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.. . `, m: R2 b/ W4 B: W) G
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) 5 i7 \4 c! Y6 ?, S+ h
) r( v7 ^* S% [1 [声明部分 . g, L& ]( Q# s. L7 G d7 Y
//二种声方法均可
' K3 u) f( E" z( J2 L6 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’; [7 r0 t V9 T
const gamestr : array [0..127] of byte = - S* ~& O0 `! `! g7 V) u5 e) J# f
($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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' c6 K1 X1 p% i; z% X3 { S代码部分
3 K) z4 \& w+ G! Efunction decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 S0 n5 d. _5 n7 e; i
var $ ^% f3 g/ Y: S- p9 \
a1, d1: byte; 2 O/ o' h" x7 y! y! s' o
j, count, di, si :integer; 1 I4 }% A8 W. A5 t/ d* P7 F
begin # p, B5 E! \7 _; W% k4 u
decode := len div 4 * 3 ; //返回解密后数据长度 . a$ L) H; I$ u' p! V7 z
j := 0; 9 a2 L) I2 N% R4 l
while i < len do q' V) ?* n4 B5 u! N# Y
begin
* I% n2 I; U5 G( U' y; L0 O: b$ \ d1 := byte (inchar[j] );
. S/ W; L+ f; [6 F if ( d1 = $3B ) or (d1 = $7A) then # F" e8 j; _. X: g" d4 S, x9 j
begin
9 Y# B9 }7 Q: c+ @ end; 5 `5 r7 b4 f; A
d1 := d1 and {&content}FF; . X. D# I, t) z
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
1 n. P/ @! ^6 E) `7 b" G) l. I, ] byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 , \0 S; P0 X- I6 c0 k( |, }' k
inc (i); / A# K4 e4 c) [2 q% [1 R
end;
# j2 M! G+ z9 M1 S# C' `$ y6 |# o5 z
di := 0 ;
# R2 ^* n; s0 N" C: d; e: } si := 0 ;
8 b' M s6 z& A' I/ O count := len div 4 ; //循环次数 + H0 s* b2 w( S8 _' d! L
for j := 1 to count do ' i4 R0 e6 K5 H o& g8 E, h! }
begin
% @% G8 z2 w' ?7 o; A. u. Z a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个)
7 Z! H, t' d9 c% r: P3 ? a1 := a1 shl 2 ;
0 o2 ?2 g& H: d" Y+ Y' W' a d1 := ord ( inchar [di + 1]);
j" s' _' J' n' M- G* A' C d1 := d1 shr 4 ;
2 ~- ^, T0 l/ B( i) } O- m, h a1 := a1 or d1 ;
' \2 m# h5 \2 R) g outchar[si] := chr (a1); & V3 C" c+ c6 w3 z0 [
7 s+ ]0 a3 ~& R7 W7 [- F6 I a1 := ord ( inchar [di + 1]); 9 i9 Q$ b5 `6 P% C1 L
a1 := a1 shl 4 ; % _7 c4 u O( v8 T; E# g5 B8 R
d1 := ord ( inchar [di + 2]); ' j, t; A! |! o$ ^- W1 T5 H3 q1 m& U8 D& Q
d1 := d1 shr 2 ;
; o3 O, q7 w( G# Y% d6 G6 i& R* M a1 := a1 or d1 ; ' a8 x4 r- Y: {0 Q! W$ t' r8 J' \% H
outchar[si + 1] := chr (a1); ! x+ G' f9 W$ T% B& G
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a1 := ord ( inchar [di + 2]);
! x5 J6 z, ^/ r5 t W8 w a1 := a1 shl 6 ;
5 T7 r/ I5 K& I1 L$ u& x+ k d1 := ord ( inchar [di + 3]);
. o. ]7 |1 x& ^, l( f6 q" } a1 := a1 or d1 ; . ~ H3 d; S7 Y
outchar[si + 2] := chr (a1);
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( _+ G4 I4 ^7 L0 F3 C, d2 f5 P di := di + 4 ;
7 S/ e% j9 O+ B1 y/ ~. R5 p si := si + 3 ; , |/ X" @ m) }/ v5 P: p
end;
0 L" j( M6 x( \% Dend; 8 R, Y1 q& k) \' M: Z
