赛时就做了两题，跟着佬的wp复现一下
task
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from sympy import primerange
import random
from collections import deque
def generate(size):
grid = [[random.randint(0, 9) for col in range(size)] for row in range(size)]
grid[0][0] = 0
return grid
def encrypt(n, a, b, mod=101):
return (a * n + b) % mod
def build_encrypted_grid(grid, a, b, mod=101):
size = 10
encry_grid = []
for y in range(size):
row = []
for x in range(size):
enc_val = encrypt(grid[y][x], a, b, mod)
row.append(str(enc_val).zfill(2))
encry_grid.append(row)
return encry_grid
def optimize(grid):
#hidden
pass
grid = generate(10)
a = random.choice(list(primerange(2, 12)))
b = random.choice(range(101))
encry_grid = build_encrypted_grid(grid, a, b, mod=101)
#nc chals1.apoorvctf.xyz 4002
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generate是正常一个10*10的迷宫，每个数字0-9，起点是0
encrypt是正常的线性同余加密
optimize没啥用
build_encrypted_grid是生成一个没个数加密后的矩阵，并且输出为字符串，比如5->05
a=[2,3,5,7,11]
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//nc情况
____ _ _
/ ___| ___ _ __ (_) _ _ | |_ ___ _ _
| | _ / _ \| '_ \ | || | | || __|/ __|| | | |
| |_| || __/| | | | | || |_| || |_ \__ \| |_| |
\____| \___||_| |_| _/ | \__,_| \__||___/ \__,_|
|__/
Welcome to Genjutsu Labyrinth!
Your goal is to navigate from the top-left to the bottom-right successfully
Note: Your current position is denoted by Pa. The first cell has a value 0
-------------------------------------------------
Pa 49 42 42 00 00 21 00 21 00
49 00 42 21 28 56 07 00 42 56
21 28 42 42 14 21 21 21 00 14
07 49 07 35 07 07 42 56 35 07
63 28 28 07 00 49 00 56 21 28
07 56 07 35 14 42 21 35 35 00
14 21 00 07 21 35 49 07 14 28
21 35 07 00 49 14 21 00 42 42
42 21 56 28 49 56 07 14 49 28
63 07 49 35 07 07 07 28 63 00
Use S/D to move down or right. Type 'exit' to quit.
Enter move (S/D):
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哎，其实我们要找的是xor为0，即从 enc_val 中穷举 a, b 的值。然后根据 a, b 的值求出 grid 的值，进行穷举路径，找到 XOR 值为 0 的路径。
exp
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from pwn import *
import itertools
from Crypto.Util.number import inverse
p = remote('chals1.apoorvctf.xyz', 4002)
for _ in range(12):
data = p.recvuntil(b'\n').strip().decode()
print(data)
# 读取加密网格，每行接收后替换'Pa'为'00'，并转换成数字列表
encry_grid = []
for _ in range(10):
data = p.recvuntil(b'\n').strip().decode()
print(data)
data = data.replace('Pa', '00')
row = list(map(int, data.split()))
encry_grid.append(row)
encs = []
for i in range(10):
for j in range(10):
if encry_grid[i][j] not in encs:
encs.append(encry_grid[i][j])
found = False
for a in [2, 3, 5, 7, 11]:
for b in range(101):
success = True
for i in range(10):
c = (a * i + b) % 101
if c not in encs:
success = False
break
if success:
found = True
break
if found:
break
grid = []
for i in range(10):
row = []
for j in range(10):
r = (encry_grid[i][j] - b) * inverse(a, 101) % 101
row.append(r)
grid.append(row)
# 构造移动指令：总共需要18步，选取9步向下（S）移动，其余为向右（D）移动，
# 利用排列枚举所有可能的路径，选择 XOR 结果为 0 的路径
seq = list(range(18))
for perm in itertools.permutations(seq, 9):
com = ''
val = 0
x = 0
y = 0
for i in range(18):
if i in perm:
com += 'S'
x += 1
val ^= grid[x][y]
else:
com += 'D'
y += 1
val ^= grid[x][y]
if val == 0:
break
for i in range(18):
data = p.recvuntil(b': ')
print(data.decode() + com[i])
p.sendline(com[i].encode())
data = p.recvuntil(b'\n').strip().decode()
print(data)
data = p.recvuntil(b'exit!\n').strip().decode()
print(data)
for _ in range(2):
data = p.recvuntil(b'\n').strip().decode()
print(data)
p.interactive()
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两个照片什么都没有，有点抽象
尝试了很多方法后，将 RGB 的值相加后除以 256 的余数，中央出现了旗帜。
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from PIL import Image
img1 = Image.open('part1.png').convert('RGB')
img2 = Image.open('part2.png').convert('RGB')
w, h = img1.size
output_img = Image.new('RGB', (w, h), (255, 255, 255))
for y in range(h):
for x in range(w):
r1, g1, b1 = img1.getpixel((x, y))
r2, g2, b2 = img2.getpixel((x, y))
r = (r1 + r2) % 256
g = (g1 + g2) % 256
b = (b1 + b2) % 256
output_img.putpixel((x, y), (r, g, b))
output_img.save('flag.png')
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task
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import hashlib
def check_hex_data(hex1, hex2, start_string):
if hex1 == hex2:
return "Error: Even a Saiyan warrior knows that true strength lies in difference! The two inputs must not be identical."
