暂时没事干，刷一下挑战
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┌──(kali㉿kali)-[/mnt/hgfs/gx]
└─$ unzip 046.zip
Archive: 046.zip
inflating: 046.txt
┌──(kali㉿kali)-[/mnt/hgfs/gx]
└─$ cat -A 046.txt
x=W0oyRmlZMlJsWm1kb2FXcHJiRzF1YjNCeGNuTjBkWFozZUhsNlFVSkRSRVZHUjBoSlNrdE1UVTVQVUZGU1UxUlZWbGRZV1ZvbklDZDZlWGgzZG5WMGMzSnhjRzl1Yld4cm;y=FtbG9aMlpsWkdOaVlWcFpXRmRXVlZSVFVsRlFUMDVOVEV0S1NVaEhSa1ZFUTBKQkp3bz0sIEhWMUR2M1FvQjNFYngyb2V5bTBQXQo=;x+y^M$
^M$
^M$
^M$
^M$
目测是base64
┌──(kali㉿kali)-[/mnt/hgfs/gx]
└─$ echo -n "W0oyRmlZMlJsWm1kb2FXcHJiRzF1YjNCeGNuTjBkWFozZUhsNlFVSkRSRVZHUjBoSlNrdE1UVTVQVUZGU1UxUlZWbGRZV1ZvbklDZDZlWGgzZG5WMGMzSnhjRzl1Yld4cm" | wc -c
130
┌──(kali㉿kali)-[/mnt/hgfs/gx]
└─$ echo -n "FtbG9aMlpsWkdOaVlWcFpXRmRXVlZSVFVsRlFUMDVOVEV0S1NVaEhSa1ZFUTBKQkp3bz0sIEhWMUR2M1FvQjNFYngyb2V5bTBQXQo=" | wc -c
102
┌──(kali㉿kali)-[/mnt/hgfs/gx]
└─$ echo -n "W0oyRmlZMlJsWm1kb2FXcHJiRzF1YjNCeGNuTjBkWFozZUhsNlFVSkRSRVZHUjBoSlNrdE1UVTVQVUZGU1UxUlZWbGRZV1ZvbklDZDZlWGgzZG5WMGMzSnhjRzl1Yld4cmFtbG9aMlpsWkdOaVlWcFpXRmRXVlZSVFVsRlFUMDVOVEV0S1NVaEhSa1ZFUTBKQkp3bz0sIEhWMUR2M1FvQjNFYngyb2V5bTBQXQo=" | base64 -d
[J2FiY2RlZmdoaWprbG1ub3BxcnN0dXZ3eHl6QUJDREVGR0hJSktMTU5PUFFSU1RVVldYWVonICd6eXh3dnV0c3JxcG9ubWxramloZ2ZlZGNiYVpZWFdWVVRTUlFQT05NTEtKSUhHRkVEQ0JBJwo=, HV1Dv3QoB3Ebx2oeym0P]
J2FiY2RlZmdoaWprbG1ub3BxcnN0dXZ3eHl6QUJDREVGR0hJSktMTU5PUFFSU1RVVldYWVonICd6eXh3dnV0c3JxcG9ubWxramloZ2ZlZGNiYVpZWFdWVVRTUlFQT05NTEtKSUhHRkVEQ0JBJwo=
HV1Dv3QoB3Ebx2oeym0P
先解第一段
┌──(kali㉿kali)-[/mnt/hgfs/gx]
└─$ echo -n "J2FiY2RlZmdoaWprbG1ub3BxcnN0dXZ3eHl6QUJDREVGR0hJSktMTU5PUFFSU1RVVldYWVonICd6eXh3dnV0c3JxcG9ubWxramloZ2ZlZGNiYVpZWFdWVVRTUlFQT05NTEtKSUhHRkVEQ0JBJwo=" | base64 -d
'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ' 'zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA'
这明显是一个 字符映射表，也就是用于 简单替换加密的
左边是明文字符集（小写+大写）：
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
右边是对应的密文字符映射（即：倒序）：
zyxwvutsrqponmlkjihgfedcbaZYXWVUTSRQPONMLKJIHGFEDCBA
换句话说，这是一个 Atbash Cipher
那你可以直接拿去CyberChef了
这样就出flag了
HMV{recursion}