UMDCTF 2025

比赛地址：UMDCTF 2025

比赛时间：26 Apr 2025 06:00 CST - 28 Apr 2025 06:00 CST

Misc

find the seeds

Challenge

can u help Alice find her seeds in the bin? She’s pretty sure the bin hasn’t been dumped since it was generated.

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import random
import time

seed = int(time.time())
random.seed(seed)

plaintext = b"UMDCTF{REDACTED}"
keystream = bytes([random.getrandbits(8) for _ in range(len(plaintext))])
ciphertext = bytes([p ^ k for p, k in zip(plaintext, keystream)])

with open("secret.bin", "wb") as f:
    f.write(ciphertext)

Solution

加密过程使用了 XOR 操作，将 plaintext 和一个由随机数生成器（random.getrandbits(8)）生成的 keystream 进行逐字节异或操作，得到 ciphertext。

已知 XOR 的性质：如果 A ^ B = C，那么 C ^ B = A。因此，如果我们知道 ciphertext 和 keystream，可以通过 ciphertext ^ keystream 还原出 plaintext。

因此可以爆破加密时的时间戳，使用该时间戳重新生成 keystream，通过 ciphertext ^ keystream 还原出 plaintext。

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import random
import time

# 读取密文文件
with open("secret.bin", "rb") as f:
    ciphertext = f.read()

# 已知明文的固定前缀
KNOWN_PREFIX = b"UMDCTF{"

# 尝试还原 plaintext
def recover_plaintext(ciphertext):
    # 获取当前时间戳
    current_time = int(time.time())
    
    # 时间设置长一点进行爆破
    for seed in range(current_time - 360000, current_time + 1):
        random.seed(seed)
        
        # 重新生成 keystream
        keystream = bytes([random.getrandbits(8) for _ in range(len(ciphertext))])
        
        # 还原 plaintext
        plaintext = bytes([c ^ k for c, k in zip(ciphertext, keystream)])
        
        # 检查 plaintext 是否以已知前缀开头
        if plaintext.startswith(KNOWN_PREFIX):
            print(f"Seed found: {seed}")
            return plaintext
    
    return None

# 还原 plaintext
plaintext = recover_plaintext(ciphertext)
print(plaintext.decode('utf-8'))
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Seed found: 1745447710
UMDCTF{pseudo_entropy_hidden_seed}
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UMDCTF{pseudo_entropy_hidden_seed}

tiktok-ban

Challenge

Oh snap! Oh crap! TikTok is banned in Ohio!

nc challs.umdctf.io 32300

Solution

服务端运行了一个 dnsmasq 实例，配置了一个 值为 flag 的 TXT 记录 tiktok.com 。

服务端首先读取 4 字节的长度信息（大端格式），然后根据该长度读取后续的数据。

如果数据中包含 tiktok\x03com（即 DNS 查询格式中的 tiktok.com），服务器会返回一段固定的错误消息，否则就返回 flag。

因此我们需要构造一个 DNS 请求，绕过检查查询 tiktok.com 的 TXT 记录，从而获取 flag。

注意到服务端使用 if b'tiktok\x03com' in req 进行检查，域名是不区分大小写的，然而这里的判断是区分大小写的，因此我们可以通过利用大小写绕过构造一个等价的域名。

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from pwn import *
import struct

HOST = "challs.umdctf.io"
PORT = 32300

# 构造 DNS 查询请求
def create_dns_query():
    # DNS Header (固定字段)
    transaction_id = b"\x12\x34" # 事务 ID，可以随意设置
    flags = b"\x01\x00"          # 标志位：标准查询
    questions = b"\x00\x01"      # 问题数：1
    answer_rrs = b"\x00\x00"     # 回答资源记录数：0
    authority_rrs = b"\x00\x00"  # 权威资源记录数：0
    additional_rrs = b"\x00\x00" # 附加资源记录数：0
    header = transaction_id + flags + questions + answer_rrs + authority_rrs + additional_rrs

    # DNS Question (查询部分)
    qname = b"\x06Tiktok\x03com\x00"  # 域名：Tiktok.com(大小写绕过)
    qtype = b"\x00\x10"               # 查询类型：TXT (16)
    qclass = b"\x00\x01"              # 查询类：IN (1)
    question = qname + qtype + qclass

    # 组合完整的 DNS 请求
    dns_query = header + question
    return dns_query

def main():
    conn = remote(HOST, PORT)
    dns_query = create_dns_query()