try:
data1 = bytes.fromhex(hex1)
data2 = bytes.fromhex(hex2)
except ValueError:
return "Error: Looks like you misfired a Ki blast! Invalid hex input detected."
start_bytes = start_string.encode()
if not (data1.startswith(start_bytes) and data2.startswith(start_bytes)):
return "Error: These aren't true warriors! Both inputs must start with the legendary sign of 'GOKU' to proceed."
def md5_hash(data):
hasher = hashlib.md5()
hasher.update(data)
return hasher.hexdigest()
hash1 = md5_hash(data1)
hash2 = md5_hash(data2)
if hash1 != hash2:
return "Error: These warriors are impostors! They wear the same armor but their Ki signatures (MD5 hashes) don't match."
try:
with open("flag.txt", "r") as flag_file:
flag = flag_file.read().strip()
return f"🔥 You have found the real Goku! Your flag is: {flag}"
except FileNotFoundError:
return "Error: The Dragon Balls couldn't summon the flag! 'flag.txt' is missing."
if __name__ == "__main__":
start_string = "GOKU"
hex1 = input("Enter first hex data: ")
hex2 = input("Enter second hex data: ")
print(check_hex_data(hex1, hex2, start_string))
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总而言之言而总之想让md5相等，直接开爆
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from pwn import *
p = remote('chals1.apoorvctf.xyz', 5002)
with open('md5_data1', 'rb') as f:
data1 = f.read().hex()
with open('md5_data2', 'rb') as f:
data2 = f.read().hex()
data = p.recvuntil(b': ')
print(data.decode() + data1)
p.sendline(data1.encode())
data = p.recvuntil(b': ')
print(data.decode() + data2)
p.sendline(data2.encode())
data = p.recvuntil(b'\n').rstrip()
print(data.decode())
p.interactive()
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task
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import sys
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from random import randbytes
def main():
key = randbytes(16)
cipher = AES.new(key, AES.MODE_ECB)
flag = b'apoorvctf{fake_flag_123}'
print("Welcome to the ECB Oracle challenge!")
print("Enter your input in hex format.")
try:
while True:
print("Enter your input: ", end="", flush=True)
userinput = sys.stdin.readline().strip()
if not userinput:
break
try:
userinput = bytes.fromhex(userinput)
ciphertext = cipher.encrypt(pad(userinput + flag + userinput, 16))
print("Ciphertext:", ciphertext.hex())
except Exception as e:
print(f"Error: {str(e)}")
except KeyboardInterrupt:
print("Server shutting down.")
if __name__ == "__main__":
main()
# nc chals1.apoorvctf.xyz 4001
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很有意思，加密格式变成x+flag+x，我们ECB的特点就是相同的明文块会生成相同的密文块，padding怎么办呢？消除影响就好了
exp
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from pwn import *
p = remote('chals1.apoorvctf.xyz', 4001)
for _ in range(2):
data = p.recvuntil(b'\n').strip().decode()
print(data)
flag = ''
for i in range(32):
for code in range(33, 127):
inp = 'X' * (31 - i) + flag + chr(code) + 'X' * (31 - i)
print('[+] input:', inp)
inp_hex = inp.encode().hex()
prompt = p.recvuntil(b': ').decode()
print(prompt + inp_hex)
p.sendline(inp_hex.encode())
data = p.recvuntil(b'\n').strip().decode()
print(data)
ct = data.split(' ')[-1]
if ct[32:64] == ct[96:128]:
flag += chr(code)
break
print('[*] flag:', flag)
p.interactive()
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