抄录下来：

%..@*>@?==%88%5
.@%#@@90-7$^=*@
17.(>()1@##-$40
~.*6?#%#8#=75+1
(*@*1%#>;0@5)%?
%*^=)&>=1%.+7&#
8681(+8*@@(.@@@
#*=#$3*#%.#%%.3
.*+7.7+@===+)61

大佬提示跟黄道十二宫杀手有关，搜索相关信息，需要用到AZdecrypt软件。

第一部分，第一个字符H，随后再向下移动一格，在向左移动两个，得到第二个字符+，以此类推。

python代码实现：

_sec = ""
with open(r"黄道十二宫杀手密码Zodiac Killer.txt", "r") as f:_sec = f.read()assert len(_sec) > 0_sec_lines = _sec.split("\n")
_line_num = len(_sec_lines)
_line_ch_cnt = len(_sec_lines[0])
_tmp = ""for _i in range(_line_ch_cnt):for _j in range(0, _line_num):_tmp += _sec_lines[_j][(2 * _j + _i) % _line_ch_cnt]_reformat_ = []
for _i in range(_line_num):_reformat_.append(_tmp[_i * _line_ch_cnt:(_i + 1) * _line_ch_cnt])
for _key in _reformat_:print("".join(_key))

 输出结果：

%%>%;.@3*.#(#0+
@#+.@)8@7@*7@@1
#5&8=.*9@=)#6#7
>0#7%%8$+@-#5?*
13@?7-+(^(*==$$
1*=+#==^4~@)8%=
%=0.*&*.+8*1*1>
@#)8@76%=@%6%..
?#1(%15@(#>%...

扔进AZdecrypt进行解码：

Score: 22983.43 IOC: 0.0654 Multiplicity: 0.1925 Seconds: 0.11
Repeats: LETH KILL ILL LL IS (2) AN (2) RE OR AS TH OU TO OP PE
PC-cycles: 136II KILLED A LOT OF PEOPLE THERE ARE ENOUGH 
SLAVES TO WORK FOR I I HOPE YOU CAN DECRYPT 
IT AS SOON AS POSSIBLE THIS IS FLAG ALPHAN 
ANKE OTHERWISE I WILL CONTINUE TO KILLL

根据大佬提示取flag后两个单词并小写：

flag{alphananke}