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Misc

p11n-trophy(签到题):

题目描述:
image-20250127165738440

我们首先会得到这样一份证书:

trophy

第一题签到题的答案就是证书下面正中间的“This certificate does not grant the rank of Master”。

trophy-plus + trophy-plus64:

这两道目描述一模一样

image-20250127165834982

其中一个flag是藏在certificate周围一圈的位置:

image-20250127170254595

image-20250127170344389

人工将这些内容识别,再翻译成二进制然后解码就会得到flag

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def decode_binary(content, mapping):
# Convert content to binary using the mapping
binary_str = ''.join(mapping[char] for char in content if char in mapping)

# Split the binary string into 8-bit chunks
bytes_list = [binary_str[i:i+8] for i in range(0, len(binary_str), 8)]

# Convert each 8-bit chunk to a character
decoded_text = ''.join(chr(int(byte, 2)) for byte in bytes_list if len(byte) == 8)

return decoded_text

# Input string
c_1 = "MVVVVMMMMMVVMMVVMVVMMMVVMVVVVMVVMVVM VMMV MV MVVVVVMVVMM VMM MMVVMMMV"
c_2 = "MVVMM VMMMVVMVVVMMM VMMVVVMVVVMV MMM VMVVVVVMVVM VMVVMVVMVVVMMMVVMMMMMVVVMVVVM VMVVVVV"
c_3 = "VMMMVVMMM VMMMVVMVVVVVM VMMV MMVVVMMMMMVVMMMVVMMVVMVVVVVM VMMM VMMVVMVVMMVVMMVVMMVVVM VMV MVVVMVVVVVM VM VM VMMVVMMV MMMVVMVVVVVMV MMMV MMVVMMMVVMVVM VMV MVVVMMMMMVVMMVVMMMVVMVVVVVM VMV MVVMVVMMVVMVVVM VMVVMVVM"
reversed_c_3 = c_3[::-1]
c_4 = "MMV MMVVMMMMMVVMMVVMMMVVMMVVVMVVMVVMMVVMMVVVVVM VMV MMVVVVMMV MMVVVMMM VMVVMMMVVVMVVM"
reversed_c_4 = c_4[::-1]
c_5 = "MMVVMMM VMVVMVVVMMVVMMVVVM VMVVVVVMVVMVVMMMMVVMMMMMVVMVVMMMVVVVVMV"

# mapping = {'M': '0', 'V': '1'}

mapping1 = {'M': '1', 'V': '0'}
mapping2 = {'M': '0', 'V': '1'}


# print(decode_binary(c_1, mapping2)+decode_binary(c_2, mapping2) + decode_binary(c_3, mapping2) + decode_binary(c_4, mapping2) + decode_binary(c_5, mapping2) )
print(decode_binary(c_1, mapping2)+decode_binary(c_2, mapping2) + decode_binary(reversed_c_3, mapping2) + decode_binary(reversed_c_4, mapping2) + decode_binary(c_5, mapping2) )

# 输出结果:x3c{i_d1dn't_kn0w_mvm_c0uld_be_us3d_f0r_b1n4ry_3nc0d1ng_l0l}

另外一个flag则是藏在右下角的勋章里:
屏幕截图 2025-01-25 210727

人工将这些内容识别出来然后用base64进行解码即可。

内容大概为:

