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..
% l% n# h& B+ x% 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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, Y3 m; G6 Y5 H6 H声明部分
9 `. W1 F: x4 l2 c7 s//二种声方法均可
: z/ N4 k0 ^. E% v# Y' Y2 Z0 \//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’;
9 A! k. G8 ^- |4 r1 m( Hconst gamestr : array [0..127] of byte = + n! r5 Q% j" }( ^2 v8 w @9 A
($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, $1A); + C4 I% I7 ` q5 |$ _5 H3 l
$ M9 m7 H6 O) n/ H代码部分
9 |0 y/ L7 r8 {: U# c1 Xfunction decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
2 `" e# C0 `8 ^var
% J7 f. i" b- |6 C6 @* C a1, d1: byte; 4 i! i+ |: I {# R
j, count, di, si :integer; , m @3 G- h7 |
begin
8 @2 ?# Q" E0 V$ W1 j: ~# h1 G4 E# A decode := len div 4 * 3 ; //返回解密后数据长度 ! W) h8 r" L W% A8 L
j := 0;
7 |: Z. G; [- T. U while i < len do 7 P/ o2 ]1 j4 t7 \( h/ F4 x
begin , |- i' X s. ], x! A1 v, Z7 |$ Z
d1 := byte (inchar[j] ); ; r+ J$ S4 y7 u$ K) q8 j
if ( d1 = $3B ) or (d1 = $7A) then
9 \) o V: h. j( i3 t% G& ~ begin
8 m# w- a% N( _, O9 D. l end;
0 n1 m: f0 {- O5 M' Z d1 := d1 and {&content}FF; 8 h0 h( H, j+ H8 J3 m) J! Q
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] ); 7 b; r0 Z" W% f- D
byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
5 W6 `. {/ i: |7 A1 L/ l, @0 S inc (i);
! [3 e0 q6 f0 G end; % q8 q J! F, a, Y' h
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di := 0 ; + q- h0 o2 B# `! y
si := 0 ; ) R9 v! A7 g( |6 B
count := len div 4 ; //循环次数 * u) }) K. i$ X, q
for j := 1 to count do
& p6 U0 a$ v% O% `; E begin
. W/ {5 \# q! R8 { a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) 6 ~; Q* q+ ^# c7 y$ T2 n9 P
a1 := a1 shl 2 ; / H8 Z9 k9 n H2 T" B7 r
d1 := ord ( inchar [di + 1]); 5 F$ e2 |" c4 Z/ m) C9 h
d1 := d1 shr 4 ;
5 j* D' Y$ |! W8 `8 W0 X4 y/ z- J1 _ a1 := a1 or d1 ; 2 O2 c$ B$ x3 l2 r4 @
outchar[si] := chr (a1);
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: P7 Z, Z: t0 w9 y a1 := ord ( inchar [di + 1]); 2 B% a; t1 S( u4 w. o
a1 := a1 shl 4 ; ( O: y6 k: l5 ]+ q7 ^) x
d1 := ord ( inchar [di + 2]); ( G* Z/ s% R1 Q
d1 := d1 shr 2 ; 5 ]8 n- b7 t% V3 `9 R9 O& {. ?
