关于加密解密问题

Question

明文:            密文:
MQ==              1
Mg==              2
Mw==              3
NA==              4
RE9T              DOS
d2luZG93CW==      windows
Zmxhc2g=          flash应该是分组加密.
三个字符进行加密,其中汉字占两上字符.
希望大家帮助.
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谢谢.

Answer 1

似乎不是什么加密算法，只是加上一些东西之后转换成BASE64码。

Answer 2

根据一些特征可以推算采用的对应BASE64码是：
0、=
1、A
2、B
……
26、Z
27、a
28、b
……
52、z
53、0
54、1
……
62、9
63、？（因为没出现过我也不知道）剩下的理论上已经可以解决了。

Answer 3

索性写了段代码测试了一下：
MQ==                               1
(13)(17)(0)(0)=3477504=0x351000    0x310000 + 0x04100=0x351000
Mg==                               2
(13)(33)(0)(0)=3543040=0x361000    0x320000 + 0x04100=0x361000
Mw==                               3
(13)(49)(0)(0)=3608576=0x371000    0x330000 + 0x04100=0x371000
NA==                               4
(14)(1)(0)(0)=3674112=0x381000     0x340000 + 0x04100=0x381000
RE9T                               DOS
(18)(5)(62)(20)=4743060=0x485f94   0x444f53 + 0x041041=0x485f94
d2lu                               win（正如搂主所说，要拆成3字节一段）
(30)(55)(38)(47)=8092079=0x7b79af  0x77696e + 0x041041=0x7b79af
ZG93                               dow
(26)(7)(62)(56)=6848440=0x687fb8   0x646f77 + 0x041041=0x687fb8
cw==                               s
(29)(49)(0)(0)=7802880=0x771000    0x730000 + 0x041000=0x771000
Zmxh                               fla
(26)(39)(50)(34)=6978722=0x6a7ca2  0x666c61 + 0x041041=0x6a7ca2
c2g=                               sh
(29)(55)(33)(0)=7829568=0x777840   0x736800 + 0x041040=0x777840其中windows那段，应该是搂主笔误：d2luZG93CW==   =>   d2luZG93cw==后面加0x041...的规则是：
明文有1字节  + 0x041000
明文有2字节  + 0x041040
明文有3字节  + 0x041041