    # 发送长度信息和 DNS 请求
    conn.send(struct.pack(">I", len(dns_query)))  # 大端格式的长度
    conn.send(dns_query)

    response = conn.recvall()
    print(response.decode())
    conn.close()

if __name__ == "__main__":
    main()

OSINT

swag-like-ohio

Challenge

swag like ohio. down in ohio. swag like ohio. down in ohio. anyway we seem to be on a bridge. what’s the address of the bridge?

flag will look like: UMDCTF

UMDCTF2025-1

Solution

UMDCTF2025-2

先识图找 ohio 找到这篇文章 Vacancies — Marietta Main Street

再用这篇文章里的图接着搜

UMDCTF2025-3

再找到维基百科上的这张图 File:Ohio - Marietta - Dime Bank.jpg - Wikimedia Commons，得知这个建筑是 Marietta Dime Bank

UMDCTF2025-4

在这里 200 Putnam Street in Historic Downtown Marietta, OH 找到地址 200 Putnam Street, Marietta, OH 45750

UMDCTF2025-5

到谷歌地图直接搜

UMDCTF2025-6

Putnam Bridge - Google 地圖

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UMDCTF{Putnam Bridge, Marietta, OH 45750}

sunshine

Challenge

what a nice sunny day. what is the full address of house number 356?

flag will look like: UMDCTF

UMDCTF2025-7

Solution

UMDCTF2025-8

搜索题目给定的这栋房子发现它是篮球篮球运动员 LeBron James 童年的住所

在第一个搜索结果 Where LeBron James Lives: A Peek Into LeBron’s Homes Interbasket 中发现 LeBron’s Childhood Home in Akron, Ohio

进一步搜索

UMDCTF2025-9

在第一个搜索结果 LeBron James’ childhood home in Akron, OH (Google Maps) 中找到题目给定的地址（甚至街景都一模一样）

356 Hillwood Dr - Google 地圖

UMDCTF2025-10

位于 356 Hillwood Dr, Akron, OH 44320美國

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UMDCTF{356 Hillwood Dr, Akron, OH 44320}

beauty

Challenge

truly a beautiful panorama. ohio is not always ugly. i really wanna know who made this pano tho. what’s their name?

flag will look like: UMDCTF

UMDCTF2025-11

Solution

UMDCTF2025-12

搜到这篇百科 List of tallest buildings in Ohio - Wikipedia

UMDCTF2025-13

找到这个建筑 AEP Building - Wikipedia，然后到谷歌地图上搜

UMDCTF2025-14

Battelle Riverfront Park - Google 地圖

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UMDCTF{Neil Larimore}

the-master

Challenge

trust me bro, i know what im talking about. im the master when it comes to these things. what street are we on?

flag will look like: UMDCTF{Campus Dr, College Park, MD 20742}

UMDCTF2025-15

Solution

UMDCTF2025-16

识图找到维基百科上的这张图片 File:Lore City UMC.jpg - Wikimedia Commons

下面给出了拍摄这张图时所在的坐标 GeoHack - File:Lore City UMC.jpg

打开谷歌地图 [39°59’02.0"N 81°27’32.0"W - Google Maps](39°59’02.0"N 81°27’32.0"W - Google Maps)

最终可以定位到这里 190 Main St - Google Maps

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UMDCTF{Main St, Lore City, OH 43755}

Nyt

the-mini

Challenge

Joel Fagliano has nothing on me. (flag is all caps)

Solution

UMDCTF2025-18

这是个填字游戏

UMDCTF2025-19

并且它的 solution 被锁住了

然而这里的 key 只有 4 位数，因此直接爆破

在 GitHub 上找了个处理 .puz 的 Python 库 alexdej/puzpy，然后就是写脚本调用接口爆破了

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import puz

def brute_force_unlock(puzzle, max_key=10000):
    """
    尝试暴力破解锁定的谜题。
    
    :param puzzle: puz.Puzzle 对象
    :param max_key: 最大密钥值，默认为 10000
    :return: 解锁后的谜题对象（如果成功），否则返回 None
    """
    for key in range(max_key):
        if puzzle.unlock_solution(key):
            print(f"成功解锁！密钥为: {key}")
            return puzzle
    print("未能找到正确的密钥。")
    return None

def main():
    # 文件路径
    file_path = "the_mini.puz"
    
    try:
        # 读取 .puz 文件
        puzzle = puz.read(file_path)
        
