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.. . O8 _& f, ?$ M/ B8 F* ]( ^, n2 j
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) / I/ ~ {3 o! `; f- N: Y8 R4 t0 e
) @/ ^/ r G+ y9 G* F, U. A* ]声明部分 - l! E. [! J$ N8 Y0 i: b3 I
//二种声方法均可 9 ^$ ^8 |# p* ]# {& l7 c
//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& Z6 x. e; V( `+ Z, Vconst gamestr : array [0..127] of byte = 1 I0 |& S* b. o5 K p. o3 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);
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8 j0 l' P6 ^& e2 ]代码部分 $ i+ M/ F/ Q! s
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 ( b5 m. A: M6 H" Z0 a% X
var ' ]5 b' \' |2 c
a1, d1: byte;
: E0 s- K( x6 I* L. [0 K j, count, di, si :integer; # i1 V5 r# R& j' t
begin
" {" x9 a% {5 I; I( i# U% y0 I3 g decode := len div 4 * 3 ; //返回解密后数据长度 ( b1 O: c0 x( r6 k* w. \
j := 0; 6 _ g6 _9 A2 v' G/ R9 N; f* u
while i < len do
* c ^2 j5 l* Q# X, D5 w8 _ Z9 V* e begin
( @9 j9 N* h4 ~& p2 k3 I+ \! k d1 := byte (inchar[j] );
$ t. A+ T* D' Y/ H3 J3 U if ( d1 = $3B ) or (d1 = $7A) then + A c9 C9 e C6 O6 `5 a- K4 {
begin 0 j+ T @8 d3 X8 u# w7 J( r- M
end;
3 `, x! `+ L( o$ E% R d1 := d1 and {&content}FF; $ ], l- I8 D* E
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
) w" c' d' r9 C' t$ R$ @ byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 ; q' I7 z0 f/ x/ B0 R, H. u
inc (i); ' s6 _/ K" Q, X% o/ k7 } l3 E
end; # H! W5 ^ `% N1 p
5 v' h: O. I/ s/ c; g' ?4 @ di := 0 ;
' `$ W5 \0 Z; Z5 \7 b+ Z5 L si := 0 ; 0 O2 P! A- v+ @8 k
count := len div 4 ; //循环次数
/ x+ B5 \4 G4 T `0 x for j := 1 to count do
6 p+ L/ `1 r5 u) a/ g begin 1 d# U$ o, ~0 c. j6 K6 @$ Q
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个)
( y$ V) p) H" R/ H a1 := a1 shl 2 ; 1 U! K, H. f! y# e' |
d1 := ord ( inchar [di + 1]);
( R9 r! K! I8 e$ H& I, U- o d1 := d1 shr 4 ; 8 ? k: p/ e& r& i9 P
a1 := a1 or d1 ;
" c! r! {+ ]/ M* {: Q outchar[si] := chr (a1); ) }3 ^/ p5 O1 A$ ?. q7 y
( R7 q1 i, q) `* V$ @ a1 := ord ( inchar [di + 1]);
' ?+ e" b( w5 R0 m* C# F0 D' y a1 := a1 shl 4 ;
5 |- J7 ~; l8 [' l) H" _$ p2 e9 f d1 := ord ( inchar [di + 2]);
2 y/ `# A [5 P. N" A. |% c& h& O d1 := d1 shr 2 ;
1 M7 Y1 I7 | N5 ~% \ a1 := a1 or d1 ;
Q0 x) I5 R3 Z {" h outchar[si + 1] := chr (a1);
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a1 := ord ( inchar [di + 2]);
7 e! g% v z0 V) Y a1 := a1 shl 6 ;
# w; z. y2 {0 P5 S/ J/ R3 h! P3 z d1 := ord ( inchar [di + 3]);
$ d: D L9 {9 O% C& I a1 := a1 or d1 ; 1 ]- x" V8 j( c& n7 h' Z" Z
outchar[si + 2] := chr (a1);
0 \+ W8 W1 W' `( x4 S. ]
" @# [5 U7 P4 p7 ]3 V7 z di := di + 4 ;
/ {6 ^4 ~# f$ E! v) P, N- ] si := si + 3 ; 0 I' R0 G& g; `% A
end;
