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: X# g9 a6 P# x
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) * K- S0 c% s6 O% ^0 o
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声明部分
( S, ~: f! O6 p% P//二种声方法均可
$ E* k4 U g2 z. }4 q2 L" D9 d//const gamestr:string[255] = ’N>TSVUJlwdcBMFjnAKb?qxvyeGzfLP=_ER@Z\am]ChgoWD;QuX[<tkpr^`iIHOYs.3" (- ’+chr(13)+’<;’+chr(12)+chr(00)+’=/!,1>#2$’+chr(39)+’89%’+chr(10)+chr(9)+’*):5&+67?40’; 1 C2 P% I3 q8 ^9 x' q1 k+ r7 c$ _
const gamestr : array [0..127] of byte =
- M& _( Y9 n- f2 e($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); # I/ F0 i; n) L; p; _8 f
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代码部分 2 l) F3 @* o+ a3 q" k
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 / o3 r X0 s4 ^1 x
var & z/ y5 \2 ~0 s) [4 b: p0 f( o# y
a1, d1: byte;
5 x' C6 g# F! G# M( V! N j, count, di, si :integer; 6 ^6 U- D6 m+ i. s% l
begin
+ V) D: M* a. C9 T' A/ X9 E decode := len div 4 * 3 ; //返回解密后数据长度 ! k+ P. ^5 N! D0 e, A/ t2 }
j := 0; ) u, k5 y% r4 X% Y. g
while i < len do
, _+ B) c% i5 r# C* w begin 8 ]4 p: t- \) g+ S. h. ]
d1 := byte (inchar[j] ); K$ Z7 u; i6 X. H9 u' P _
if ( d1 = $3B ) or (d1 = $7A) then
% p7 O9 L9 w+ Z, a6 G5 a4 j begin
2 [; o/ ~; b- m/ Q/ x* l end; 1 a% |9 |( N3 X0 V! h- }, L
d1 := d1 and {&content}FF;
, O+ v/ t0 F7 R. u, u d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
1 r: u" r. J+ i, {# a byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
6 \$ W: \1 [% X1 D: G! A inc (i);
0 l( K6 N" @; u$ A# }1 | end; 0 X. o8 x2 W9 K# j" x9 _
8 Y8 b* s# x, n% y di := 0 ; 3 i3 r& \. w0 v6 y5 e' v
si := 0 ;
8 w+ j( A! V* Y, N count := len div 4 ; //循环次数
& T5 T- |4 B) a3 M P8 O for j := 1 to count do ! X8 y' X6 k" K3 w
begin ' U. Y+ v) G4 L+ s* M
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) , j2 P# ]; g0 e V
a1 := a1 shl 2 ;
4 K. M. s1 V$ |9 ]. ?8 x$ c d1 := ord ( inchar [di + 1]);
+ }. e9 B$ a# [0 k+ P; ? d1 := d1 shr 4 ;
6 c1 q. i e& D; L a1 := a1 or d1 ; & }' P, [; U0 F1 F: z: u* [
outchar[si] := chr (a1);
