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标题: 千年加密解密代码(delphi) [打印本页]
作者: 快要发癫啦 时间: 2023-2-14 11:29
标题: 千年加密解密代码(delphi)
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..
$ Q5 K) M+ I. U% T6 `- ?& n9 ?/ W* j$ ^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 v5 H. s, \; s3 a* Y
& U/ r9 R3 R0 _5 D m声明部分 $ T( x2 a- H4 e
//二种声方法均可 : \5 Q# t" s$ p3 H
//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’;
4 m* s. |: h' i* B* `+ y" Pconst gamestr : array [0..127] of byte = * g9 p4 j1 } s: [& M2 s
($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, $1A); : a1 r' R1 F1 Z* |" @9 A8 `
- J. J# J' c3 R4 k+ s0 ` t# A代码部分 L2 R0 x0 X) }# |# I% j
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
: }" U" R! e/ |/ w, @. Bvar ) o m* F2 q- }; U$ G
a1, d1: byte;
9 `0 a S: Z" H8 {, N& k/ \ j, count, di, si :integer; ( H. ]# Q7 z$ U* M4 ?
begin 6 [$ [. F! U; l4 E g) A: k
decode := len div 4 * 3 ; //返回解密后数据长度
# Y$ H& s8 D! U0 ]. k' _7 B! r j := 0;
/ l0 f& M" A( Y% v while i < len do 3 l m b: R/ J& h/ n3 g
begin
) m, s5 h: H. b% J, D! ` d1 := byte (inchar[j] );
& j: ^* A/ r! T$ n if ( d1 = $3B ) or (d1 = $7A) then
) b7 X8 B! q' z" [0 I @ begin . I: U- e) V) I
end; & m- G& }' f$ U! e1 v o$ Q3 ^
d1 := d1 and {&content}FF; 7 h M, a; \: d4 c* \# m% d
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] );
$ r/ r' ]/ w5 ^3 I! g3 u" ?9 a6 p byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法
2 m0 ?2 J/ @# q4 N/ D( `' Z$ x inc (i); 7 c2 O5 j1 t5 ]
end; : q/ w# ?' X0 c; @9 e9 v
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di := 0 ;
$ l. L. ]9 l: R! |5 m7 I! o si := 0 ; ' |" G2 L ]/ ~( y( y( I, X
count := len div 4 ; //循环次数
% Y0 _5 x4 Z( i: Q h9 ?9 l6 A! { for j := 1 to count do 9 l% t! s/ D+ q7 T9 h* W
begin 9 O& a# k, \; S% c
a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) / N& u6 {8 ?* _
a1 := a1 shl 2 ;
+ @1 O. m2 M$ Z/ h; k8 W d1 := ord ( inchar [di + 1]); 1 l/ W. A& f( r* @/ M
d1 := d1 shr 4 ; 8 j9 @8 x2 L0 ^5 V; k
a1 := a1 or d1 ; 2 x( D4 \6 }$ _ ]' u
outchar[si] := chr (a1);
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4 z$ g: w* @: t: _) i. _ a1 := ord ( inchar [di + 1]); % E2 T" ~: u' g) I4 O
a1 := a1 shl 4 ; 7 m5 Q$ b6 C' m" [; u# u
d1 := ord ( inchar [di + 2]);
9 v! G H) ^5 f6 f- I d1 := d1 shr 2 ; ! a( U1 X+ ?! v1 J6 Y: F& Z
a1 := a1 or d1 ;
V+ o* y. q) k2 }6 r1 X1 C outchar[si + 1] := chr (a1); - n) z1 o) p0 [, [# B% F
. Z* s& A6 I) ^7 }3 q b* { a1 := ord ( inchar [di + 2]); 1 x1 ]" Y7 c" {: J/ _3 C0 w- h
a1 := a1 shl 6 ; . }6 \. d. ]' S
d1 := ord ( inchar [di + 3]);
, t8 c' p9 v3 w5 ~; J5 d a1 := a1 or d1 ;
( U0 }; u; I8 {+ D/ k2 C! M$ j outchar[si + 2] := chr (a1);
2 q. l. I& R9 t1 U( y& ?
