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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..
5 ~; ]$ y' B! X4 Q. D, w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)
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声明部分
, P/ P2 ?1 m0 m8 f o//二种声方法均可 - O- C* o9 S8 T
//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 s5 O2 l! o) l4 wconst gamestr : array [0..127] of byte = 8 L2 w/ ^4 a2 n, t
($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, $1A); ) t4 \% ^1 Y/ W$ O7 u$ q7 ?
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代码部分
' V7 h$ C* ^$ r; m& z. rfunction decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密
* m) L& L! W: t: `3 p5 ?var 0 w% V- m! o G7 ?
a1, d1: byte; ' F0 r0 \4 `1 s& J" \' T* W
j, count, di, si :integer;
# p2 u& D( i! s r1 h# D" zbegin . x- |( k: n P) d; X2 w. G
decode := len div 4 * 3 ; //返回解密后数据长度 * O6 v" W: h) s9 S& W
j := 0; ) O, @+ ]' w$ J3 G& c! o
while i < len do % B4 C0 g; p0 {# q
begin # f6 s" s' f6 h% u8 Q
d1 := byte (inchar[j] );
X* D Y2 a+ E! g* } if ( d1 = $3B ) or (d1 = $7A) then , ~# o3 t/ T/ @
begin 4 F6 w) A; D) T
end;
1 N) C2 N6 Q8 c" A8 K. w d1 := d1 and {&content}FF;
) d4 A! A; t7 K" z6 [# |- g d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] ); ; e0 k' K, s2 _' y
byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 / @: a4 ^( z1 J" u
inc (i); - Q( f( Y8 f& Q! w7 V
end; 8 y0 `+ C5 y7 \. u- [& }8 |1 o( _
, e1 W+ u% n7 `( e) q( q8 y di := 0 ; ) h. f2 s4 N, j* S5 G- D
si := 0 ; & w+ S0 j; t, V0 Q2 k
count := len div 4 ; //循环次数
7 m6 @! O$ l, Y2 i/ a0 s7 R" S for j := 1 to count do 1 i* v3 p. S- D$ a
begin
: @5 {' X% d2 G1 X9 q1 P a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) ( G# I! E U% T" G) Y) Y
a1 := a1 shl 2 ;
d+ E! T3 T8 f- B d1 := ord ( inchar [di + 1]); : g( m4 V% l' c' ^5 j `, E
d1 := d1 shr 4 ;
5 h& t( M; n1 n) ~7 p a1 := a1 or d1 ;
+ L( k" S7 Q; D+ C% M" E) p+ x9 {. N outchar[si] := chr (a1);
% Q/ m6 E7 v" ~' t9 q7 p, E# S& Y( B( e% G0 j
a1 := ord ( inchar [di + 1]);
! [3 j, m" A$ F! {- } a1 := a1 shl 4 ;
% x- k6 N' q% } r# T d1 := ord ( inchar [di + 2]); 6 f/ g, U# q3 {& f4 c# z
d1 := d1 shr 2 ;
/ ?& i4 v9 X2 i+ C4 o; ? a1 := a1 or d1 ; 3 S8 G" H% k0 `6 O/ a3 p6 g2 g0 x
outchar[si + 1] := chr (a1); 5 s+ v/ c/ k9 q' N) U" o6 ?7 }
[. G9 _, E; ?2 c$ {- l3 F
a1 := ord ( inchar [di + 2]); + T5 H+ B/ i: f5 G( j5 f: }, d
a1 := a1 shl 6 ;
7 Q" N$ ~3 g, O0 Q; H. m9 ^6 u d1 := ord ( inchar [di + 3]);
7 b2 R0 j3 v: v/ _" W8 F# d- X7 g6 ~ a1 := a1 or d1 ; . _) Z% V! `1 j8 f( K
outchar[si + 2] := chr (a1); ' ?; K- G) ]1 g b7 I: d
9 K! p( u& ?0 H& l( R di := di + 4 ;