# G3 ^6 f% {9 E6 c. [% T5 l# Z7 D$ _1 i- M$ ^0 b, ?
a1 := ord ( inchar [di + 1]); & \! C8 t2 b* t0 x& r8 a8 l
a1 := a1 shl 4 ;
" X- V5 o6 Q; b% P( `+ R4 Y d1 := ord ( inchar [di + 2]); - H |! A1 X# W b
d1 := d1 shr 2 ; 4 K% h/ S: ?& u- E
a1 := a1 or d1 ;
7 J7 }1 i; A' R& L4 ] outchar[si + 1] := chr (a1);
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, J! ?, G0 U& F! B4 H a1 := ord ( inchar [di + 2]);
" M# ]) l k( v a1 := a1 shl 6 ; 9 F; @) _0 B4 t2 Q
d1 := ord ( inchar [di + 3]);
, U" u: ?( I9 l) P8 v* J a1 := a1 or d1 ;
0 L' m% h' A; D9 }! O outchar[si + 2] := chr (a1);
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# a6 D% t4 X$ ] di := di + 4 ; 9 E/ _9 B- A" T' e8 o
si := si + 3 ;
( i, l) f* }) w' g! T end;
) Z; @1 [$ U0 V6 rend;
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 7 T, ~3 Q, x. J8 a1 J) V4 ]: a$ h
var
1 N7 [% l8 G; e# U a1, d1 :byte;
3 O3 K/ _9 ]& p( ? j,count,di,si :integer;
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6 p! w+ Q2 F! B6 w9 M) tbegin
7 \; Y& \$ l: m( d$ q encode := len div 3 * 4 ; //返回加密后数据长度
: H( w2 ]& M4 t! V4 w di := 0 ;
. ~2 u; ?3 C( a0 L6 C9 e si := 0 ;
4 b; L' h k; A8 v; w7 L! b count := len div 3; //定义循环次数
2 ]7 U' ?3 A8 Y. C for j := 1 to count do
% h% R' Y/ D+ `: y+ x, `9 i( z% y& v begin
3 H+ _/ U ]" V a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个 " i; A9 h3 [5 P+ R0 @( i
a1 := a1 shr 2 ; ) `* m1 s/ k; W( z: O
outchar [si] := chr (a1);
2 M0 R& E& m, t
# K/ W7 ~ W4 Q5 f: ~ a1 := ord (inchar[di]);
4 O: w1 I' Y6 d6 B a1 := a1 and 3 ;
, w$ ~ I. F6 T0 i4 C a1 := a1 shl 4 ;
) }7 W9 S! { c+ E d1 := ord (inchar[di +1]); " l" n1 H: c8 |) N+ E* ?
d1 := d1 shr 4 ;
$ m4 d0 u( \. [! K, L; R; p+ Z: ` a1 := a1 or d1 ;
2 `+ N' q3 L8 d& N. P* `' t( H outchar[si +1] := chr (a1) ;
+ r" {9 Q" ]6 w
- k8 t- o1 m# o* Q" b; Q a1 := ord ( inchar[di + 1]) ;
9 I( R" A& ~3 U$ ~1 x a1 := a1 and {&content}F ; ( D j! {" s! g' z( N
a1 := a1 shl 2 ; * J1 `& ]* s( ^( a& ^
d1 := ord ( inchar[di + 2]) ;
/ e- q$ ^3 o( ~/ Z8 ?5 k4 \8 E d1 := d1 shr 6 ; x4 G) A; B! P4 D" }
a1 := a1 or d1 ;
p. g: A1 _1 {, O# V( X outchar[si + 2] := chr (a1);
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a1 := ord ( inchar[di + 2]) ;
. k8 j- X$ x& b a1 := a1 and $3F;
5 k1 X% g0 L' T outchar[si + 3] := chr (a1) ; 1 R* d& {% g2 e' z& |
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//第二次转换
* _6 d) L$ y8 Z d1 := ord ( outchar [si]); //第 1 个 e" U7 Z. f! F' q+ [
d1 := gamestr [d1];
0 |) B& O& I4 d8 _! w3 t/ [, v outchar [si] := chr (d1); & I2 h0 ~9 y1 S1 A( B1 k$ p
# n# P& b. e3 q1 u7 o d1 := ord ( outchar [si + 1]); //第 2 个 7 z. a* S6 ]- z. e% { p
d1 := gamestr [d1]; 8 N, X: n7 f, c7 X/ Y6 `* i+ x
outchar [si + 1] := chr (d1); 2 w& H2 L' P! n6 P' A6 \
+ ?0 \& N% E6 O4 h- B+ D d1 := ord ( outchar [si + 2 ]); //第 3 个 8 a: P9 D V8 G) p4 s5 s& ^; k$ I
d1 := gamestr [d1];
1 X) R4 p7 R5 `0 p# R outchar [si + 2] := chr (d1);
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d1 := ord ( outchar [si + 3]); //第 4 个 . {5 a1 y: j. Y
d1 := gamestr [d1];
9 ?" J' q, U+ i3 k outchar [si + 3] := chr (d1); $ f9 g; T% c3 E! g9 |! Z# ^% B9 X
7 l% v! g& e: h R% k9 z7 y di := di + 3 ;
" s( l$ F; m& Q7 K- m si := si + 4 ; ' N s x ^- ]( ^* s& e
end;
0 N- K" D( N) ?, A0 Qend; + F! Q, w" F y% r: E
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