2 J3 D) P) g5 _4 efunction encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 % H# E1 t6 y" x+ @0 F% w d' b+ b9 a
var
3 G& o. a( c' A" {* c: Z" }0 v' X a1, d1 :byte; * o+ k m& @# J6 z0 D- T( |* J. V
j,count,di,si :integer; ! q, e1 X/ J7 S8 O! j
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begin
8 n) K" }1 |' B" X% \( j4 k( D encode := len div 3 * 4 ; //返回加密后数据长度
. S: a3 M! j' Y8 f6 |9 R9 u di := 0 ; d3 p* w% H) t4 t
si := 0 ; + o6 w: R3 ]% d( h7 r) f
count := len div 3; //定义循环次数 5 X2 r9 u" Y% D: x) U$ s
for j := 1 to count do
9 G; ]+ R2 I$ m; L' E7 T! k2 E; U P begin
* k, ^. e: L" i: ?1 ~ a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个 ( H2 m# q# S+ [+ D1 Q( M( S
a1 := a1 shr 2 ; S0 R9 f5 d8 ]3 o- g. C2 M
outchar [si] := chr (a1);
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/ z7 b. O: {! V, M2 \ a1 := ord (inchar[di]); + c7 O9 Q/ L8 n) a1 r! t
a1 := a1 and 3 ; 6 r1 @! }( z8 Y! R @3 `
a1 := a1 shl 4 ; $ t' R* K3 P- |8 Y2 _8 k$ O
d1 := ord (inchar[di +1]); 1 A$ r" D7 ]7 o3 ` k+ x9 r
d1 := d1 shr 4 ; , I$ o* e4 U9 M, ], K
a1 := a1 or d1 ; % \( D) n' f& U; W% i% P& `# J/ a
outchar[si +1] := chr (a1) ; ! h6 r9 ~ u" b& u3 a( b/ k' b
5 G+ X& C# y$ \2 p a1 := ord ( inchar[di + 1]) ; & Z5 k* X7 Q( R a. m9 t+ Y
a1 := a1 and {&content}F ; 1 V$ E+ S4 l# f3 o$ C' e
a1 := a1 shl 2 ;
0 C4 N, f6 n$ `4 X1 N: s) o, V d1 := ord ( inchar[di + 2]) ;
4 Z. G+ ]2 ?" W6 ]( \: ^8 J' G: O d1 := d1 shr 6 ;
5 X- v! y5 D9 e- F1 Z0 d, Z a1 := a1 or d1 ; 9 v/ C0 u( D( z; c
outchar[si + 2] := chr (a1);
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a1 := ord ( inchar[di + 2]) ;
* A* m4 B8 T, q+ W8 |" ] a1 := a1 and $3F;
& `8 k! t% z! e outchar[si + 3] := chr (a1) ; 2 Z& Z% M: N b/ e, w. B0 }
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//第二次转换 % ^# R( K1 H( Z8 ^ t
d1 := ord ( outchar [si]); //第 1 个 + n( v9 c) M& L1 B. `7 [; J
d1 := gamestr [d1];
3 o; c5 s9 f( r& y outchar [si] := chr (d1); 1 q7 j, W* d3 N0 `$ S
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d1 := ord ( outchar [si + 1]); //第 2 个
! ?2 Z! V( P, A# ]2 N d1 := gamestr [d1];
- x' V( N6 j6 `: K outchar [si + 1] := chr (d1);
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/ V) u2 o0 c$ I6 H. t d1 := ord ( outchar [si + 2 ]); //第 3 个 , o% Z1 K0 b: v
d1 := gamestr [d1];
+ `/ N8 ^$ r6 C9 { outchar [si + 2] := chr (d1); # M& {- j# D5 h5 a
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d1 := ord ( outchar [si + 3]); //第 4 个 4 r1 V+ E! G" | |$ |0 ?
d1 := gamestr [d1]; 1 B! A$ R' {6 f8 W. o
outchar [si + 3] := chr (d1);
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di := di + 3 ; / ?5 _% }( @, u* |1 }
si := si + 4 ; 8 r& ?6 V( P! X! u. U
end; " P+ i% X, }/ [- ]+ a4 g3 x$ n
end;
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