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-----BEGIN CERTIFICATE-----
MIIDyjCCAlCgAwIBAgISBKmF/S4TYSXpTzcor9eZJ/GrMAoGCC
qGSM49BAMDMDIxCzAJBgNVBAYTAlVTMRYwFAYDVQQKEwlMZXQn
cyBFbmNyeXB0MQswOQYDVQQDEwJFNjAeFw0yNTAxMDYyMDM2MD
FaFw0yNTA0MDYyMDM2MDBaMEAxPjA8BgNVBAMMNXgzY3ttdTV0
X2IzX2Zlbl90eXAxbmdfdGgxcl9ieV9oNG5kXzEzNzUxMDUzMD
QyNDgzNjF9MFkwEwYHKoZIzj0CAQYIKoZIzj0DAQcDQgAEcYu3
flnEI2dttI5lQQmzRld72SDdBqCDtfto9pg5t/NFFIolkY8W8C
ryM9XlJEx3NAOGTgBoeUNTuWgiCseQeaOCAjYwggIyMA4GA1Ud
DwEB/wQEAwIHgDAdBgNVHSUEFjAUBggrBgEFBQcDAQYIKwYBBQ
UHAwIwDAYDVR0TAQH/BAIwADAdBgNVRQ4EFgQUrbtyF28hjw8o
IqwXpakw8t7J9jQwHwYDVR0jBBgwFoAUkydGmAOpUWiOmNbEQk
jbI79YlNIwVQYIKwYBBQUHAQEESTBHMCEGCCsGAQUFBzABhhVo
dHRwOi8vZTYuby5sZW5jci5vcmcwIgYIKwYBBQUHMAKGFmh0dH
A6Ly91Ni5pLmxlbmNyLm9yZy8wQAYDVR0RBDkwN4IleDNje2ll
NXRfYjNfZnVuX3R5cDFuZl90aDFzX2J5X2g0bmRfMTM3NTEwNT
MwNDI0ODM2MX0wEwYDVR0gBAwwCjAIBgZngQwBAgEwggEDBgor
BgEEAdZ5AgQCBIH0BIHxAO8AdQB9WR4S4XggexxhZ3xe/fjQhl
w0oE6VnrkDL9kOjC55uAAAAZQ9inTEAAAEAwBGMEQCIENpWRg9
8SQo5JdzyjgnyFeUY0WnNVzn5NkdDA3bzeKbAiBsAkk3fe5esm
7A0efsLN/EyFjEK/NBGqYxgOucgZheQwB2ABNK3xglmEIjeAxv
70x6kaQWtyNJzlhXat+u2qfCq+AiAAAB1D2KdXoAAAQDAEcwRQ
IgBfU4pkiNyNsl+I6skjXz6qqu+mNoI4JvtDsoYxoI+ZoCIQCR
iMQSCEwahN0ImXu3cwDeyM+AbNeve0VgSLMSUBdxvTAKBggghk
jOPQQDAwNoADBlAjEAvxa6nSpUMl7NuDB/+LJfzTskR498vLoe
tnZuHo14J6d9zuFRGQ8Dk4w2aQNsbuVsAjB9fE6GJYBiebb4aH
u/J2amych3KP//D951/CdmiV5PKZqXWWdpaQZL+pbmsXRa8rM=
-----END CERTIFICATE-----

会有一些误差,所以最后提交flag时需要多试几次。

foundations (Osint):

题目描述:

image-20250127171132744

使用https://archive.org/来搜索这个比赛网站的历史纪录内容

image-20250127171253342

可以在这里发现最早的纪录是在2024年7月14日:

image-20250127171328034

点进去会发现:
image-20250127171419684

x3CTF{m4yb3_w3ll_m4ke_4_ch4ll3nge_0u7_0f_7h1s}

mvm:

image-20250127171618699

打开下载文件会得到

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MMVMVMVVMMVMVMVVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVMVMMVMVVVMMMVMVMVVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVMVVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVVVMMMVVVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMMVMVVMVMVMVVMVVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMVMVVMVVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVVVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVMVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMMVMVVVMMMVMVMVVMMVMVMVVMVMVVMVVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVVVVVMMMVMVVMVMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVM

跟之前一样,将其转成二进制再解码会得到

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++[---------->+<]>.+++++++++.---------.-[->+++++<]>-.+[----->+<]>+.+++++++++.---------.-[---->+++++<]>.+[--->++<]>++.>-[--->+<]>---.--[->++++<]>+.++++++++.+++++.[-->+++++++++<]>.[--->+++++<]>.++++++++++.++++++++++++.-[----->+<]>.>-[--->+<]>.-[----->+<]>-.++++++++.------.-.++[->+++++<]>+.[----->++++<]>+.+++++++++.---------.>--[-->+++<]>.

很显然这是Brainfuck,所以找个在线的intepreter运行一下就可以得到flag:

image-20250127172039130

MVM{MVM_BRAIN_IS_FUCKED_MVM}

count-the-mvms

主要是数背景的mvm个数,发现它们的像素点是一样的。所以写个图像匹配脚本即可.

