5 k
inc (i);
2 z$ H4 ]$ n2 { end; H% M7 J5 T& {. J' H# n
& Q5 c% K+ A+ x+ o" e5 b di := 0 ;
5 U4 ~$ a0 D5 W) u0 P si := 0 ; 0 i( V/ K" M% m+ P: V6 @1 z
count := len div 4 ; //循环次数
9 a1 S) M7 b+ G! q for j := 1 to count do ' d; ]5 [' l5 X b& U
begin
( O5 @! ] Y8 R- H a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) ! }- `- h5 l) o- f: R$ T- i
a1 := a1 shl 2 ; " ^- b9 q; `* F/ ~
d1 := ord ( inchar [di + 1]);
7 x9 m# W9 q( q5 R- C1 W d1 := d1 shr 4 ; ! Y S9 M# J) ]% Y: d2 c5 F
a1 := a1 or d1 ;
5 h3 a/ j+ l- t$ B) `3 J7 J outchar[si] := chr (a1); $ F; Q& \( ~/ _' A9 f& ^
2 _4 h0 F- [9 x% k2 l- v" m% @
a1 := ord ( inchar [di + 1]);
7 x; X: j: \; Y5 | u1 l9 ] a1 := a1 shl 4 ; ( l$ g4 x- k( ^0 v- S& q
d1 := ord ( inchar [di + 2]); ( a; F1 T$ ?% o& p* Y3 i( e) b
d1 := d1 shr 2 ; w2 t2 C8 h7 W- E$ K" k, f5 B0 g
a1 := a1 or d1 ;
/ @3 P& k" W# X outchar[si + 1] := chr (a1); 5 t9 O; h) r8 d
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a1 := ord ( inchar [di + 2]);
* ]' p7 K) I u a1 := a1 shl 6 ;
! o% }5 [; z, P0 d) G7 S, @ d1 := ord ( inchar [di + 3]); 3 P. m* t6 r0 C" G
a1 := a1 or d1 ;
5 |; B) w; C8 `6 {; S& C outchar[si + 2] := chr (a1);
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di := di + 4 ;
, z% U# V3 x" H3 k' }7 c si := si + 3 ;
, V& Y4 i7 x. H+ \" E end;
6 ~! O/ X: U' h' l6 _8 I9 e7 }end; 8 i* X, B) c/ e8 B- k, \7 ~: s
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 9 L+ ^! e. I9 u
var ) W1 ~7 {9 d0 l* l R" m" u1 K
a1, d1 :byte;
0 `' N5 O; H# e j,count,di,si :integer;
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. p6 _' u- D+ o1 Cbegin
4 s+ B# T% o; i0 \9 j encode := len div 3 * 4 ; //返回加密后数据长度
7 @1 U! `' ?5 a0 R- u8 f di := 0 ; 4 f9 L* ^' u+ r8 E( B! \
si := 0 ;
+ I- u( P. D5 w; X count := len div 3; //定义循环次数 , p- I' v3 _; k/ k `
for j := 1 to count do
' y2 ~; g& R( ^3 Z4 k' w begin 1 P1 i7 f/ I, l9 o
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
3 T. { ~- _ @- i a1 := a1 shr 2 ;
- v i7 Q# {+ m. c' b outchar [si] := chr (a1);
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a1 := ord (inchar[di]); ( ]4 y, t+ m. A
a1 := a1 and 3 ;
2 G, Q" q1 z5 X8 d% t- W% _9 Y( ` a1 := a1 shl 4 ;
# {6 Q. H# _6 P7 m; Q0 D d1 := ord (inchar[di +1]); # j3 S8 ^9 U J5 Z1 L
d1 := d1 shr 4 ;
; i& a% S( ^- N) I a1 := a1 or d1 ; 8 t! j D1 L% P: J' v" d
outchar[si +1] := chr (a1) ; 4 V" h F% b& y0 f1 x) `
; V3 Z- p4 I2 U r1 F0 O a1 := ord ( inchar[di + 1]) ;
' K* e a) @! O2 e a1 := a1 and {&content}F ;
3 e! W" I8 R+ \+ }4 q* g a1 := a1 shl 2 ;
: Q2 F! P( w& Q- R7 X/ U d1 := ord ( inchar[di + 2]) ;
: q6 l& u. p8 V5 T, { d1 := d1 shr 6 ; 5 ^9 _4 r- X) P6 C) {: F$ ?. K
a1 := a1 or d1 ; ( K8 }- d. n( A, I+ g Z
outchar[si + 2] := chr (a1);
* I% M. Z& U' W4 C( x1 i1 |4 {2 A5 }; `# q! u
a1 := ord ( inchar[di + 2]) ;
* H& N( s7 J! u2 p, y W% d0 t/ p a1 := a1 and $3F;
8 T# W. g! C, U7 N9 } outchar[si + 3] := chr (a1) ; ) W. I2 [0 @5 H& G
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//第二次转换 3 {# r/ k6 H& A) j$ s* D: B; G
d1 := ord ( outchar [si]); //第 1 个
9 _ ?! j$ f/ q d1 := gamestr [d1];
- k# p4 J5 f* V9 I6 r) ?4 g+ g2 N: c outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个
: t$ m- m- U1 {" @: e5 {! ?4 N2 f2 r& T d1 := gamestr [d1];
: r9 c. U3 b# _# r9 Q outchar [si + 1] := chr (d1); 2 x* S/ X3 X- U
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d1 := ord ( outchar [si + 2 ]); //第 3 个 9 c, Q1 d+ l8 E/ M# a
d1 := gamestr [d1];
+ l) f5 Q* T! M- ? outchar [si + 2] := chr (d1); $ z" L4 Z* T! M; u8 G q
: W7 F ?' J# f9 g d1 := ord ( outchar [si + 3]); //第 4 个 2 m& y: ^( O) \) w. I2 n0 _4 c. K; C7 }. r
d1 := gamestr [d1]; - P! W8 a y) O$ ~
outchar [si + 3] := chr (d1);
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di := di + 3 ;
7 j7 {- [/ p3 t1 U si := si + 4 ; 2 @+ b8 W4 }. b' F
end; * a5 \* J/ X8 {) K" q2 ?7 h% i5 E* q! U& K' j
end;
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