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..
% `% b) h( Q, R6 i& h; 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) - v" S. [2 P6 \& e& V
1 n+ ]% u5 P8 p1 t+ G7 c a# p声明部分
5 ?4 x I, Z) j0 i; f2 [//二种声方法均可
0 G# _) a0 Y. X; g( G//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’;
$ Y1 `# m1 J7 Wconst gamestr : array [0..127] of byte = ) Y- r1 ^7 R+ G) j2 }
($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, $1A);
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8 S4 M A( p1 j$ L9 f3 u代码部分 & K$ j1 m2 F4 j1 o+ H2 E
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
$ C6 E& T( B$ kvar
& ]3 Z4 m; a& d* _ a1, d1: byte; 0 g4 ?2 s# N" v; |
j, count, di, si :integer; . g( D0 }2 P7 j2 A9 i F/ @: I
begin ; w* L' I' Y G0 q7 P5 ~
decode := len div 4 * 3 ; //返回解密后数据长度 & i5 D/ W& o# ~5 i; }; Y4 ~- W( w
j := 0; / B1 ^* b& t) X3 e
while i < len do
1 W7 H0 \/ r2 }/ d5 a begin 2 G! \) D/ e V2 v
d1 := byte (inchar[j] ); 1 k% @* d3 O8 x$ d H7 p
if ( d1 = $3B ) or (d1 = $7A) then
& Z# o% ]8 @( s3 u, @9 N begin
! v/ P. g$ w' |2 D7 v end;
' P0 R. u/ P7 B- Z4 }& X8 K d1 := d1 and {&content}FF;
f6 i# X/ M% H& @) Y, A. {9 r# w d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
) T' A& N# k: I$ A* g; b byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
0 T" |$ T8 D( ~2 g. m& S9 e inc (i);
$ A" Y8 [' o7 Y A! A, m2 \- ]' ` end; % X6 Q' X. O) r
4 `8 o' [$ x; u di := 0 ; % Y* T- Q9 A; C: M; [& l! o
si := 0 ;
$ d, K9 j. T9 G6 m8 Y) \+ e( F6 J count := len div 4 ; //循环次数 4 @0 L5 d0 x. M
for j := 1 to count do , ?+ _' \; H; i4 s0 f- P
begin
- i! e9 d8 q8 r2 T1 _- r( g% ] a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) + M* X7 `8 p1 w7 U6 A+ k9 m$ o
a1 := a1 shl 2 ; * Q& c4 l: }# Y4 D4 V$ K5 J
d1 := ord ( inchar [di + 1]);
2 n z2 R# s* o# Y9 P! d- f d1 := d1 shr 4 ;
& |5 J5 B O0 [* T a1 := a1 or d1 ; ' j# @) N( _* w u' t* ]
outchar[si] := chr (a1);
* z* |3 U5 M8 D; {- S. E Q$ z7 ]
$ c1 o+ i& y7 _& B s. w7 n y6 B a1 := ord ( inchar [di + 1]); 6 E' T& B0 f8 N
a1 := a1 shl 4 ; 7 i+ s* @- }) \: o
d1 := ord ( inchar [di + 2]); + c3 c/ y8 G9 k' r& d! D% ~
d1 := d1 shr 2 ;
' ]9 }: ?& E4 l a1 := a1 or d1 ;
# Q7 j+ X3 \9 p outchar[si + 1] := chr (a1); o8 o/ J7 d) N0 \4 z1 m
# l4 f( s+ F% U a1 := ord ( inchar [di + 2]);
1 \/ t0 [6 `0 a- J; H* j3 X a1 := a1 shl 6 ;
/ y$ }! e5 O7 F0 m; D0 A1 y( X/ G d1 := ord ( inchar [di + 3]); & Q9 E2 _& e+ m! m6 X
a1 := a1 or d1 ;
0 @ v# S G# d" B8 ?3 s outchar[si + 2] := chr (a1);
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9 w# V5 I; S7 b% ^: |7 V3 D di := di + 4 ;
8 h8 |7 |" x' ^/ z* X9 ]8 i si := si + 3 ; 7 w+ _+ @+ }5 a( z8 T) C' h
end; ( `# l; `* d" a5 \% W, r2 O