a1 := a1 or d1 ;
* T$ q0 {5 d# ?! s outchar[si + 1] := chr (a1); 3 t3 @0 L- B8 C9 [$ \0 J C; @
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a1 := ord ( inchar [di + 2]); : j/ f; E" J/ I9 _$ ^# e
a1 := a1 shl 6 ; / R: z# o' F8 P0 W% J, W
d1 := ord ( inchar [di + 3]); # Z1 b+ W/ ~; U
a1 := a1 or d1 ;
: Q) y+ E( N8 E5 v. e7 c4 V' [ outchar[si + 2] := chr (a1);
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% t* W8 u! d) W& C6 i: ~0 V& D di := di + 4 ; " l0 z" ?& J, P' K- H' T5 h
si := si + 3 ;
8 m# C) f8 y* P; [+ v end;
/ H0 M g h0 v, \0 Y& o, h; I, xend;
0 X% a5 D8 R# b3 i6 k- e: q! C0 q; M1 _ }
function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 5 r; L( ]3 Z8 `- l1 b
var
% j1 `# R2 h+ s7 {" `' ? a1, d1 :byte;
6 o* p& v, r' N$ I; ? j,count,di,si :integer; 4 q1 m3 h% o& C, @
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begin 0 j% R" R# V# N" c& T- y
encode := len div 3 * 4 ; //返回加密后数据长度
5 ^) i) A& w' T4 C di := 0 ;
4 f& |; k% ?$ b4 v# g. n9 a9 n si := 0 ;
, N9 T x _1 n; k( m& ^ count := len div 3; //定义循环次数
$ o: L7 d, g e; F& u+ S for j := 1 to count do
! A+ p7 d! ^; e5 z" W1 a" W begin - @5 y6 `# ]$ K% f
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
9 D/ O' t9 D! z& Z2 }5 } a1 := a1 shr 2 ;
# i* S( {2 ]* q" s6 L$ q) q. x outchar [si] := chr (a1); S5 ^7 Y% w' @3 ]5 |! ]- Q
/ C8 G) r1 d2 s, `$ y8 m a1 := ord (inchar[di]); ! L5 n5 @$ _% g' ]
a1 := a1 and 3 ; $ H. n0 Y9 W" r$ b0 Z) Y
a1 := a1 shl 4 ; 8 k# P( [& d( @1 z1 f
d1 := ord (inchar[di +1]); Q7 `, O6 m8 r$ L2 z# I: y
d1 := d1 shr 4 ; : y9 N: p/ u4 N( g4 h
a1 := a1 or d1 ;
& w6 Q& a* Q O/ | outchar[si +1] := chr (a1) ; ' R/ a1 u7 ]0 L
' F/ y7 l1 P1 C0 X& g1 F& ^ a1 := ord ( inchar[di + 1]) ; 1 c3 C7 S/ l* ]
a1 := a1 and {&content}F ;
# X) H+ O1 Z. l5 }8 B a1 := a1 shl 2 ;
R( m6 | e% p! B d1 := ord ( inchar[di + 2]) ; - y5 {6 w/ O( b; Y
d1 := d1 shr 6 ; , y4 z3 q1 U c! u# |% a* f
a1 := a1 or d1 ; 9 \; `7 G: m% A, U* {% i
outchar[si + 2] := chr (a1); 0 {' J+ }& G" z' x+ p
" u; U Q( M3 Z2 I: G+ H) R a1 := ord ( inchar[di + 2]) ;
0 U F# I. t/ @) `8 W# X1 ~6 B1 n2 P) x a1 := a1 and $3F; 8 F" Y9 R/ m0 h. t6 Z
outchar[si + 3] := chr (a1) ; . K% A) z `# N: ~2 v; F8 M2 o9 k
9 t( A% e* N" y //第二次转换
% S& u6 K( h% j0 k+ M d1 := ord ( outchar [si]); //第 1 个 ' } b. O2 H9 _: Q
d1 := gamestr [d1];
" h9 _4 m. Q4 p2 w7 A outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个 4 Z& _! ^; p; P- G4 K9 s
d1 := gamestr [d1]; * ?3 r H3 ]+ J, H9 l4 ^! Z
outchar [si + 1] := chr (d1); ( k5 x% R1 T1 }
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d1 := ord ( outchar [si + 2 ]); //第 3 个
; p; m/ O9 a; r! I d1 := gamestr [d1];
|/ M6 D& U' S7 w- ` outchar [si + 2] := chr (d1); ) T2 d: |" e& n9 e& C# R; S6 m7 D
# P$ K% J8 _& m( b6 I d1 := ord ( outchar [si + 3]); //第 4 个 ( M; Z, z7 M& [5 R
d1 := gamestr [d1];
0 C; a' J. q1 d5 d outchar [si + 3] := chr (d1); ) F+ X1 o: X: h& `3 c
3 t6 Y+ I: n5 J% K7 j di := di + 3 ;
' ~! G9 @% k0 f) `& u; O si := si + 4 ;
0 K u" b: g3 x+ u1 f; a, r end; . B# L) K- t4 Y9 D; x2 @9 n
end; ! R! @, {+ [; Z% M) f9 Q
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