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import sys
import math
from Crypto.Util.number import long_to_bytes, inverse
# --------------------------------------------------------------------------
# 纯整数 LLL 算法 (Integer LLL)
# --------------------------------------------------------------------------
def lll_reduction_integer(basis, delta_num=99, delta_den=100):
"""
纯整数 LLL 算法,避免浮点数精度问题。
"""
n = len(basis)
m = len(basis[0])
b = [list(row) for row in basis] # 复制基
# 内部使用大整数维护 Gram-Schmidt 系数
# 为了效率,我们使用简化的 "Rational LLL" 思想的变体
# 这里使用一个非常稳健的定点数模拟
PREC = 4000
ONE = 1 << PREC
# 维护定点数正交基
def update_gs(b_int):
n = len(b_int)
b_star = []
mu = [[0]*n for _ in range(n)]
sq_norms = []
for i in range(n):
vec = list(b_int[i])
for j in range(i):
# dot(b[i], b_star[j])
dot_val = sum(b_int[i][k] * b_star[j][k] for k in range(m))
norm_j = sq_norms[j]
if norm_j == 0: c = 0
else: c = (dot_val * ONE) // norm_j
mu[i][j] = c
for k in range(m):
vec[k] -= (c * b_star[j][k]) // ONE
b_star.append(vec)
sq_norms.append(sum(x*x for x in vec))
return b_star, mu, sq_norms
# 初始化
b_shifted = [[x * ONE for x in row] for row in b]
b_star, mu, sq_norms = update_gs(b_shifted)
k = 1
while k < n:
# Size Reduction
for j in range(k - 1, -1, -1):
if abs(mu[k][j]) > (ONE // 2):
q = (mu[k][j] + (ONE // 2)) // ONE
for idx in range(m):
b[k][idx] -= q * b[j][idx]
b_shifted[k][idx] -= q * b_shifted[j][idx]
b_star, mu, sq_norms = update_gs(b_shifted)
# Lovasz Condition
# |b*_k|^2 >= (delta - mu[k][k-1]^2) |b*_{k-1}|^2
norm_k = sq_norms[k]
norm_k1 = sq_norms[k-1]
mu_val = mu[k][k-1]
# logical check: norm_k * ONE^2 >= (delta * ONE^2 - mu^2) * norm_k1
lhs = norm_k * ONE * ONE
rhs = (delta_num * ONE * ONE // delta_den - mu_val * mu_val) * norm_k1 // (ONE * ONE)
rhs = rhs * ONE * ONE
if lhs < rhs:
b[k], b[k-1] = b[k-1], b[k]
b_shifted[k], b_shifted[k-1] = b_shifted[k-1], b_shifted[k]
b_star, mu, sq_norms = update_gs(b_shifted)
k = max(k - 1, 1)
else:
k += 1
return b
# --------------------------------------------------------------------------
# 工具函数
# --------------------------------------------------------------------------
def solve_pell(D):
m = 0
d = 1
a = int(math.isqrt(D))
if a * a == D: return None
a0 = a
h_prev, h_curr = 0, 1
k_prev, k_curr = 1, 0
while True:
h_next = a * h_curr + h_prev
k_next = a * k_curr + k_prev
if h_next**2 - D * k_next**2 == 1:
return h_next, k_next
h_prev, h_curr = h_curr, h_next
k_prev, k_curr = k_curr, k_next
m = d * a - m
d = (D - m * m) // d
a = (a0 + m) // d
# --------------------------------------------------------------------------
# 主逻辑
# --------------------------------------------------------------------------
def main():
# 题目数据
n_val = 18443962106578943927922829208562388331564422618353954662348987125496135728205879853444693999188714508145409575298801277623433658530589571956301880815632542860363148763704636874275223979061507756787642735086825973011622866458454405794279633717255674221895468734500735123736684346340314680683830866884050311047424068122453972745273167956795195575475691048908906061023817574695902603984554911326264947716547564759877947888574515784489778380086664649338093680740990860192640619047071160362288611331225632270531304525264824445326394068892806774552310748255977040249822464839809344521107040968321810533993659358229305320413
c_val = 8176283809770578639445916571748890916863681496488338436815389781344271720445865752568007651231910205530735296305471880971422173915403956857863330698931559658909826642456860761540607878553228782799635976463090037022164739976302533892173751687781100980039065722082091714141141136171701360981540040678479802206949078162548124224838019262997441233919136963696523351831737708850863538007579105976954619102728135600542584651031405327214877358323388674864043740117718200022790892542634633918493245432384562983429810936975869853596007429259749282607844407676244954057886824475948603911174707176467261179324130051317766768127