end;
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( B7 y7 `3 J! n6 c' ffunction encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
3 j8 S: l$ B! n& A! W, z8 V- `var
: X* \9 A( T0 v a1, d1 :byte; 2 D E3 x) @1 k3 B6 i$ c
j,count,di,si :integer;
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begin
! i w8 F" f- g T, q encode := len div 3 * 4 ; //返回加密后数据长度 1 M' Y) F, q: `3 W( y* P
di := 0 ;
+ T, y8 W* P9 Y: o; F: W si := 0 ; 8 ]4 [/ A$ y6 J, N8 F5 B) j$ _
count := len div 3; //定义循环次数
% i/ ~ V' T4 O) j! k for j := 1 to count do & F+ l& v% C4 m: q- T* N
begin
! l$ R" } r* ^+ @ a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
8 n7 o8 o; ~6 @ R* A9 E3 x a1 := a1 shr 2 ;
1 R3 l" d T7 m; Y- g6 k, O& k outchar [si] := chr (a1); - m7 P' z. X" i0 K' d# O
2 F0 Y; S* Z4 H5 _* P$ ]* O a1 := ord (inchar[di]);
- K6 C! `# y( G p0 Y( N- o& K) c a1 := a1 and 3 ; / J! Z' m( c; V. J4 V" p
a1 := a1 shl 4 ;
7 T P m/ [; c% Q4 y) X d1 := ord (inchar[di +1]);
) o4 y; v2 n5 S! o d1 := d1 shr 4 ; 8 {, q- D8 m/ ?
a1 := a1 or d1 ; / k$ i; q6 F6 D8 f0 O
outchar[si +1] := chr (a1) ;
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2 Q s5 `; ~! ?% Z+ y. V a1 := ord ( inchar[di + 1]) ; 4 ^8 D+ [4 b1 o2 d
a1 := a1 and {&content}F ; 6 }& Y8 x- v6 @2 n
a1 := a1 shl 2 ; : d+ Q; p: g8 [, z. E( h- }" b
d1 := ord ( inchar[di + 2]) ; 4 R# \; H; s/ p9 n
d1 := d1 shr 6 ; 5 F% H& N; r% s' H
a1 := a1 or d1 ;
1 A) h5 J i6 S2 @( g outchar[si + 2] := chr (a1);
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a1 := ord ( inchar[di + 2]) ;
$ q5 U l- E$ f9 O: |; }, g a1 := a1 and $3F;
+ B* h! ?, H5 U outchar[si + 3] := chr (a1) ;
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//第二次转换 , y# f3 k! @1 A) r1 [3 y$ k
d1 := ord ( outchar [si]); //第 1 个 & e( `7 @7 N! c# e
d1 := gamestr [d1]; , S. M$ \6 I: y6 H; P7 ^, Z
outchar [si] := chr (d1);
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k0 Z2 S4 n8 K% v5 d d1 := ord ( outchar [si + 1]); //第 2 个 ' t4 Z% z3 W8 [: B
d1 := gamestr [d1]; : ^+ r/ r; O& I" m7 f
outchar [si + 1] := chr (d1); " S, y% K- e: j6 A: L `
/ n7 {# F! Y! y% O d1 := ord ( outchar [si + 2 ]); //第 3 个
) X3 R1 n2 L5 n% J- Y/ I d1 := gamestr [d1]; , r$ G! O; ?# c! h
outchar [si + 2] := chr (d1);
" [- N# ^' |1 h) b9 W6 U+ ]: g7 ^% o8 ^5 J n, F/ `, @
d1 := ord ( outchar [si + 3]); //第 4 个 9 I7 q! G7 @/ C, ~5 T7 n, B
d1 := gamestr [d1];
2 J; `4 H m1 G outchar [si + 3] := chr (d1);
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( t7 u# B3 P1 k; B di := di + 3 ; + l9 l2 O R( ] h
si := si + 4 ; ) H" C7 P+ E/ @4 Q
end;
+ F; O! |* n) l' {0 Dend; - ]& P+ z2 b! d: k- W
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