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..
: k+ j; i, r4 y, q; A& j% J9 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)
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% b* Y' Q9 Q5 s- B; M声明部分 % h7 y, U. s% G6 E: O; I- H7 s
//二种声方法均可 P, J( T: \; N5 E5 ^$ d- B
//const gamestr:string[255] = ’N>TSVUJlwdcBMFjnAKb?qxvyeGzfLP=_ER@Z\am]ChgoWD;QuX[<tkpr^`iIHOYs.3" (- ’+chr(13)+’<;’+chr(12)+chr(00)+’=/!,1>#2$’+chr(39)+’89%’+chr(10)+chr(9)+’*):5&+67?40’; / w; e& v% W2 M1 ]. x
const gamestr : array [0..127] of byte =
. M$ c* W( s3 B' b$ j) K( ~($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, $1A);
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代码部分 * u2 r* Y; V9 Y9 T$ Y" U
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 # N" s5 K) }+ `8 ^/ Z& Z: h% z
var
, \% y# I! T0 P1 P a1, d1: byte;
$ h, Z& k# K$ O0 ~' M* A j, count, di, si :integer;
$ }. x+ q/ W* a3 M6 Z- j7 K) ?6 Ybegin 6 r% S2 ~4 b3 {3 y- {% X
decode := len div 4 * 3 ; //返回解密后数据长度 # E" y- I I% J6 v/ L$ \
j := 0; + l, f0 k. _0 J9 K ]
while i < len do
( v6 w; Y5 ~4 ]1 j& G# j+ Q begin ! l' F/ ]1 C. B# n/ @
d1 := byte (inchar[j] ); 6 K+ k0 d7 t/ \" @/ u
if ( d1 = $3B ) or (d1 = $7A) then ( W' }3 `! y: k
begin
; ^& L; ~% u4 m8 t" c/ A" ]* j( V end; # G; Z% o" h5 V* P; V' u
d1 := d1 and {&content}FF; : d4 ^+ ]( N, v9 g
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
6 j9 W9 Z$ E5 G7 C3 z5 D8 [' x2 j byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 5 ^5 m/ V7 S8 r. X6 m @# S7 R9 n
inc (i);
; m6 _& b V, V: E end; 9 X! ?! Z- g! E! M6 C* n9 ]; K
: X5 j g3 z) z
di := 0 ;
: s" z" f" s6 r6 a7 O2 o' Y si := 0 ; d8 M1 B( u- h V
count := len div 4 ; //循环次数 / S# t- t8 B7 \, x
for j := 1 to count do
- i. ~1 u i. c, n: @* u begin 5 x$ F* j4 A: @$ I y# I
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) . z8 A/ w) H, l+ a8 D6 I/ ]* j8 _
a1 := a1 shl 2 ;
; O9 I- s) _% Q& ^- l d1 := ord ( inchar [di + 1]);
1 I' `- i o e! ]7 H d1 := d1 shr 4 ;
* N8 ]9 i- S/ d3 z5 P8 ~ a1 := a1 or d1 ; . {' b$ b+ I- @, _! W" P
outchar[si] := chr (a1); , s. z& W5 T7 a$ V. [9 y, k
3 ~$ |7 A! Q! m* N+ n
a1 := ord ( inchar [di + 1]);
) f6 M" z4 a6 G4 I! }9 t a1 := a1 shl 4 ; ( k, D m' @/ i+ L
d1 := ord ( inchar [di + 2]);
# w. V# q# i8 ^ d1 := d1 shr 2 ; # z% R! |2 _+ W- h+ x
a1 := a1 or d1 ; 7 f. p `, i, {* e$ t) Y( g
outchar[si + 1] := chr (a1); ' D; `6 ?) G) D1 H+ V
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a1 := ord ( inchar [di + 2]);
+ T1 H* x) d; T% y$ d( O/ O a1 := a1 shl 6 ; 9 w/ F" E0 b u' N7 l! O
d1 := ord ( inchar [di + 3]); : } p: B" W+ p5 I
a1 := a1 or d1 ;
( D) `6 Q f3 X' r outchar[si + 2] := chr (a1); + ^4 K1 V9 e" L2 m; F! s
" n3 i! Q' ?1 w7 J di := di + 4 ; ' A5 `! p! [# R- Z) _, C: h
