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..
7 O& V6 P8 _: Q1 u9 y# J" r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) ) u6 D8 R/ s( D T" Q0 Q
) Y4 @/ i6 k, c+ Y i声明部分
3 I) |. Z2 |3 X% K% V, P//二种声方法均可
8 H+ V2 v0 g% h$ R//const gamestr:string[255] = ’N>TSVUJlwdcBMFjnAKb?qxvyeGzfLP=_ER@Z\am]ChgoWD;QuX[<tkpr^`iIHOYs.3" (- ’+chr(13)+’<;’+chr(12)+chr(00)+’=/!,1>#2$’+chr(39)+’89%’+chr(10)+chr(9)+’*):5&+67?40’;
: M' G6 l/ t7 d" @/ r7 V6 jconst gamestr : array [0..127] of byte = - t9 Z$ F e3 }; ~: q
($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, $1A); ' H" R; y0 W5 v$ }
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代码部分 . s$ [2 y& _: X' I) {
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
) ]0 U( m" @1 d& t1 Lvar 3 h2 `! M4 M& {% U! |1 G& V; Y
a1, d1: byte; " Z7 _5 H4 B- O& p
j, count, di, si :integer;
) y1 M3 o3 q$ \, [, n5 Ibegin % f% d. e3 q, C) e/ p
decode := len div 4 * 3 ; //返回解密后数据长度
& s- P+ ?$ |4 D3 L j := 0; 2 L. m6 i; T7 z$ A. b
while i < len do 3 N! m: k+ f* B2 @0 M# a& u: ^% n
begin
! x" b+ h8 a' \1 ]4 D: ~ h% c d1 := byte (inchar[j] ); ! K0 }- F1 h. x7 }8 M7 T
if ( d1 = $3B ) or (d1 = $7A) then ' L7 a0 ?9 V7 K
begin ; x0 G" f! r3 b+ q' t
end; , C7 l+ U' ?3 q+ j. g
d1 := d1 and {&content}FF;
" e. d5 C W } d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
' B( e, b% o4 ~' r7 M3 ]$ O byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 * `% [6 g, y9 w) F( Y
inc (i); " f( F5 M+ i8 q: b$ G" `
end; 8 Z1 ]7 `5 E0 N, c2 @
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di := 0 ;
# l3 l' W* R/ e/ r( K si := 0 ; 6 d7 |+ E4 L- t
count := len div 4 ; //循环次数
" ?0 k0 w" l- ]% l% b9 K! I- x7 T! f for j := 1 to count do
$ Q7 I3 o! O y h' p g, S begin # T# G! Z. C( t% G
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) + A( y! a8 o; ^& @* y5 }
a1 := a1 shl 2 ; / M9 z- n& M7 R, e. c
d1 := ord ( inchar [di + 1]); 2 I5 I- [' Z7 T u3 D4 H
d1 := d1 shr 4 ;
. \# k& j( ?1 z( b a1 := a1 or d1 ; / @ j5 l% ?6 A0 s) ?
outchar[si] := chr (a1); & v6 s/ F9 t' I4 T
" V9 D ~0 |9 @5 l# d/ O3 w a1 := ord ( inchar [di + 1]); $ [4 @5 i# {8 l' ?% _
a1 := a1 shl 4 ; ; I& u8 ~6 \7 L; a0 j6 O) y& D' ]
d1 := ord ( inchar [di + 2]);
6 R/ B4 ]' k$ x' i, N d1 := d1 shr 2 ; . L5 k% V* ?4 Q& D$ S
a1 := a1 or d1 ;
2 ~6 |: L. \. l# D9 t3 Q6 |7 r outchar[si + 1] := chr (a1); 1 z# u. @1 {* w" S" V
! h4 \/ B/ I9 I ~3 D. j a1 := ord ( inchar [di + 2]);
4 z( m9 J9 D" Y2 N8 i( ] a1 := a1 shl 6 ;
" R: n0 |& ` f! j( O d1 := ord ( inchar [di + 3]); : _& X8 G8 F/ k, E: e
a1 := a1 or d1 ; & j0 L0 O" }! h
outchar[si + 2] := chr (a1);
/ X8 i7 s6 @2 B! g- f, ^
) ^- y% c# L! {0 | di := di + 4 ;
