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 ~ m( ]1 u" @& y
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) , K- i% e% z6 z$ }
& Z* u8 x9 F$ r3 t* \( R声明部分
2 J# [' O5 E b* W% M//二种声方法均可 5 e/ ^$ _3 P& b2 l I* F7 h) J& m; s
//const gamestr:string[255] = ’N>TSVUJlwdcBMFjnAKb?qxvyeGzfLP=_ER@Z\am]ChgoWD;QuX[<tkpr^`iIHOYs.3" (- ’+chr(13)+’<;’+chr(12)+chr(00)+’=/!,1>#2$’+chr(39)+’89%’+chr(10)+chr(9)+’*):5&+67?40’; - y9 s: W! U& ~4 z
const gamestr : array [0..127] of byte =
/ \6 T7 M! x: j8 r" e$ d+ }($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, $1A);
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4 E) D& B$ Z. P6 A; t8 z代码部分 , H" ?* D K( m
function decode (inchar:pchar; len:integer ; outchar:pchar):integer; //解密 8 w3 A, i& a# n5 m, g- a- u1 ~
var
, [( R3 `3 U) L. t2 {6 B a1, d1: byte;
3 Z0 p( @+ i+ i j, count, di, si :integer;
+ J4 C/ O* x1 g9 H) X2 V! ]begin : x/ ? i7 |* [8 z
decode := len div 4 * 3 ; //返回解密后数据长度 1 |% H) f, u8 R: Q$ R3 b8 V0 O
j := 0; ' e" l- v! y9 X" F! o- F7 X% T
while i < len do
5 w% ?, e' U/ o5 G& j# |6 H begin / Z! v5 r @9 E5 U
d1 := byte (inchar[j] ); . l+ A2 {1 I' }0 q! f4 p
if ( d1 = $3B ) or (d1 = $7A) then
! e7 O, t$ z# o4 c; X1 `% } begin . T0 f" W" t: c1 W i7 _
end;
! }% s( c1 q! g5 W- V/ N d1 := d1 and {&content}FF; + z1 U% |* i& I. s4 j7 W
d1 := gamestr [d1 + 05]; //d1 := byte ( gamestr [ 1 + d1 ] ); " c( q# w2 D8 y" ~. O; w! {
byte (inchar [j] ) := d1; // 根据不同的gamestr数据定义选用不现的方法 + A: V" Z& _) ^7 z6 d
inc (i);
" O+ L5 _3 r/ i1 t1 H# Z9 Y9 q end; 5 l/ {/ n% H: G4 v( J$ a: X$ H( ?% ?
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di := 0 ;
% Z% B% D7 @4 T si := 0 ;
8 ^$ q, v7 k+ ?0 }, T+ C+ T count := len div 4 ; //循环次数 - j& D) R& v0 Q7 \" C( x
for j := 1 to count do
' L: d. ^9 [/ M* f8 o& Z begin
) I% T6 l$ N& c; E' _ a1 := ord ( inchar [di]); //解密数据 (取4个.转化3个) 5 A% e3 g0 A; ]# X/ l! V3 m
a1 := a1 shl 2 ; . ?; N) m$ s3 w1 y: ]
d1 := ord ( inchar [di + 1]);
( q! F6 H K: t d1 := d1 shr 4 ;
7 K' l2 ^5 p6 [7 f8 M a1 := a1 or d1 ; # U& v& j- l9 Z" R
outchar[si] := chr (a1); / g/ m% S1 d: G1 L' y: T! g% y6 L/ [( E
. h' ^# C2 A# p2 c. Z, s' Z( n
a1 := ord ( inchar [di + 1]); : N9 i; A2 F8 [8 N( _0 t8 _
a1 := a1 shl 4 ;
/ ]- z9 H% ? ?# X7 C d1 := ord ( inchar [di + 2]);
5 o8 z/ A/ K+ c6 f- V d1 := d1 shr 2 ; + }/ P* Q# `+ ~+ z
a1 := a1 or d1 ; 2 ^9 |0 M5 q% J+ a+ ]. L1 k5 G
outchar[si + 1] := chr (a1);
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& [1 F" }- b) E a1 := ord ( inchar [di + 2]); P7 X( c% C ^& D$ Q& ^
a1 := a1 shl 6 ;
9 C5 J3 N, {9 i; U7 f/ A( F- u d1 := ord ( inchar [di + 3]);
& M) w' L& v( a a1 := a1 or d1 ; 5 _8 _$ q( A+ \" q# C% {9 d3 G
outchar[si + 2] := chr (a1); + Q+ j: u5 L4 q ?