si := si + 3 ;
0 ]) ^$ M9 C% z. y* U P! x end;
# k2 a! w; d4 {6 F* Y" ?end; 8 z' p1 |5 _, q- W& w0 C
: {8 s/ r( z% o& ofunction encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
2 v0 t4 F U3 P0 X; K( R1 c: x1 kvar : S1 Z4 b8 p# {; a2 h2 }; d# U
a1, d1 :byte;
: x/ Q U5 l3 h4 V5 e; _ j,count,di,si :integer;
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begin ! M, F; r+ X6 J- [$ {9 P* P
encode := len div 3 * 4 ; //返回加密后数据长度
$ w; w9 m9 D, Z di := 0 ; 0 G/ O$ D0 v3 W; D$ N
si := 0 ; 6 _2 t5 a9 \. _
count := len div 3; //定义循环次数
2 D8 I2 n$ ]% T0 H; u% w for j := 1 to count do 4 Z" Z f z% o6 `
begin
1 ` Z7 L8 i4 T c7 ?! h a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
! k- y+ i' r& n( o" d2 n6 Q7 g a1 := a1 shr 2 ; * p/ g: Z/ u0 K& j' {
outchar [si] := chr (a1); 6 m, {( j0 ?6 Y+ ~5 v" ^% R- b
! ]' O/ u9 N. T7 t# J8 n a1 := ord (inchar[di]); % U5 H7 A4 u. {2 y
a1 := a1 and 3 ;
2 N; L$ V) ^6 B/ R a a1 := a1 shl 4 ;
, S" ]4 R( C; ?* N: c d1 := ord (inchar[di +1]); + z9 M+ O+ D2 }! z( c9 ^
d1 := d1 shr 4 ; * Z& R3 r- h2 S1 Z7 p" d
a1 := a1 or d1 ; , I8 G% d: `- Q
outchar[si +1] := chr (a1) ; 5 [4 H' ` f6 Q6 b K) H
% r6 G! {4 c b2 [; ?
a1 := ord ( inchar[di + 1]) ;
! E+ m9 @0 z( J, o6 B6 j; C0 W a1 := a1 and {&content}F ;
( Y: v% a9 v: r( d7 e6 m9 m a1 := a1 shl 2 ; - P. }7 P: ~' T+ B$ n+ Z7 T
d1 := ord ( inchar[di + 2]) ; ) D8 W6 E0 @! M2 |9 K' h4 x
d1 := d1 shr 6 ;
) }% w8 b8 |9 F! h a1 := a1 or d1 ;
- m7 ]3 Z. a6 d& S- J/ g6 F outchar[si + 2] := chr (a1);
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a1 := ord ( inchar[di + 2]) ;
4 J: S0 X$ c4 G" W a1 := a1 and $3F; " ]3 X5 O5 H7 e' Y$ A0 f% K
outchar[si + 3] := chr (a1) ; 5 A- h; j8 l, s( `) W6 ^
* ?$ l& E4 C9 _ //第二次转换 8 Q& ~9 P/ Z) W( M" q
d1 := ord ( outchar [si]); //第 1 个
" o& L* }6 V& _" n$ d d1 := gamestr [d1]; U8 t) q; I$ N) X i
outchar [si] := chr (d1); 7 b' y! y. j# Q9 u2 g# h9 A
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d1 := ord ( outchar [si + 1]); //第 2 个 L$ C. U3 ^8 E% C
d1 := gamestr [d1];
! \+ L+ x/ h& U7 { outchar [si + 1] := chr (d1); ' g% E5 Q5 x' e% j( E
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d1 := ord ( outchar [si + 2 ]); //第 3 个 4 L% e7 Z. {3 q" M3 A+ Z: H! u
d1 := gamestr [d1];
; v9 E* N' Q7 w$ c3 x% W outchar [si + 2] := chr (d1); - y: D& ]. P* a3 a" Q
. w8 u8 z# n ~+ N1 u' V
d1 := ord ( outchar [si + 3]); //第 4 个
: V; l! M7 q+ p, C, z( j- W d1 := gamestr [d1];
' a# }/ f; n% H- z/ q& p* s outchar [si + 3] := chr (d1); & m; k& S$ i& j8 O- P
2 T. e5 B$ Z/ ?% X di := di + 3 ;
8 q/ E9 Y9 G" B. L' E9 V4 g7 S6 { si := si + 4 ;
5 O1 y: S* i5 a. h+ B+ J' N! a end;
4 \& v# @# x% h# h+ I8 ?7 S2 ~% oend;
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