        # 检查是否被锁定
        if puzzle.is_solution_locked():
            print("谜题被锁定，尝试暴力破解...")
            unlocked_puzzle = brute_force_unlock(puzzle)
            
            if unlocked_puzzle:
                # 输出解锁后的解决方案
                print("解锁后的解决方案:")
                print(unlocked_puzzle.solution)
            else:
                print("暴力破解失败。")
        else:
            print("谜题未被锁定，直接输出解决方案:")
            print(puzzle.solution)
    
    except puz.PuzzleFormatError as e:
        print(f"读取谜题文件时发生错误: {e}")
    except Exception as e:
        print(f"发生未知错误: {e}")

if __name__ == "__main__":
    main()

运行得到输出

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谜题被锁定，尝试暴力破解...
成功解锁！密钥为: 5727
解锁后的解决方案:
UMDCTFCANYOUBEATMYTIME...
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UMDCTF{CANYOUBEATMYTIME}

Reverse

deobfuscation

Challenge

the chall is not that complex. the key is to read ASSEMBLY!

Solution

UMDCTF2025-17

入口函数的分析如图，只要提取固定数组 byte_402000 和 byte_402034，然后通过异或运算就可以还原出正确的输入，下面是提取出的 byte_402000 和 byte_402034 的内容

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.data:0000000000402000 20                            byte_402000 db 20h                      ; DATA XREF: LOAD:00000000004000C0↑o
.data:0000000000402000                                                                       ; start:loc_40105D↑r
.data:0000000000402001 22                            db  22h ; "
.data:0000000000402002 20                            db  20h
.data:0000000000402003 26                            db  26h ; &
.data:0000000000402004 35                            db  35h ; 5
.data:0000000000402005 37                            db  37h ; 7
.data:0000000000402006 14                            db  14h
.data:0000000000402007 07                            db    7
.data:0000000000402008 46                            db  46h ; F
.data:0000000000402009 00                            db    0
.data:000000000040200A 5A                            db  5Ah ; Z
.data:000000000040200B 17                            db  17h
.data:000000000040200C 44                            db  44h ; D
.data:000000000040200D 35                            db  35h ; 5
.data:000000000040200E 52                            db  52h ; R
.data:000000000040200F 0C                            db  0Ch
.data:0000000000402010 70                            db  70h ; p
.data:0000000000402011 28                            db  28h ; (
.data:0000000000402012 37                            db  37h ; 7
.data:0000000000402013 1C                            db  1Ch
.data:0000000000402014 5B                            db  5Bh ; [
.data:0000000000402015 1D                            db  1Dh
.data:0000000000402016 70                            db  70h ; p
.data:0000000000402017 16                            db  16h
.data:0000000000402018 76                            db  76h ; v
.data:0000000000402019 50                            db  50h ; P
.data:000000000040201A 69                            db  69h ; i
.data:000000000040201B 5C                            db  5Ch ; \
.data:000000000040201C 6E                            db  6Eh ; n
.data:000000000040201D 6C                            db  6Ch ; l
.data:000000000040201E 1B                            db  1Bh
.data:000000000040201F 12                            db  12h
.data:0000000000402020 54                            db  54h ; T
.data:0000000000402021 69                            db  69h ; i
.data:0000000000402022 2D                            db  2Dh ; -
.data:0000000000402023 38                            db  38h ; 8
.data:0000000000402024 06                            db    6
.data:0000000000402025 23                            db  23h ; #
.data:0000000000402026 11                            db  11h
.data:0000000000402027 3D                            db  3Dh ; =
.data:0000000000402028 2F                            db  2Fh ; /
.data:0000000000402029 00                            db    0
.data:000000000040202A 02                            db    2
.data:000000000040202B 4A                            db  4Ah ; J
.data:000000000040202C 68                            db  68h ; h
.data:000000000040202D 45                            db  45h ; E
.data:000000000040202E 3B                            db  3Bh ; ;
.data:000000000040202F 64                            db  64h ; d
.data:0000000000402030 1A                            db  1Ah
.data:0000000000402031 20                            db  20h
.data:0000000000402032 55                            db  55h ; U
.data:0000000000402033 05                            db    5
.data:0000000000402034                               ; char byte_402034[52]
.data:0000000000402034 75                            byte_402034 db 75h                      ; DATA XREF: start+43↑r
.data:0000000000402035 6F                            db  6Fh ; o
.data:0000000000402036 64                            db  64h ; d
.data:0000000000402037 65                            db  65h ; e
.data:0000000000402038 61                            db  61h ; a
.data:0000000000402039 71                            db  71h ; q
.data:000000000040203A 6F                            db  6Fh ; o
.data:000000000040203B 75                            db  75h ; u
.data:000000000040203C 75                            db  75h ; u
.data:000000000040203D 76                            db  76h ; v
.data:000000000040203E 69                            db  69h ; i
.data:000000000040203F 45                            db  45h ; E
.data:0000000000402040 60                            db  60h ; `
.data:0000000000402041 70                            db  70h ; p
.data:0000000000402042 7F                            db  7Fh ; 
.data:0000000000402043 65                            db  65h ; e
.data:0000000000402044 54                            db  54h ; T
.data:0000000000402045 77                            db  77h ; w
.data:0000000000402046 63                            db  63h ; c
.data:0000000000402047 74                            db  74h ; t
.data:0000000000402048 68                            db  68h ; h
.data:0000000000402049 42                            db  42h ; B
.data:000000000040204A 53                            db  53h ; S
.data:000000000040204B 54                            db  54h ; T
.data:000000000040204C 45                            db  45h ; E
.data:000000000040204D 03                            db    3
.data:000000000040204E 3D                            db  3Dh ; =
.data:000000000040204F 7F                            db  7Fh ; 
.data:0000000000402050 31                            db  31h ; 1
.data:0000000000402051 58                            db  58h ; X
.data:0000000000402052 75                            db  75h ; u
.data:0000000000402053 46                            db  46h ; F
.data:0000000000402054 75                            db  75h ; u
.data:0000000000402055 44                            db  44h ; D
.data:0000000000402056 60                            db  60h ; `
.data:0000000000402057 78                            db  78h ; x
.data:0000000000402058 6A                            db  6Ah ; j
.data:0000000000402059 74                            db  74h ; t
.data:000000000040205A 51                            db  51h ; Q
.data:000000000040205B 4F                            db  4Fh ; O
.data:000000000040205C 1C                            db  1Ch
.data:000000000040205D 5F                            db  5Fh ; _
.data:000000000040205E 76                            db  76h ; v
.data:000000000040205F 79                            db  79h ; y
.data:0000000000402060 0B                            db  0Bh
.data:0000000000402061 2D                            db  2Dh ; -
.data:0000000000402062 75                            db  75h ; u
.data:0000000000402063 45                            db  45h ; E
.data:0000000000402064 4B                            db  4Bh ; K
.data:0000000000402065 55                            db  55h ; U
.data:0000000000402066 66                            db  66h ; f
.data:0000000000402067 78                            db  78h ; x