: D# h8 y' q! f- b; u di := di + 4 ; * M. s- ?- F% K. Y( } Z
si := si + 3 ;
q) N* y6 ?, K { end;
) Z0 v( C" C- Q% F) b. q* @) uend;
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
9 ?; J7 p3 i8 L+ Wvar # b- U) s; F+ D" ?6 U* k' l
a1, d1 :byte;
3 a6 J9 K, S6 a/ B j,count,di,si :integer; 5 G+ E' L4 ~4 f8 D3 u8 V+ E. H g
2 ^# w; x9 }5 ~. N4 N. dbegin
/ Z* Z. O) Y8 t7 [0 T# H9 a encode := len div 3 * 4 ; //返回加密后数据长度 ) R3 K, L: ~- [5 B, H
di := 0 ; % S' M7 f) _9 P
si := 0 ; 5 i0 D% C! R1 c
count := len div 3; //定义循环次数
0 \; X! c; K% D. [! r+ H6 r# o for j := 1 to count do $ F$ e! y ~4 r0 W
begin * }3 j0 [; q4 M. A
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个 ' X+ A; R u ~& H% I
a1 := a1 shr 2 ;
4 Q& L* Q6 _# k0 X8 t% V outchar [si] := chr (a1); - @% f) s7 a) n. Q8 p9 D, n$ `
. l( @, Z& i( b% \& Z a1 := ord (inchar[di]);
, }8 |# a) x7 ]8 i4 D( h& b a1 := a1 and 3 ;
: G8 E2 A P0 q. j a1 := a1 shl 4 ;
" t' {# ]$ W- G$ G# C d1 := ord (inchar[di +1]);
: L' R1 b/ R k8 n, G- ? d1 := d1 shr 4 ;
0 \' ^/ [1 R9 A% U: X5 P a1 := a1 or d1 ;
# Y2 O8 v" }, |6 I% x/ N outchar[si +1] := chr (a1) ;
3 v6 q% y8 r; P. r# B- ~: f1 }% Z' f8 e2 E4 Z/ w
a1 := ord ( inchar[di + 1]) ;
+ L3 f8 l% F0 o7 M" a7 v) G a1 := a1 and {&content}F ; 7 S$ }/ Q e! P& s) Z
a1 := a1 shl 2 ; " l; q0 g" J2 c% ]
d1 := ord ( inchar[di + 2]) ;
0 i. l" d/ K. L: z7 d9 C d1 := d1 shr 6 ;
! S* p" A% p2 s) \ a1 := a1 or d1 ;
\$ H7 }: \- J# \ outchar[si + 2] := chr (a1);
, E- D6 x$ A; ^/ m9 {
" g; M0 `: r# _$ j a1 := ord ( inchar[di + 2]) ; & [# \" W2 X, x( R, @
a1 := a1 and $3F;
- e# U; ?8 ?7 z* j2 s2 p* i outchar[si + 3] := chr (a1) ; i3 S* O) t( ?0 W" J* S0 J
8 a# {9 M+ M+ _ //第二次转换
4 E) i: W/ d# a. o d1 := ord ( outchar [si]); //第 1 个 0 o) Q& @8 z$ A* B: U( N& a
d1 := gamestr [d1]; , u3 w6 c2 T) X+ i' ~$ D
outchar [si] := chr (d1);
, ?; U# L2 ] V2 I7 `) @3 |2 W% V! C0 m$ g' I$ j7 V
d1 := ord ( outchar [si + 1]); //第 2 个 ( R- m9 H; Q: l+ B/ p' A
d1 := gamestr [d1];
' G: |: O. {3 ~ @- a3 h) |+ q4 i outchar [si + 1] := chr (d1); 4 P2 h0 y) o' a& z3 `7 @
5 f( s& |' N& a d1 := ord ( outchar [si + 2 ]); //第 3 个
2 |) l1 g. L9 M8 |6 c d1 := gamestr [d1];
! M) d" e& t) E6 [( D9 g outchar [si + 2] := chr (d1);
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: Z& S3 H4 O: l. s d1 := ord ( outchar [si + 3]); //第 4 个
: k- Q. a5 \! n+ l H/ h# T d1 := gamestr [d1]; * f! T/ }+ J5 [7 E
outchar [si + 3] := chr (d1); - E5 ?; `% i, U
: u: }0 p0 Z& S, j
di := di + 3 ; 8 v. s8 C9 w% T# y T! Q* U: f
si := si + 4 ; + Z' n! R3 h# o# [# A& I2 ~- v. I
end;
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