beginCTF 2024

2024/2/6 题解 CTF

BeginCTF 2024 Writeup

MISC

1. real check in

base64解码即可

2. tupper

好多文件，先拼接起来

for i in range(0, 673, 4):
    filename = str(i) + '.txt'
    with open(filename, 'r', encoding='utf-8') as f:
        data = f.readline()
        print(data, end='')

“=”结尾，base64解码得到一串数字

1427819343272802604929857457555753432106234935254356265676646
9704092874688354679371212444382298821342093450398907096976002
4588075985357351721266575041311716849071730866595051439203000
8580880964725679038437855378028289423975189862004114338331706
4727136903634770936398518547900512548419486364915399253941245
9112052624935911584977082191264535874566373028887013033822107
4862980008182168428318736854360155977843173500679476154241300
6621219207322808449232050578852431361678745355776921132352419
931907838205001184

搜索Tupper，发现是Tupper’s Self-Referential Formula（塔珀自指公式）

网站：Tuppers-Formula-Tools

3. devil’s word

一眼温州话+16进制数

leng -> 0 lia -> 2 sa -> 3 sii -> 4 ng -> 5 leu -> 6 cai -> 7 bo -> 8 jau -> 9

构造字典

s = 'leu lia leu ng leu cai leu jau leu e cai b cai jau sa leng cai 
ng ng f leu b leu e sa leng cai cai ng f cai cai sa sa leu e cai a leu 
bo leu f cai ng ng f leu sii leu jau sa sii leu c leu ng leu sa cai sii cai d'
lib = {'leng':'0','lia':'2','sa':'3','sii':'4','ng':'5','leu':'6','cai':'7',
'bo':'8','jau':'9','a':'A','b':'B','c':'C','d':'D','e':'E','f':'F'}

str_list = s.split()
n = len(str_list)
for i in range(n):
    str_list[i] = lib[str_list[i]]
str_new = []
for i in range(0, n, 2):
    str_new.append(str_list[i]+str_list[i+1])
for i in str_new:
    print(chr(int(i,16)),end='')

得到flag

4. 下一站上岸

010editor打开，末尾发现垃圾字符

5o+Q56S6OuaRqeaWr+WvhueggQ==

base64解码，提示是摩斯密码

按照交点数为012得到： 221022201122120120111011110222012101

0一定是间隔符（因为只有0成单出现）两种情况替换’-‘和’.’得到flag

5. where is crazyman v1.0

谷歌识图

6. where is crazyman v2.0