编写脚本还原

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byte_402000 = [
    0x20, 0x22, 0x20, 0x26, 0x35, 0x37, 0x14, 0x07,
    0x46, 0x00, 0x5A, 0x17, 0x44, 0x35, 0x52, 0x0C,
    0x70, 0x28, 0x37, 0x1C, 0x5B, 0x1D, 0x70, 0x16,
    0x76, 0x50, 0x69, 0x5C, 0x6E, 0x6C, 0x1B, 0x12,
    0x54, 0x69, 0x2D, 0x38, 0x06, 0x23, 0x11, 0x3D,
    0x2F, 0x00, 0x02, 0x4A, 0x68, 0x45, 0x3B, 0x64,
    0x1A, 0x20, 0x55, 0x05
]

byte_402034 = [
    0x75, 0x6F, 0x64, 0x65, 0x61, 0x71, 0x6F, 0x75,
    0x75, 0x76, 0x69, 0x45, 0x60, 0x70, 0x7F, 0x65,
    0x54, 0x77, 0x63, 0x74, 0x68, 0x42, 0x53, 0x54,
    0x45, 0x03, 0x3D, 0x7F, 0x31, 0x58, 0x75, 0x46,
    0x75, 0x44, 0x60, 0x78, 0x6A, 0x74, 0x51, 0x4F,
    0x1C, 0x5F, 0x76, 0x79, 0x0B, 0x2D, 0x75, 0x45,
    0x4B, 0x55, 0x66, 0x78
]

correct_input = []
for i in range(len(byte_402000)):
    correct_input.append(byte_402000[i] ^ byte_402034[i])

print("".join(chr(c) for c in correct_input))
1
UMDCTF{r3v3R$E-i$_Th3_#B3ST#_4nT!-M@lW@r3_t3chN!Qu3}