# x3ctf Write Up

# 1. Misc

# p11n-trophy（签到题）:

题目描述：

我们首先会得到这样一份证书：

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

#

其中一个 flag 是藏在 certificate 周围一圈的位置：

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

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：

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：

-----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：

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

可以在这里发现最早的纪录是在 2024 年 7 月 14 日：

点进去会发现：

x3CTF

# mvm：

打开下载文件会得到

MMVMVMVVMMVMVMVVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVMVMMVMVVVMMMVMVMVVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVMVVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVVVMMMVVVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMMVMVVMVMVMVVMVVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMVMVVMVVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVVVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVMMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVMVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVMVVMVMMVMVVVMMMVMVMVVMMVMVMVVMVMVVMVVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMVMVVMVVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVMVVMMVMVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVMVVMMVMVVVMMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVMVMMVMVVVMMMVVVVVMMMVMVVMVMMVMVVMVMVMVVMVVMMVMVVMVMMVMVVMVMMVVVVVMMMVMVMVVMMVMVMVVMMVMVMVVMMVVVVMMMVMVVVMVMMVVVVVMMMVMVVVM

跟之前一样，将其转成二进制再解码会得到

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

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

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

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")

# 2. Crypto

# man-vs-matrix:

题目描述：

打开下载文件会看到：

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

$$(M \cdot state) \cdot new\_state$$

(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

$$new\_state[i] = 2^{state[i]}.$$

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：

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}'