結果
| 問題 | No.2272 多項式乗算 mod 258280327 |
| コンテスト | |
| ユーザー |
 qwewe
|
| 提出日時 | 2025-05-14 12:59:51 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 8,903 bytes |
| 記録 | |
| コンパイル時間 | 339 ms |
| コンパイル使用メモリ | 82,228 KB |
| 実行使用メモリ | 238,472 KB |
| 最終ジャッジ日時 | 2025-05-14 13:01:53 |
| 合計ジャッジ時間 | 10,032 ms |
|
ジャッジサーバーID （参考情報） |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 27 WA * 3 TLE * 1 -- * 2 |
ソースコード
import sys
# Fast modular exponentiation
def fast_pow(base, power, modulus):
"""Computes (base^power) % modulus efficiently."""
result = 1
base %= modulus
while power > 0:
if power % 2 == 1:
result = (result * base) % modulus
base = (base * base) % modulus
power >>= 1 # Use right shift for integer division by 2
return result
# Modular inverse using Fermat's Little Theorem
def mod_inverse(a, m):
"""Computes the modular inverse of a modulo m using Fermat's Little Theorem.
Assumes m is prime and a is not divisible by m."""
# Check if inverse exists (a is not 0 mod m)
if a % m == 0:
# This case should not happen if used correctly with NTT primes
# and non-zero elements required for inverse (like 1/n).
# For CRT, M_i % mod_i != 0 should hold because NTT moduli are primes.
raise ValueError("Modular inverse does not exist for a divisible by m")
return fast_pow(a, m - 2, m)
# Number Theoretic Transform (NTT)
def ntt(a, mod, root, inverse=False):
"""Performs Number Theoretic Transform (NTT) or its inverse.
Modifies the input list 'a' in-place.
'mod' must be prime.
'root' must be a primitive root modulo 'mod'.
'(mod - 1)' must be divisible by n = len(a), and n must be a power of 2.
"""
n = len(a)
if n == 1:
return a
# Bit-reversal permutation - ensures elements are in the correct order for iterative FFT
# Precompute or compute on the fly; this is standard iterative method
rev = [0] * n
for i in range(n):
# Calculate reverse bit order index
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1))
# Swap elements if the current index is less than the reversed index
if i < rev[i]:
a[i], a[rev[i]] = a[rev[i]], a[i]
# Cooley-Tukey butterfly - iterative version
len_ = 2
while len_ <= n:
# Calculate the principal n/len_-th root of unity for this level
wlen = fast_pow(root, (mod - 1) // len_, mod)
if inverse:
# For inverse NTT, use the inverse of the root
wlen = mod_inverse(wlen, mod)
half_len = len_ >> 1
# Iterate through blocks of size len_
for i in range(0, n, len_):
w = 1 # Current power of the principal root for this block
# Iterate through the first half of the block
for j in range(half_len):
u = a[i + j]
v = (w * a[i + j + half_len]) % mod
# Butterfly operation
a[i + j] = (u + v) % mod
# Ensure the result of subtraction is non-negative before modulo
a[i + j + half_len] = (u - v + mod) % mod
# Update the power of the root for the next element
w = (w * wlen) % mod
len_ <<= 1 # Double the block size for the next level
# Scale by 1/n for inverse NTT
if inverse:
n_inv = mod_inverse(n, mod)
for i in range(n):
a[i] = (a[i] * n_inv) % mod
# The function modifies 'a' in-place, but returning it is conventional
return a
# Main function to solve the problem
def solve():
P = 258280327 # The target modulus given in the problem
# Choose NTT-friendly primes and their primitive roots.
# Their product should be larger than the maximum possible coefficient
# of the product polynomial (F mod P) * (G mod P), which is roughly (N+M+1)*P^2.
# With N, M <= 2e5 and P ~ 2.6e8, max coeff ~ 4e5 * (2.6e8)^2 ~ 2.7e22.
# The product of three primes (~10^9 each) is ~10^27, which is sufficient.
MOD1 = 998244353 # Prime, P-1 = 2^23 * 7 * 17
ROOT1 = 3 # Primitive root for MOD1
MOD2 = 1004535809 # Prime, P-1 = 2^21 * 479
ROOT2 = 3 # Primitive root for MOD2
MOD3 = 1012924417 # Prime, P-1 = 2^24 * 483
ROOT3 = 5 # Primitive root for MOD3
MODS = [MOD1, MOD2, MOD3]
ROOTS = [ROOT1, ROOT2, ROOT3]
# Read input polynomials F(x) and G(x) using fast I/O
N = int(sys.stdin.readline())
F_coeffs = list(map(int, sys.stdin.readline().split()))
M = int(sys.stdin.readline())
G_coeffs = list(map(int, sys.stdin.readline().split()))
