結果

問題 No.1962 Not Divide
ユーザー 👑 rin204rin204
提出日時 2022-05-28 15:43:36
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 3,689 bytes
コンパイル時間 222 ms
コンパイル使用メモリ 82,288 KB
実行使用メモリ 99,700 KB
最終ジャッジ日時 2024-09-20 22:24:29
合計ジャッジ時間 6,814 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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ファイルパターン 結果
sample AC * 3
other AC * 19 RE * 2
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ソースコード

diff #
プレゼンテーションモードにする

from collections import deque
MOD = 998244353
class FFT:
inv_ = [1]
def __init__(self, MOD=998244353):
FFT.MOD = MOD
g = self.primitive_root_constexpr()
ig = pow(g, FFT.MOD - 2, FFT.MOD)
FFT.W = [pow(g, (FFT.MOD - 1) >> i, FFT.MOD) for i in range(30)]
FFT.iW = [pow(ig, (FFT.MOD - 1) >> i, FFT.MOD) for i in range(30)]
def primitive_root_constexpr(self):
if FFT.MOD == 998244353:
return 3
elif FFT.MOD == 200003:
return 2
elif FFT.MOD == 167772161:
return 3
elif FFT.MOD == 469762049:
return 3
elif FFT.MOD == 754974721:
return 11
divs = [0] * 20
divs[0] = 2
cnt = 1
x = (FFT.MOD - 1) // 2
while x % 2 == 0:
x //= 2
i = 3
while i * i <= x:
if x % i == 0:
divs[cnt] = i
cnt += 1
while x % i == 0:
x //= i
i += 2
if x > 1:
divs[cnt] = x
cnt += 1
g = 2
while 1:
ok = True
for i in range(cnt):
if pow(g, (FFT.MOD - 1) // divs[i], FFT.MOD) == 1:
ok = False
break
if ok:
return g
g += 1
def fft(self, k, f):
for l in range(k, 0, -1):
d = 1 << l - 1
U = [1]
for i in range(d):
U.append(U[-1] * FFT.W[l] % FFT.MOD)
for i in range(1 << k - l):
for j in range(d):
s = i * 2 * d + j
f[s], f[s + d] = (f[s] + f[s + d]) % FFT.MOD, U[j] * (f[s] - f[s + d]) % FFT.MOD
def ifft(self, k, f):
for l in range(1, k + 1):
d = 1 << l - 1
for i in range(1 << k - l):
u = 1
for j in range(i * 2 * d, (i * 2 + 1) * d):
f[j+d] *= u
f[j], f[j + d] = (f[j] + f[j + d]) % FFT.MOD, (f[j] - f[j + d]) % FFT.MOD
u = u * FFT.iW[l] % FFT.MOD
def convolve(self, A, B):
n0 = len(A) + len(B) - 1
k = (n0).bit_length()
n = 1 << k
A += [0] * (n - len(A))
B += [0] * (n - len(B))
self.fft(k, A)
self.fft(k, B)
A = [a * b % FFT.MOD for a, b in zip(A, B)]
self.ifft(k, A)
inv = pow(n, FFT.MOD - 2, FFT.MOD)
A = [a * inv % FFT.MOD for a in A]
del A[n0:]
return A
# [x ^ n] P(x) / Q(x)
def BostanMori(P, Q, n):
fft = FFT()
while n:
R = [(x * (-1) ** (i % 2)) % MOD for i, x in enumerate(Q)]
Q = fft.convolve(Q, R[:])[::2]
P = fft.convolve(P, R[:])[n % 2::2]
n >>= 1
return P[0] * pow(Q[0], MOD - 2, MOD) % MOD
n, m = map(int, input().split())
queue = deque()
fft = FFT()
for i in range(2, m + 1):
q = [1] * (i + 1)
q[i] = -1
p = [1] * i
p[0] = 0
queue.append((p, q))
while len(queue) >= 2:
p1, q1 = queue.popleft()
p2, q2 = queue.popleft()
l1 = len(q1)
l2 = len(q2)
nq = fft.convolve(q1[:], q2[:])
np1 = fft.convolve(p1, q2)
np2 = fft.convolve(p2, q1)
assert len(np1) == len(np2)
np = [(p1 + p2) % MOD for p1, p2 in zip(np1, np2)]
queue.append((np, nq))
P, Q = queue.popleft()
Q_P = Q[:] + [0] * (len(P) - len(Q))
for i, p in enumerate(P):
Q_P[i] -= p
Q_P[i] %= MOD
ans = BostanMori(Q, Q_P, n)
print(ans)
"""
(1 + x - x^2) / (1 - x^2)
(1 + x - x^2) / (2 + x - 2x^2)
(1 + x + x^2 - x^3) / (1 - x^3)
(1 + x + x^2 - x^3) / (2 + x + x^2 - 2x^2)
"""
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