A_val = 1293023064232431070902426583269468463
B_val = 105279230912868770223946474836383391725923
gift2 = 26161714402997656593966327522661504448812191236385246127313450633226841096347099194721417620572738092514050785292503472019045698167235604357096118735431692892202119807587271344465029467089266358735895706496467947787464475365718387614
e = 65537
# --- Step 1 ---
print("[*] Step 1: Pell Equation")
D = B_val // A_val
x_sol, y_sol = solve_pell(D)
# --- Step 2 ---
print("[*] Step 2: Hensel Lifting")
bits_target = 777
candidates = [1]
for k in range(1, bits_target):
mod = 2**(k + 1)
new_candidates = []
for cand in candidates:
p0 = cand
if (pow(p0, 20, mod) - (gift2 * pow(p0, 13, mod)) + pow(n_val, 13, mod)) % mod == 0:
new_candidates.append(p0)
p1 = cand | (1 << k)
if (pow(p1, 20, mod) - (gift2 * pow(p1, 13, mod)) + pow(n_val, 13, mod)) % mod == 0:
new_candidates.append(p1)
candidates = new_candidates
# --- Step 3: Coppersmith Attack (Dimension 3) ---
print("[*] Step 3: Coppersmith Attack (3x3 Lattice)")
p = 0
q = 0
for p_low in candidates:
inv_scale = inverse(2**777, n_val)
A = (p_low * inv_scale) % n_val
X = 2**250
# 构造 3维格
# Polynomials: N, (x+A), x(x+A)
# Shifted vectors (coeffs of 1, x, x^2):
# 1. N(1) -> [N, 0, 0]
# 2. (x+A) -> [A, X, 0]
# 3. x(x+A) -> [0, AX, X^2] <-- 注意这里的系数位置
# x(x+A) = x^2 + Ax. Coeffs: const=0, x=A, x^2=1. Scaled by X^k:
# x coeff is scaled by X. x^2 coeff is scaled by X^2.
# Vector: [0, A*X, 1*X*X]
M = [
[n_val, 0, 0],
[A, X, 0],
[0, A*X, X*X]
]
reduced = lll_reduction_integer(M)
# 检查所有行
for row in reduced:
# P(x) = v0 + v1(x/X) + v2(x/X)^2
# P(x) = row[0] + (row[1]//X)*x + (row[2]//X^2)*x^2
c0 = row[0]
c1 = row[1] // X
c2 = row[2] // (X*X)
# 我们在寻找整数根 x。
# 如果 c2 == 0, 则是线性方程: x = -c0/c1
if c2 == 0 and c1 != 0:
if c0 % c1 == 0:
x_cand = -c0 // c1
p_cand = x_cand * (2**777) + p_low
if p_cand > 1 and n_val % p_cand == 0:
p = int(p_cand)
q = n_val // p
print(f" [+] Found p = {p}")
break
# 如果 c2 != 0, 解二次方程: c2*x^2 + c1*x + c0 = 0
elif c2 != 0:
delta = c1*c1 - 4*c2*c0
if delta >= 0:
sq = math.isqrt(delta)
if sq * sq == delta:
# roots = (-c1 +/- sq) / 2c2
for s in [1, -1]:
num = -c1 + s*sq
den = 2*c2
if den != 0 and num % den == 0:
x_cand = num // den
p_cand = x_cand * (2**777) + p_low
if p_cand > 1 and n_val % p_cand == 0:
p = int(p_cand)
q = n_val // p
print(f" [+] Found p = {p}")
break
if p: break
if p: break
if not p:
print("[-] Step 3 Failed.")
return
# --- Step 4 ---
phi = (p - 1) * (q - 1)
d = inverse(e, phi)
m_val = pow(c_val, d, n_val)
print(f" [+] m recovered")
# --- Step 5: Flag Lattice (Scaled) ---
print("[*] Step 5: Flag Lattice (Scaled)")
# Scale last column by S to balance the lattice
# f0 ~ 2^256, diff ~ 2^99.
# We want f0 approx S * diff. => S approx 2^157.
S = 2**160
M_flag = [
[1, 0, x_sol * S],
[0, 1, y_sol * S],
[0, 0, m_val * S]
]
reduced = lll_reduction_integer(M_flag)
for row in reduced:
for sign in [1, -1]:
try:
# row[0], row[1] correspond to f0, f1
part1 = long_to_bytes(abs(row[0]))
part2 = long_to_bytes(abs(row[1]))
cands = [part1 + part2, part2 + part1]
for c in cands:
if b'flag{' in c or b'}' in c:
print("\n[SUCCESS] Flag: " + c.decode(errors='ignore'))
return
except: pass
print("[-] Flag auto-check failed.")
if __name__ == "__main__":
main()
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