# Handle the edge case where one or both polynomials are the zero polynomial.
# The problem defines the degree of the zero polynomial as 0.
# A polynomial is zero if its degree is 0 and its only coefficient is 0.
is_F_zero = (N == 0 and len(F_coeffs) == 1 and F_coeffs[0] == 0)
is_G_zero = (M == 0 and len(G_coeffs) == 1 and G_coeffs[0] == 0)
if is_F_zero or is_G_zero:
print(0) # Degree of the resulting zero polynomial
print(0) # Coefficients list is just [0]
return
# Reduce input coefficients modulo P. We want to compute (F(x) * G(x)) mod P.
# This is equivalent to computing (F(x) mod P) * (G(x) mod P) mod P.
F_mod_P = [c % P for c in F_coeffs]
G_mod_P = [c % P for c in G_coeffs]
# The degree of the product polynomial H(x) = F(x) * G(x) is L = N + M.
L = N + M
# Determine the size for NTT: smallest power of 2 that is >= L + 1.
# The result polynomial has L+1 coefficients (from degree 0 to L).
n = 1
while n <= L:
n <<= 1
# Store the results of polynomial multiplication modulo each NTT prime
results_mod_pi = []
# Perform NTT-based polynomial multiplication for each modulus
for i in range(3):
mod = MODS[i]
root = ROOTS[i]
# Prepare coefficient lists modulo the current NTT prime 'mod'.
# Start with coefficients already reduced modulo P.
F_pi = list(F_mod_P)
G_pi = list(G_mod_P)
# Reduce coefficients further modulo the current NTT prime 'mod'.
# This is only necessary if mod < P, but good practice.
F_pi = [c % mod for c in F_pi]
G_pi = [c % mod for c in G_pi]
# Pad coefficient lists with zeros to length n for NTT.
F_pi.extend([0] * (n - len(F_pi)))
G_pi.extend([0] * (n - len(G_pi)))
# Perform NTT. Pass copies because ntt modifies lists in-place.
ntt_F = ntt(list(F_pi), mod, root, inverse=False)
ntt_G = ntt(list(G_pi), mod, root, inverse=False)
# Pointwise multiplication in the NTT domain (convolution theorem)
ntt_H = [(ntt_F[k] * ntt_G[k]) % mod for k in range(n)]
# Perform inverse NTT to get the coefficients of the product polynomial H(x) modulo 'mod'.
# ntt function modifies ntt_H in-place.
H_pi = ntt(ntt_H, mod, root, inverse=True)
# Store the resulting coefficients (we only need up to degree L).
results_mod_pi.append(H_pi[:L+1])
# Combine the results using the Chinese Remainder Theorem (CRT)
# We have H(x) mod MOD1, H(x) mod MOD2, H(x) mod MOD3.
# We want to find H(x) mod P.
# First, find H(x) mod (MOD1*MOD2*MOD3) using CRT.
H_final = [0] * (L + 1) # Stores the final coefficients modulo P
# Precompute CRT constants
M_prod = MOD1 * MOD2 * MOD3 # Product of NTT moduli
M1 = M_prod // MOD1
M2 = M_prod // MOD2
M3 = M_prod // MOD3
# Compute modular inverses required for CRT. Check for errors.
try:
y1 = mod_inverse(M1, MOD1)
y2 = mod_inverse(M2, MOD2)
y3 = mod_inverse(M3, MOD3)
except ValueError:
# This error should not occur if MOD1, MOD2, MOD3 are distinct primes.
print("Error: Modular inverse failed in CRT setup.", file=sys.stderr)
return
# Precompute terms for the CRT formula: Ci = (M/Mi) * mod_inverse(M/Mi, Mi)
C1 = M1 * y1
C2 = M2 * y2
C3 = M3 * y3
# Apply CRT for each coefficient k from 0 to L
for k in range(L + 1):
# Get the k-th coefficient modulo each NTT prime
r1 = results_mod_pi[0][k]
r2 = results_mod_pi[1][k]
r3 = results_mod_pi[2][k]
# Apply CRT formula: H_k = sum(ri * Ci) mod M_prod
# Python handles large integers required for intermediate calculations.
H_k_mod_Mprod = (r1 * C1 + r2 * C2 + r3 * C3) % M_prod
# The final coefficient is H_k mod P.
H_final[k] = H_k_mod_Mprod % P
# Determine the actual degree of the resulting polynomial H(x) modulo P.
# This is the index of the highest non-zero coefficient.
L_actual = L
while L_actual >= 0 and H_final[L_actual] == 0:
L_actual -= 1
# Handle the case where the product polynomial is the zero polynomial modulo P.
if L_actual < 0:
# All coefficients are zero modulo P.
print(0) # Degree is 0
print(0) # Coefficients list is [0]
else:
# Print the actual degree L_actual
print(L_actual)
# Print the coefficients from degree 0 up to L_actual, space-separated.
print(*(H_final[:L_actual + 1]))
# Execute the solver function
solve()