! [: W4 q( [' C0 {# } si := si + 3 ; ! ]; [9 u, Z4 b z* B
end; # F6 W5 x. b5 V. Y& U
end; 4 k+ l/ S$ c4 l$ {
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function encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密 2 A b, |' U) [' Y4 M
var
7 [4 ]9 Y5 e! r% `/ l# n7 G a1, d1 :byte; - ]2 s& V/ s% w0 ~ E9 k. v7 O
j,count,di,si :integer;
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begin
6 |/ T8 Y8 ^# K2 M encode := len div 3 * 4 ; //返回加密后数据长度
, U$ X! ^5 o, z& A) o3 ~ di := 0 ;
( r% K: V+ E6 S& W5 ?6 _ si := 0 ; / }7 }2 |/ j' s \- ]
count := len div 3; //定义循环次数 4 b/ o2 H3 r4 w" I6 z
for j := 1 to count do
7 d" Z: m/ ]6 Q begin ' Y: \5 @1 K! f0 m3 N7 d W% L
a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
; M* t" [$ k' Z8 d' i0 E a1 := a1 shr 2 ;
2 K) {4 @7 _* V8 j4 L4 p- N outchar [si] := chr (a1);
, c& W2 l2 J) o- V0 W, d$ e$ \. n6 ?8 i1 Z+ M+ V8 {
a1 := ord (inchar[di]); 7 ~% Y) a: h& F" n( k' m
a1 := a1 and 3 ;
2 M- b" Z# `* }) D G5 h a1 := a1 shl 4 ;
- P9 I; \+ w4 X4 y1 N7 M0 u d1 := ord (inchar[di +1]); / J" u( N) [$ T
d1 := d1 shr 4 ;
+ c% p$ [1 x$ ]: k5 M/ o! {: z a1 := a1 or d1 ;
- ^4 ^ [, W+ P- P* I8 Q' D outchar[si +1] := chr (a1) ; : T% l* f- r5 n/ Z2 z
+ r) V( U# m) h7 Y a1 := ord ( inchar[di + 1]) ; 9 s9 K( x+ | S$ b- r
a1 := a1 and {&content}F ; 9 v6 Q% t `( O( d b/ N
a1 := a1 shl 2 ;
5 J7 k. I6 `* Q6 H7 C) E& y( B# E d1 := ord ( inchar[di + 2]) ;
$ V# n' {( r+ t% l. [; w3 B d1 := d1 shr 6 ;
) X1 E/ y7 u7 C a1 := a1 or d1 ;
9 I; M% ?! U- Z0 M/ b outchar[si + 2] := chr (a1);
2 ?- l; j& u* x- s8 l: y
7 C* y4 _+ v+ j$ k. x a1 := ord ( inchar[di + 2]) ;
) |+ n5 H+ t ~" w4 a a1 := a1 and $3F;
& l. q& ?3 M' p: v outchar[si + 3] := chr (a1) ; # R+ @4 b, Z' h2 b8 Q. r* m
' M+ g$ _5 j) s- w
//第二次转换 3 W; y" [0 f5 a9 V0 n
d1 := ord ( outchar [si]); //第 1 个
' O8 `" D; n0 |& f d1 := gamestr [d1]; 7 _# l% E0 H; N, V9 e
outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个 9 f: y- ?; s2 h, a9 z% w6 W- d
d1 := gamestr [d1]; ; N `+ X* |) D5 G( q4 p0 H
outchar [si + 1] := chr (d1); . v6 Q7 o) d, ^% S
* Y( s( Y$ o& L. ^
d1 := ord ( outchar [si + 2 ]); //第 3 个
; v) f( n8 P8 n) E. d6 t9 B- O d1 := gamestr [d1];
" K' ~6 _& g& s) G9 O outchar [si + 2] := chr (d1);
9 h# d4 z4 ~- f1 Y r9 X9 Y
8 G3 D& k. `: P# q; {7 W' t) f d1 := ord ( outchar [si + 3]); //第 4 个
4 T2 `9 K6 h% h8 c: F d1 := gamestr [d1];
( @% S. E4 h! o3 [ outchar [si + 3] := chr (d1); 3 F. r, o* Z8 F2 ]: @) D
7 K8 J4 i" y* U2 T9 p8 ] di := di + 3 ;
/ [, o' g" I3 u: Q5 R/ w& s! W si := si + 4 ; ( {' V- e) y3 n5 U- X7 t- C
end;
. G0 D3 i, W9 w! A: b6 Eend;
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