+ A8 R" i* T" g; p; U7 o0 F si := si + 3 ;
0 F. J' _$ c; ]7 N) S, i end; + D5 t8 D1 g3 o! F1 p* @
end;
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 7 k! d* ?" O2 i
var 7 W5 g, D9 J# W: H. g; T$ I: A
a1, d1 :byte; - |. j! I T6 C8 D3 G2 |
j,count,di,si :integer; 5 l0 [9 M" U" z6 ~7 n
( ^) A, n" o/ [" c
begin
- t& b1 ~; q6 ~: b$ g encode := len div 3 * 4 ; //返回加密后数据长度 ! \4 s& {& F7 |, `. i2 N- t
di := 0 ; 5 p, z7 W" O8 M; X7 l( D( ~- |
si := 0 ;
% c) j$ {: Z( J count := len div 3; //定义循环次数 + {. H# {' d$ x/ O
for j := 1 to count do 4 S8 T+ B( ~) F
begin " j* z- s) x& A) L6 Q
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
9 N; _& n( U+ l5 k. M& l4 I a1 := a1 shr 2 ;
, {, U- u/ f8 @) H" u outchar [si] := chr (a1);
7 r& G2 u/ u6 J& l
& K: ^* n$ H ^; T8 z. u/ | a1 := ord (inchar[di]);
8 w# E! U% i5 A+ o. } q; |# s a1 := a1 and 3 ;
" ?3 k" R5 J/ c7 ^% t" a a1 := a1 shl 4 ;
! q( l* Q' V, u3 Q( F! W9 ^% H d1 := ord (inchar[di +1]); ' Z9 `. `, M8 Z* H8 E4 e5 e
d1 := d1 shr 4 ; ; O. R) D2 O$ {) J* c
a1 := a1 or d1 ;
+ }' K, L# A4 O. e9 e outchar[si +1] := chr (a1) ;
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a1 := ord ( inchar[di + 1]) ;
( l# O3 [: {' G: C/ `. { I a1 := a1 and {&content}F ;
' x3 M. N i. Z a1 := a1 shl 2 ;
) U& Q( b8 Y9 w* x# N& c( A d1 := ord ( inchar[di + 2]) ; 3 X) z% Q" ?8 H
d1 := d1 shr 6 ; # l$ R0 Z/ y+ G" i8 z
a1 := a1 or d1 ; * K% u% f! S* f" ]) j
outchar[si + 2] := chr (a1);
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a1 := ord ( inchar[di + 2]) ;
+ B# `. G0 \# M$ b( M: @ a1 := a1 and $3F;
) ?/ I0 G+ Y6 o2 R' V3 y% Y# H. I outchar[si + 3] := chr (a1) ; + h9 M+ |9 O, S$ o8 O, g% F6 [
4 F4 |. |9 L' Y, U //第二次转换 - T; o* Y. A+ _+ J. x
d1 := ord ( outchar [si]); //第 1 个 8 |7 ^! G5 Z* A4 P5 r6 J
d1 := gamestr [d1];
+ g- e4 b% N1 i$ g2 }0 q outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个
# V+ a( c* D3 o; D d1 := gamestr [d1];
! H2 P8 _1 f6 T9 t3 m( T7 {6 c outchar [si + 1] := chr (d1); ^- Y" f" o& z$ t( B9 @; T
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d1 := ord ( outchar [si + 2 ]); //第 3 个
) Q4 [( J: ^& f% K- e2 x: V/ l d1 := gamestr [d1]; 5 G# c1 Q; ~1 {9 x- f
outchar [si + 2] := chr (d1);
p ^( d4 L$ t# m$ u0 U6 G/ ?/ n( N( E# B7 \5 A, f4 _
d1 := ord ( outchar [si + 3]); //第 4 个
7 ?: ], p* X3 G6 O6 ^) A; p% u d1 := gamestr [d1]; ! M* O6 E5 I$ I, F9 K) F
outchar [si + 3] := chr (d1);
8 |6 l s9 g* i# W4 ?- ~. h! e2 M6 O; [5 I4 Q5 c) y6 t
di := di + 3 ;
( o- |6 B2 m/ R/ Q& w" C si := si + 4 ; 1 x. r4 o e$ Y7 r# w- O9 K; ]4 Q
end; 2 w" e$ K2 O8 k6 b" b
end; $ O, A/ `6 {" R. G5 [# z8 {
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