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..
4 V' v9 X$ Z- t9 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) / `* I9 Z: i& u
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声明部分
6 j" I( E- W! v" a//二种声方法均可 & H Z4 {, I0 r7 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’; / g( e1 }9 j8 y, |' @* L. ~% r
const gamestr : array [0..127] of byte = / V/ Q9 \ L8 v7 X1 j
($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, $1A); 9 F; a, q, n# R3 u& ?
& T" i# H* J( P代码部分
# k2 j- w3 K; s9 J: c% m) yfunction decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 & K( @( }6 \* v9 u5 {
var
( R! u3 m b Y/ L; t6 L1 H a1, d1: byte;
# ^6 @2 `* C/ u& s9 ]/ o j, count, di, si :integer;
9 m7 Y9 _4 K# O! M Ebegin
; i p' z' I" Z. ~3 g0 c decode := len div 4 * 3 ; //返回解密后数据长度
# S$ `( t; n3 p# R1 n7 P j := 0; " l8 O( g- f' ?: O& n% u' o1 d3 R5 w
while i < len do 9 k0 x: Z1 h# m; C, o( B
begin : ~1 I! }+ _1 _" L, [) r! h! E
d1 := byte (inchar[j] );
: q7 M: r4 C6 \. ]7 F6 z if ( d1 = $3B ) or (d1 = $7A) then
7 i2 G$ |) E J+ y& ^) l/ l: n begin
. {/ h; e7 Q/ a! i6 @' T5 c end; . x8 n, @# ^: M' Q' G
d1 := d1 and {&content}FF;
& c5 x9 O7 ~3 l8 ^/ p3 ^ d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] ); 0 @8 U2 _* s7 |' K& p
byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
* ?# t* R x. j, A) k5 y8 u inc (i); * V2 S5 m8 k- |$ T8 K
end;
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di := 0 ;
0 K7 {* \0 D7 v' R9 ^ si := 0 ; 0 N, ^: _+ k- S: r' t5 I& i/ M
count := len div 4 ; //循环次数
( J$ X" m- s1 [* n% G! L* M for j := 1 to count do
. F1 e3 s/ F0 q begin
7 o' D0 v9 T0 m( | a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个)
+ K7 `, f3 H/ N: ^* O7 y a1 := a1 shl 2 ;
2 H# Y" ?& S# @ d1 := ord ( inchar [di + 1]); 5 j; _# h) h0 G+ _$ B, V
d1 := d1 shr 4 ;
1 ]- f( R! V) Z C" Z2 L/ d a1 := a1 or d1 ; - b& D5 d$ K6 U/ H
outchar[si] := chr (a1);
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a1 := ord ( inchar [di + 1]); , N9 D3 u) n1 s. H/ E5 d
a1 := a1 shl 4 ;
: g$ f: T8 T5 {- D& M7 K6 [9 W/ s d1 := ord ( inchar [di + 2]);
6 U8 ~( y# J1 r/ i) W' l d1 := d1 shr 2 ; ( G/ P% O0 P; X
a1 := a1 or d1 ;
0 z4 |: W5 T7 a, S8 u! A& q outchar[si + 1] := chr (a1); h% S. ?. v5 ^2 m1 Q! \# I
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a1 := ord ( inchar [di + 2]); 4 k5 y+ ?8 S8 p* ?