首先要把pdf转换成 png,推荐 adobe acrobat

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import cvlib
# cvlib 是自己写的库
im = cvlib.read_img("certificate_h4tum.png")
sim = cvlib.subrectimg(im, 605, 516, 837, 665)
sim2 = cvlib.subrectimg(sim, 44, 32, 79, 48)
mvm = cvlib.subrectimg(sim2, 2, 2, 32, 13)
print("read success")


def match(im, x,y):
if x+len(mvm) > len(im):
return False
if y+len(mvm[0]) > len(im[0]):
return False

for i in range(len(mvm)):
for j in range(len(mvm[i])):
[r,g,b] = mvm[i][j]
[ri,gi,bi] = im[x+i][y+j]
if r != ri or g != gi or b != bi:
return False
return True

def count_matches(im):
cnt = 0
for i in range(len(im)):
print(i)
for j in range(len(im[i])):
if match(im, i,j):
cnt += 1
j += len(mvm[0]) - 1
return cnt

print(count_matches(im))
print("finish")

Crypto

man-vs-matrix:

题目描述:

image-20250127172638257

打开下载文件会看到:

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from sage.all import *
from Crypto.Util.number import bytes_to_long

class RNG:

def __init__(self, seed):
self.p = next_prime(2**24) # 16777259
self.F = GF(self.p)
self.M = matrix(self.F, 3,3, [bytes_to_long(seed[i:i+3]) for i in range(0, len(seed), 3)])
self.state = vector(self.F, map(ord, "Mvm")) # [77, 118, 109]
self.gen = self.F(2)

def get_random_num(self):
out = self.M * self.state

for i in range(len(self.state)): # len = 3
self.state[i] = self.gen**self.state[i]

return out * self.state

flag = b"MVM{???????????????????????????}"
seed = flag[4:-1] # 27位,27/3=9

rng = RNG(seed)
samples = []

for i in range(9):
samples.append(rng.get_random_num())

print(f"{samples = }")
# samples = [6192533, 82371, 86024, 4218430, 12259879, 16442850, 6736271, 7418630, 15483781]

是几个随机数的生成器(RNG),但生成逻辑非常简单。

每次会计算

(括号外的乘法是内积。在sage里,矩阵与矩阵/向量的乘法和向量与向量的内积都是用*)。并且有

这里的初始state是已知的,所以我们只需要建立一个9元1次线性方程组即可。

我们可以写一段sagemath的代码来通过解方程逆推出matrix以及flag内容:

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p = 16777259		# = next_prime(2**24)
F = GF(p)

samples = [6192533, 82371, 86024, 4218430, 12259879, 16442850, 6736271, 7418630, 15483781]

# 初始状态 S0 = [77, 118, 109]
S0 = vector(F, [77, 118, 109])

def next_state(st):
return vector(F, [F(2)^int(x) for x in st])

# 求出 S0 ~ S9
S = [None]*10
S[0] = S0
for i in range(1, 10):
S[i] = next_state(S[i-1])

# 构造线性方程组 X * M_vec = Y
X = matrix(F, 9, 9)
Y = vector(F, 9)

for i in range(9):
row_coeffs = []
# M_vec 的顺序: M[0,0], M[0,1], M[0,2], M[1,0], ..., M[2,2]
for k in range(3):
for j in range(3):
row_coeffs.append(S[i][j] * S[i+1][k])
X[i] = row_coeffs
Y[i] = samples[i]

# 求解 9 个未知量
M_vec = X.solve_right(Y)
M_mat = matrix(F, 3, 3, M_vec)
print("Recovered M =")
print(M_mat)


# 将 3x3 矩阵以行优先顺序(row-major)展开成 9 个元素
# 对应当初 [bytes_to_long(seed[0:3]), bytes_to_long(seed[3:6]), ..., bytes_to_long(seed[24:27])]
m_ints = []
for i in range(3):
for j in range(3):
# Sage 返回的是 GF(p) 的元素,先转成普通整型
val = int(M_mat[i, j])
m_ints.append(val)

# 将每个 val 转成 3 字节后依次拼接
seed_recovered = b"".join(val.to_bytes(3, "big") for val in m_ints)

# 最终还原 flag = b"MVM{" + seed_recovered + b"}"
flag_recovered = b"MVM{" + seed_recovered + b"}"

print("Recovered seed =", seed_recovered)
print("Recovered flag =", flag_recovered)

#最后得到的结果:
# Recovered M =
# [7090542 3355762 6252149]
# [5137236 3223662 3497780]
# [7484255 7174495 6698102]
# Recovered seed = b'l1n34r_fuNcT10n5_4r3_my_f4v'
# Recovered flag = b'MVM{l1n34r_fuNcT10n5_4r3_my_f4v}'