% S3 @! E( R! Q% y1 | Q: iend; 5 a* X# e' M$ n4 |3 _
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
; b* P' ]/ g2 T8 a0 f1 Vvar
* N! v( e+ w3 x% n# T9 H a1, d1 :byte;
4 f% o1 ?7 N5 R8 l$ \# u; V j,count,di,si :integer; # C1 s$ q8 p+ n& k
/ F* ], J8 o3 }. Pbegin 2 B" b5 J/ i j5 b4 s7 H
encode := len div 3 * 4 ; //返回加密后数据长度 " a, O8 |, {7 B# R' m/ \+ |1 @
di := 0 ;
0 d+ J- j H( U0 h6 r si := 0 ;
Q$ `; _& `8 T1 A; K! g0 ~& D4 x count := len div 3; //定义循环次数 7 e. p1 D! o7 g+ V% l
for j := 1 to count do
5 A4 e8 l s$ r# @, ^# W& ]2 R" _ begin
" Q% w/ e* s" ^! [. u a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个 ! l4 a8 k9 B/ F( }4 B
a1 := a1 shr 2 ;
8 A7 z2 A6 ?; L8 r# q! I outchar [si] := chr (a1);
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% o2 O6 U J2 A& U; P f J a1 := ord (inchar[di]);
+ Q# ?, r( h7 G a1 := a1 and 3 ; 0 U" F$ o5 h0 Z; T. f; ^
a1 := a1 shl 4 ;
8 w: x. ]- _# R ?9 E7 ^3 o+ l d1 := ord (inchar[di +1]);
1 y" U4 M& F; E; _' {% J" K( E d1 := d1 shr 4 ;
* z" j# a- ^$ x a1 := a1 or d1 ; ) l3 s' X! C+ T% ^9 f
outchar[si +1] := chr (a1) ; `5 x& K# q0 D- y
" d( H1 @ X/ U1 t
a1 := ord ( inchar[di + 1]) ;
0 y$ z1 r6 H: }, n( y. q3 W- y* A6 x. B a1 := a1 and {&content}F ;
4 H5 S4 { a# K+ h7 X0 H4 F a1 := a1 shl 2 ; 3 y3 A! }6 d( B
d1 := ord ( inchar[di + 2]) ; ' s3 S8 L3 q9 P9 H7 |
d1 := d1 shr 6 ;
) x9 R- S- h' O1 u; W2 k$ X2 W% D a1 := a1 or d1 ; + ]6 ~% i1 Y" P/ E1 c) a! J0 \
outchar[si + 2] := chr (a1); & V( z/ J; ]8 R. R3 y1 |. H
* e) _5 h5 L! g/ w3 r( I( ] a1 := ord ( inchar[di + 2]) ; 9 f L! j& @6 W/ K' w0 f9 d, e
a1 := a1 and $3F; 4 ?1 C$ c5 E$ t$ g: O4 j
outchar[si + 3] := chr (a1) ; % k, m- `. s J! o$ p# r
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//第二次转换
8 o K2 z5 [) g* @ d1 := ord ( outchar [si]); //第 1 个
* O. `/ W. m, d- [ _4 G d1 := gamestr [d1]; ; G1 H0 ?1 @) {( x% |, g" E
outchar [si] := chr (d1); 6 L8 u: E8 v' u
" h4 k0 ?- C4 D/ E0 s8 h
d1 := ord ( outchar [si + 1]); //第 2 个 9 E4 _6 {5 ^& H6 D% F/ S$ B
d1 := gamestr [d1]; " V3 k" \& o- K, o% c6 E
outchar [si + 1] := chr (d1);
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) j5 y1 w1 Y2 i8 H6 n d1 := ord ( outchar [si + 2 ]); //第 3 个 # U/ _' U* C D7 k
d1 := gamestr [d1];
; M: T8 W' v O3 K$ Q6 ~+ m outchar [si + 2] := chr (d1); 4 s8 N2 s! \3 u$ X4 J- s9 p
7 V" |& V% T, h! m. o8 t( n' h d1 := ord ( outchar [si + 3]); //第 4 个
* @# q9 L4 s& d5 x, e d1 := gamestr [d1];
* N% f( d. A2 D; y% h, j/ f \ outchar [si + 3] := chr (d1);
0 U% K. h5 S$ y2 ~/ M+ ~: N) t2 q5 B0 N3 ^3 T. K3 Q( h; b
di := di + 3 ;
3 b$ X) G K' m- h si := si + 4 ; # P0 a$ v) q1 K5 e
end;
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