& L, L% J3 t) ~ T5 G' ?) [. m o di := di + 4 ; 9 ~0 P: e( o+ ]2 V, _5 |8 N8 c8 m
si := si + 3 ; ' Y+ M& l) R! Z7 R K( z' e% h5 c
end; 4 Q7 G& D4 v/ r# o' d( }
end; . b% @) u% U! u. _5 N5 ^% T" h+ Y. G
9 _ L, u1 ?9 F7 s. r0 xfunction encode (inchar:pchar; len:integer ; outchar:pchar):integer; //加密
2 V9 H/ }- m% b! j9 ]2 X& t$ Dvar ; w4 f; y( H; d# h
a1, d1 :byte; 2 s$ e7 ]' J6 b' q, ]9 b4 v
j,count,di,si :integer;
# b' | B8 B5 j, k) g7 ]
0 K. x9 i/ X0 E. i- o, gbegin
) U2 A" W1 s! F! I( c encode := len div 3 * 4 ; //返回加密后数据长度
$ w5 e* s0 N0 A' o; A/ K9 d4 Q di := 0 ;
" E3 _- w: R- ] e si := 0 ;
8 {; a+ s) U# A2 u count := len div 3; //定义循环次数 . u4 l& [2 @6 n3 z" ^: D
for j := 1 to count do , j) u4 S7 ^- M3 {
begin
( H5 r. i3 r& x$ V8 u2 l a1 := ord ( inchar [di]); //第一次转换 取3个 输出4个
8 t' k6 ^, C9 d$ w9 E a1 := a1 shr 2 ; - h; Y2 t% t) T
outchar [si] := chr (a1); - |" Q. ~( V2 q* W
8 @4 R7 `* U% J) u a1 := ord (inchar[di]); 5 J$ L: F9 ]8 |4 C/ D
a1 := a1 and 3 ; * ]3 m: {" y0 v" o2 f
a1 := a1 shl 4 ; ! p, C7 b6 Y/ n w: C; ~
d1 := ord (inchar[di +1]); ! M1 Q7 ?2 B" K3 Y
d1 := d1 shr 4 ;
/ F1 O) @- s/ x& Z5 i' y O3 F a1 := a1 or d1 ;
6 J# b7 R$ @" X. A: m% L0 S% ~ outchar[si +1] := chr (a1) ;
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a1 := ord ( inchar[di + 1]) ;
8 a5 h8 U0 R$ _: y; E2 S# ? a1 := a1 and {&content}F ; 2 V; p" W8 t l+ K! \$ `! O
a1 := a1 shl 2 ;
0 r. ]: ~6 D' x; R( t d1 := ord ( inchar[di + 2]) ;
5 K* E2 X( S, t. P- w" C" t d1 := d1 shr 6 ; ' v: J7 J8 r1 \# h: M
a1 := a1 or d1 ;
. l# t0 S( u+ T' `. L outchar[si + 2] := chr (a1); , ~# s& d& n3 \! u! y$ G, f
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a1 := ord ( inchar[di + 2]) ; # D8 B$ c+ l6 B. l) o [# L9 F
a1 := a1 and $3F; 3 D+ ]! t& n, g; j: m1 V) g
outchar[si + 3] := chr (a1) ;
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//第二次转换 8 r) l) N& F, Z
d1 := ord ( outchar [si]); //第 1 个 * W0 U: m0 U, g( G. |* a" I( z$ m
d1 := gamestr [d1]; 9 l( l* J$ p' ?& A
outchar [si] := chr (d1);
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d1 := ord ( outchar [si + 1]); //第 2 个
8 |- x/ x5 R. p' I/ i. ^ d1 := gamestr [d1];
! ~* N t1 w/ M1 A outchar [si + 1] := chr (d1);
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d1 := ord ( outchar [si + 2 ]); //第 3 个 $ b! c$ \' V) q% S
d1 := gamestr [d1]; 7 w9 f _. r1 j- O" d [( G
outchar [si + 2] := chr (d1);
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7 a% \. ~9 ^) e! k. r! S d1 := ord ( outchar [si + 3]); //第 4 个 1 k& c& c1 W" Y$ P6 c
d1 := gamestr [d1]; * D0 T# `0 |0 a' ~2 y
outchar [si + 3] := chr (d1);
2 g# W# \0 n! k: M! {
+ a2 Q+ Z" `6 C, E1 w0 S# @! ]& z di := di + 3 ; ; `6 O2 b7 S! l5 @0 Q" R. V
si := si + 4 ;
e* T! B- v; @- s/ [) w8 T end; ( [$ v/ n0 K% w
end; [( l1 t% Z; Y k0 b8 G- @
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