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..
* C' i1 ~7 o5 L3 W. 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) - I7 |6 m- Z: D" j" D
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声明部分 , q/ @3 [# R: T- y1 ^2 v4 \
//二种声方法均可
. Y: z' Q; p6 r; M+ I# a4 S2 P3 f//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’; 2 u9 u+ o* L. x% M+ t
const gamestr : array [0..127] of byte = & F# [' H, H# M: f; r+ W! I4 P# U
($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, $1A); , ^( t8 m$ R+ ?2 w/ Y2 q! ]
0 x" t/ y! \4 V2 K5 R; ^代码部分 & U& K- r9 e' |1 ]; Y3 v
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
# ]& J) N! u% R, ~2 e. {3 c! gvar
0 Z) q8 \2 Z4 h a1, d1: byte; 1 F* e; w; n5 @8 |
j, count, di, si :integer;
6 ]% P! h. f. H8 } }" N O- Xbegin
! [% N2 L0 i0 x1 ]3 d0 ~3 @ decode := len div 4 * 3 ; //返回解密后数据长度 % q3 i6 k4 V' k' v0 N9 V2 k) `, m
j := 0; 9 T4 O$ }2 R+ F7 c% |1 e
while i < len do
9 s# J' G/ U0 k5 \4 }9 q) j% B& m- Q! } begin
# g, q' |) g+ p3 h d1 := byte (inchar[j] );
0 X0 c8 Y5 f# E$ R& u m if ( d1 = $3B ) or (d1 = $7A) then ( Y/ v+ r6 O, n+ r, _
begin " |' e& X f9 U/ E! m* B0 Y! t+ x
end;
& p" F# T: `7 f; k d1 := d1 and {&content}FF;
V M. ~, U- a3 c$ F, K4 V0 N d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] ); ' m8 B" Y1 J: Z" M+ ?# v
byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 4 q1 ~3 U2 E' v5 q K0 ?# r
inc (i);
# i l# L: b9 ?7 h end;
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di := 0 ;
) J. c. [# ~* o4 S/ m; G8 ` si := 0 ; * q% i7 K, J5 k. _2 S3 i. ~
count := len div 4 ; //循环次数 0 d& m3 S6 r& J4 O
for j := 1 to count do
" e `" V. T4 G( w begin , k% d- f, v6 `1 d* ^" J( E5 |
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) ' ^6 v% k% T) A1 m# S F
a1 := a1 shl 2 ; 5 D# g! k; b, X! {* ^. Z0 `: d3 T
d1 := ord ( inchar [di + 1]); 4 }# \1 {1 D# g5 M+ l
d1 := d1 shr 4 ; * O# t5 Z/ [5 w8 x4 W/ l0 ^
a1 := a1 or d1 ; " \% \! b4 h1 A7 d4 {+ _
outchar[si] := chr (a1); " t6 m- r+ K9 v% f) R
: T7 f6 `, ?/ @5 @7 f7 m( S a1 := ord ( inchar [di + 1]); # X. ^- ^+ r. J! C
a1 := a1 shl 4 ; @( w6 X2 D" h% R, \3 V
d1 := ord ( inchar [di + 2]);
- R' F" W$ p' H- X$ [ B d1 := d1 shr 2 ; . h# O% A$ ^. P$ D- p, ]4 Y
a1 := a1 or d1 ;
$ M, L9 Y+ l; m$ n7 ^9 s' c$ @ outchar[si + 1] := chr (a1); 5 T# X8 z0 K8 e! } i1 t* K
1 y; n/ z% S' z1 W' y
a1 := ord ( inchar [di + 2]);
1 @2 _& x& B* I, i: r a1 := a1 shl 6 ; 4 Z1 Y- ~9 e/ c
d1 := ord ( inchar [di + 3]); ! z- u2 K0 z; P8 c0 ]2 H
a1 := a1 or d1 ;
9 E5 F' Q+ f9 |& ?. v0 o outchar[si + 2] := chr (a1); ' U9 R0 x9 ?) `* l