a1 := a1 shl 6 ;
4 z) g3 w' |. K) e d1 := ord ( inchar [di + 3]); * t4 r& e% k+ ]8 m6 e
a1 := a1 or d1 ;
: ^0 X3 r5 j, R outchar[si + 2] := chr (a1);
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di := di + 4 ; 3 Y1 b4 H' f- Y) E6 Q* s5 S( S
si := si + 3 ;
0 Y9 h5 |" `* @) o# J end;
8 u6 I! e( w! K) y7 Lend; $ X. \+ `" y: a3 E
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 5 ]/ S- r! o0 g/ `
var * t7 m: Q3 X; ]. v
a1, d1 :byte;
, X; v- a" g- R& _( r& F, B j,count,di,si :integer;
# S- B& l/ q) t# o9 Z% V
2 E3 k2 n( h) z+ C7 d" ]0 fbegin / q7 ?! L& m. U4 s2 `
encode := len div 3 * 4 ; //返回加密后数据长度
! T& y# |; u+ I/ X di := 0 ;
1 ` h' `' S+ U% k# L5 e8 r si := 0 ; * _ `7 Y! v# |( D. n
count := len div 3; //定义循环次数 5 a4 e; j5 R0 V' B4 t& i6 o
for j := 1 to count do # L5 G# c0 I( n$ ]3 M% u7 ]: S. S
begin ( c$ @0 b; D$ w& q$ h: l
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个 & a3 [7 v L) E. `
a1 := a1 shr 2 ; + F5 ]4 {* `2 v/ @ k
outchar [si] := chr (a1);
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! `9 u" A. T: F% A+ g3 Z a1 := ord (inchar[di]);
& z3 U' E) i, S5 P9 E, q8 e a1 := a1 and 3 ;
+ }2 c; `% T7 v, v2 R$ P a1 := a1 shl 4 ; ( Y! K% `$ t, y) W) \8 r% G$ Y
d1 := ord (inchar[di +1]);
$ j* p/ {0 N1 i" U& i d1 := d1 shr 4 ; # c) Q( |1 D3 B* u; f: t. T9 m
a1 := a1 or d1 ;
: h: N, [2 u/ P; Y outchar[si +1] := chr (a1) ; " `2 a0 c. k9 `/ N9 O
L( @! P7 }; i1 b% C5 P1 K2 ? a1 := ord ( inchar[di + 1]) ;
. G. k8 n/ f& u8 L- Y+ v a1 := a1 and {&content}F ;
$ S {" T& v" x, U" o a1 := a1 shl 2 ;
5 \6 z3 U7 e# E9 ^! _8 i2 O) [ d1 := ord ( inchar[di + 2]) ; / \( e& n: ?, k7 c: b
d1 := d1 shr 6 ;
. `5 r i" k7 a a1 := a1 or d1 ; , ]0 ^; p" {1 o5 q' E Y
outchar[si + 2] := chr (a1);
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( L7 x" b; f0 H; S& Y* X! L a1 := ord ( inchar[di + 2]) ;
& m- z+ u8 M2 i8 d. I4 i a1 := a1 and $3F;
; o4 k# \# p+ E8 E0 [ outchar[si + 3] := chr (a1) ; 4 M; K% e5 f1 Q8 {1 _! o( z8 e( o
9 H2 z r4 a$ _8 U" Y% H- B( Q- i //第二次转换 # H4 Q0 ^ n! J: b- [7 l1 v
d1 := ord ( outchar [si]); //第 1 个 }' V/ I4 a2 y+ K/ |2 `
d1 := gamestr [d1]; 0 k }3 y3 c7 I( M4 \7 |2 x
outchar [si] := chr (d1); $ x) _* I& o P
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d1 := ord ( outchar [si + 1]); //第 2 个
- F+ `) ]3 X3 f* Q# U1 [, m3 r d1 := gamestr [d1];
/ ?' Z+ D% b6 J; y- R outchar [si + 1] := chr (d1); % N5 \* `, L$ V
9 C2 g% T2 }+ ]+ P d1 := ord ( outchar [si + 2 ]); //第 3 个
( j) E. J+ `& Q- b$ E d1 := gamestr [d1]; ; D; e# y$ O" x7 d
outchar [si + 2] := chr (d1); 1 g+ S6 x+ }) x* F: p
( _1 I8 v9 q' p d1 := ord ( outchar [si + 3]); //第 4 个
; ^( V" Q; ?0 E$ j d1 := gamestr [d1];
/ s/ i! g( L% l outchar [si + 3] := chr (d1);
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di := di + 3 ; ) q- n2 Y& J4 T0 x1 q2 M" B8 x
si := si + 4 ; ' q$ j9 a) A5 t. F' x$ x' x
end; - v7 ]3 K) @( H G* ^& [* x
end;
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