【出题】批处理实现16进制数据最小体积存储

plp626

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设计一种编码批处理变量存放方案，将上面的字符信息用尽量少的字符个数储存，当使用的时候，又可以用相应的方案还原。
简单的理解就是类似于数据压缩与解压缩操作。

要求，用cmd的内部命令实现。
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PS: 这个“设计方案”所含的批处理代码不要过长，体积的计算要把这个“设计方案”所含代码和储存信息一起计算。

当然，如果方案能通用的话，可以适当放宽设计方案所含代码的长度

plp626

PS: 这个“设计方案”所含的批处理代码不要过长，体积的计算要把这个“设计方案”所含代码和储存信息一起计算。

当然，如果方案能通用的话，可以适当放宽设计方案所含代码的长度

-----------------------------------

有人也许会问这个能干什么？ 我说下应用，当然还是娱乐，但“娱乐度”比较高，呵呵。

有个霍夫曼编码，可以对所发报文以最短编码存储，我们把他应用到bat2any(利用debug写文件)那个程序里，可以最大限度的压缩文件的体积，

假如有人想把三方工具嵌入到bat中，这样体积可以压缩很多多多。。。。

hanyeguxing

如果源文件是 32 位命令行，直接使用 UPX 压缩

plp626

3# hanyeguxing


上面这个exe文件体积就1K多，选择编译的时候我加了/align:16，体积最小模式，upx没法再压缩了

caruko

16*16 =256
一个简单的办法就是使用ACSII码。

DEBUG即可互相转换。

CrLf

同上，相信大家第一反映都差不多，不过我想，无论用ansi码或者什么码来以字典的形式与原始数据一一对应，其根本思路无非是把16进制转换为更高位的进制，以填补用1~2个字节只记录一个十六进制数时浪费的容量。
对于调用第三方、debug、vbs等方法不做评论，因为想必都是可行的，不过我想楼主的意思大概是尽量用纯批的吧。
再说一个大众看法，大家肯定都想到了，只是没说。首先，字典肯定要用到的，再用A个字典里的字符与B个十六进制数对应，不过要知道A:B的最佳比例，就要做一个计算：将能在批处理中识别为单字符的单字节ansi码数量设为N，然后求N与16的最小公倍数，然后...你知道的，不解释

terse

转 Base64码

hanyeguxing

楼主提供的本身就是源文件的 ANSI 。。。

terse

转 Base64码 后文件原来的 2/3

523066680

我看0出现很多次，要不用N个0表示  恩恩

plp626

我目前，理论上，

可以把一楼的数据压缩到1100 字节左右（不含解压代码）

也就是压缩为原来体积的37%左右。

不知大家的结果是怎样的？

CrLf

如果引入有损压缩的概念，压缩比必定更高

batman

思路：
0-9的数字就用原值存储(原来用两个字符存储压缩一半)，a-z和A-Z全用a-z表示(压缩比视大小写字母比)，非数值非字母可见字符用原值存储(原来用两个字符存储压缩一半，)，回车换行退格分别用A、B、C存储（压缩一半）。

@echo off&setlocal enabledelayedexpansion
for /f "skip=8 tokens=1*" %%a in (%~fs0) do set "%%a=%%b"
set "str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
:lp
set "var=!var!!%str:~,2%!"
if defined str set "str=%str:~2%"&goto lp
set "var=!var:空格= !"
echo !var:跳格=	!
pause>nul&goto :eof
09 跳格
20 空格
21 !
22 "
23 #
24 $
25 %
26 &
27 '
28 (
29 )
2A *
2B +
2C ,
2D -
2E .
2F /
30 0
31 1
32 2
33 3
34 4
35 5
36 6
37 7
38 8
39 9
3A :
3B ;
3C <
3D =
3E >
3F ?
40 @
41 a
42 b
43 C
44 d
45 e
46 f
47 g
48 h
49 i
4A j
4B k
4C l
4D m
4E n
4F o
50 p
51 q
52 r
53 s
54 t
55 u
56 v
57 w
58 x
59 y
5A z
5B [
5C \
5D ]
5E ^
5F _
60 `
61 a
62 b
63 c
64 d
65 e
66 f
67 g
68 h
69 i
6A j
6B k
6C l
6D m
6E n
6F o
70 p
71 q
72 r
73 s
74 t
75 u
76 v
77 w
78 x
79 y
7A z
7B {
7C |
7D }
7E ~
0A A
0D B
08 C 
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hanyeguxing

11# plp626


用批处理自身来存储、解压，怎么能不算解压代码呢？

terse

另一思路  把一定长度字符（转为10进制） 除一固定数 储存这个信息