9 j9 P6 Z; O/ r% V% D) h [. g di := di + 4 ; * X2 r7 u% J. q8 g; [
si := si + 3 ;
4 e v5 O) Y5 p8 b end;
. W( `. H% ~1 Z% ^) T+ bend;
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
3 t: K3 b5 ^% e3 Z4 x( x! w& uvar
4 W: G8 [/ n3 O6 x* X5 e a1, d1 :byte;
% y2 i# D% r, i/ [! d0 Q j,count,di,si :integer;
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5 ?4 c: n( k3 J1 r$ j: dbegin # J( h1 J( _: J$ H$ c* d
encode := len div 3 * 4 ; //返回加密后数据长度
, J4 y% ^9 ^& ^. ?( D di := 0 ;
0 @* a+ ]; H+ R; Q9 M* Q4 P si := 0 ;
9 _2 P2 D8 V0 ~! d' t count := len div 3; //定义循环次数
# n) s; n+ e# x& N& j9 ? for j := 1 to count do $ c1 d1 S- ?& I8 B
begin 5 ?5 R$ G1 L" }* L o
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
5 ^% c7 @/ \" f2 K. Y a1 := a1 shr 2 ; 6 Q: O+ b8 P7 s, ?0 [; S4 ^. o
outchar [si] := chr (a1); 5 }0 |; M _2 ?/ y O
" \* x2 n1 f2 z% ~1 N# e
a1 := ord (inchar[di]);
5 g) s" l' ?3 i) o3 d9 b N a1 := a1 and 3 ; / \" N0 J6 ~: ]/ k
a1 := a1 shl 4 ;
; J" B6 x6 K. e! t, C d1 := ord (inchar[di +1]);
& O6 t1 j9 T8 G d1 := d1 shr 4 ; " A- |$ |7 N' R0 P. b! `
a1 := a1 or d1 ; - @) J7 w9 }+ C3 d3 w* `
outchar[si +1] := chr (a1) ; 2 f( V& O& V' f: i) Y' B2 u
5 d/ L7 X3 N. B0 w) _
a1 := ord ( inchar[di + 1]) ;
; q! |$ ]" T1 V8 f a1 := a1 and {&content}F ;
$ r" [( t+ [5 j. Y& Q a1 := a1 shl 2 ;
6 x% F% p, a6 ?; `3 ~ d1 := ord ( inchar[di + 2]) ;
( z& L' t& n% l2 r* S& L d1 := d1 shr 6 ;
9 G7 t: ^; g- ~, F6 [, K( E4 K a1 := a1 or d1 ; 5 Z N* \7 [- j7 |4 Q
outchar[si + 2] := chr (a1);
3 d+ o' v9 v4 M4 |+ b! {
" e* R) A+ |# i+ e a1 := ord ( inchar[di + 2]) ;
' o% ^2 z8 W8 r a1 := a1 and $3F;
, [# _9 p8 i6 s% p: | outchar[si + 3] := chr (a1) ;
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//第二次转换
4 L- @ z- o. ^ d1 := ord ( outchar [si]); //第 1 个
3 w' H7 U! b$ l2 F d1 := gamestr [d1];
8 z: ?9 h" L5 L1 ^6 X; }5 w5 C! l outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个 ; Y! R( \% ^; h/ r0 G6 U
d1 := gamestr [d1];
8 R4 b. ?+ G/ h0 A$ }1 W( c outchar [si + 1] := chr (d1); 2 y9 X3 Z% F& z, U+ W* t( y
) ^ u5 T, E* [& e d1 := ord ( outchar [si + 2 ]); //第 3 个 - S+ Q& U, D, S1 W
d1 := gamestr [d1]; ( O# |- a8 U& S9 j& R6 s
outchar [si + 2] := chr (d1);
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; U2 X4 D* s! s+ l- w3 }1 a d1 := ord ( outchar [si + 3]); //第 4 个 : v( N, f- D2 s" A2 U+ ?
d1 := gamestr [d1];
/ S4 T- d4 r- ?5 e/ R outchar [si + 3] := chr (d1);
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di := di + 3 ;
" {+ d' \& t3 l* M( D$ Y/ L- ` si := si + 4 ; 1 }! D; b! q: O/ V9 M9 a; n
end;
$ v% v+ J3 C" oend;
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