結果
| 問題 |
No.2660 Sweep Cards (Easy)
|
| コンテスト | |
| ユーザー |
 chineristAC
|
| 提出日時 | 2024-01-15 00:56:15 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
RE
(最新)
AC
(最初)
|
| 実行時間 | - |
| コード長 | 9,081 bytes |
| コンパイル時間 | 322 ms |
| コンパイル使用メモリ | 82,124 KB |
| 実行使用メモリ | 84,508 KB |
| 最終ジャッジ日時 | 2024-09-29 11:03:14 |
| 合計ジャッジ時間 | 5,909 ms |
|
ジャッジサーバーID （参考情報） |
judge5 / judge3 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | RE * 3 |
| other | RE * 32 |
ソースコード
import sys
from itertools import permutations
from heapq import heappop,heappush
from collections import deque
import random
import bisect
from math import gcd
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
mod = 998244353
omega = pow(3,119,mod)
rev_omega = pow(omega,mod-2,mod)
N = 5*10**5
g1 = [1]*(N+1) # 元テーブル
g2 = [1]*(N+1) #逆元テーブル
inv = [1]*(N+1) #逆元テーブル計算用テーブル
for i in range( 2, N + 1 ):
g1[i]=( ( g1[i-1] * i ) % mod )
inv[i]=( ( -inv[mod % i] * (mod//i) ) % mod )
g2[i]=( (g2[i-1] * inv[i]) % mod )
inv[0]=0
_fft_mod = 998244353
_fft_imag = 911660635
_fft_iimag = 86583718
_fft_rate2 = (911660635, 509520358, 369330050, 332049552, 983190778, 123842337, 238493703, 975955924, 603855026, 856644456, 131300601,
842657263, 730768835, 942482514, 806263778, 151565301, 510815449, 503497456, 743006876, 741047443, 56250497, 867605899)
_fft_irate2 = (86583718, 372528824, 373294451, 645684063, 112220581, 692852209, 155456985, 797128860, 90816748, 860285882, 927414960,
354738543, 109331171, 293255632, 535113200, 308540755, 121186627, 608385704, 438932459, 359477183, 824071951, 103369235)
_fft_rate3 = (372528824, 337190230, 454590761, 816400692, 578227951, 180142363, 83780245, 6597683, 70046822, 623238099,
183021267, 402682409, 631680428, 344509872, 689220186, 365017329, 774342554, 729444058, 102986190, 128751033, 395565204)
_fft_irate3 = (509520358, 929031873, 170256584, 839780419, 282974284, 395914482, 444904435, 72135471, 638914820, 66769500,
771127074, 985925487, 262319669, 262341272, 625870173, 768022760, 859816005, 914661783, 430819711, 272774365, 530924681)
def _butterfly(a):
n = len(a)
h = (n - 1).bit_length()
len_ = 0
while len_ < h:
if h - len_ == 1:
p = 1 << (h - len_ - 1)
rot = 1
for s in range(1 << len_):
offset = s << (h - len_)
for i in range(p):
l = a[i + offset]
r = a[i + offset + p] * rot % _fft_mod
a[i + offset] = (l + r) % _fft_mod
a[i + offset + p] = (l - r) % _fft_mod
if s + 1 != (1 << len_):
rot *= _fft_rate2[(~s & -~s).bit_length() - 1]
rot %= _fft_mod
len_ += 1
else:
p = 1 << (h - len_ - 2)
rot = 1
for s in range(1 << len_):
rot2 = rot * rot % _fft_mod
rot3 = rot2 * rot % _fft_mod
offset = s << (h - len_)
for i in range(p):
a0 = a[i + offset]
a1 = a[i + offset + p] * rot
a2 = a[i + offset + p * 2] * rot2
a3 = a[i + offset + p * 3] * rot3
a1na3imag = (a1 - a3) % _fft_mod * _fft_imag
a[i + offset] = (a0 + a2 + a1 + a3) % _fft_mod
a[i + offset + p] = (a0 + a2 - a1 - a3) % _fft_mod
a[i + offset + p * 2] = (a0 - a2 + a1na3imag) % _fft_mod
a[i + offset + p * 3] = (a0 - a2 - a1na3imag) % _fft_mod
if s + 1 != (1 << len_):
rot *= _fft_rate3[(~s & -~s).bit_length() - 1]
rot %= _fft_mod
len_ += 2
def _butterfly_inv(a):
n = len(a)
h = (n - 1).bit_length()
len_ = h
while len_:
if len_ == 1:
p = 1 << (h - len_)
irot = 1
for s in range(1 << (len_ - 1)):
offset = s << (h - len_ + 1)
for i in range(p):
l = a[i + offset]
r = a[i + offset + p]
a[i + offset] = (l + r) % _fft_mod
a[i + offset + p] = (l - r) * irot % _fft_mod
if s + 1 != (1 << (len_ - 1)):
irot *= _fft_irate2[(~s & -~s).bit_length() - 1]
irot %= _fft_mod
len_ -= 1
else:
p = 1 << (h - len_)
irot = 1
for s in range(1 << (len_ - 2)):
irot2 = irot * irot % _fft_mod
irot3 = irot2 * irot % _fft_mod
offset = s << (h - len_ + 2)
for i in range(p):
a0 = a[i + offset]
a1 = a[i + offset + p]
a2 = a[i + offset + p * 2]
a3 = a[i + offset + p * 3]
a2na3iimag = (a2 - a3) * _fft_iimag % _fft_mod
a[i + offset] = (a0 + a1 + a2 + a3) % _fft_mod
a[i + offset + p] = (a0 - a1 +
a2na3iimag) * irot % _fft_mod
a[i + offset + p * 2] = (a0 + a1 -
a2 - a3) * irot2 % _fft_mod
a[i + offset + p * 3] = (a0 - a1 -
a2na3iimag) * irot3 % _fft_mod
if s + 1 != (1 << (len_ - 1)):
irot *= _fft_irate3[(~s & -~s).bit_length() - 1]
irot %= _fft_mod
len_ -= 2
def _convolution_naive(a, b):
n = len(a)
m = len(b)
ans = [0] * (n + m - 1)
if n < m:
for j in range(m):
for i in range(n):
ans[i + j] = (ans[i + j] + a[i] * b[j]) % _fft_mod
else:
for i in range(n):
for j in range(m):
ans[i + j] = (ans[i + j] + a[i] * b[j]) % _fft_mod
return ans
def _convolution_fft(a, b):
a = a.copy()
b = b.copy()
n = len(a)
m = len(b)
z = 1 << (n + m - 2).bit_length()
a += [0] * (z - n)
_butterfly(a)
b += [0] * (z - m)
_butterfly(b)
for i in range(z):
a[i] = a[i] * b[i] % _fft_mod
_butterfly_inv(a)
a = a[:n + m - 1]
iz = pow(z, _fft_mod - 2, _fft_mod)
for i in range(n + m - 1):
a[i] = a[i] * iz % _fft_mod
return a
def _convolution_square(a):
a = a.copy()
n = len(a)
z = 1 << (2 * n - 2).bit_length()
a += [0] * (z - n)
_butterfly(a)
for i in range(z):
a[i] = a[i] * a[i] % _fft_mod
_butterfly_inv(a)
a = a[:2 * n - 1]
iz = pow(z, _fft_mod - 2, _fft_mod)
for i in range(2 * n - 1):
a[i] = a[i] * iz % _fft_mod
return a
def convolution(a, b):
"""It calculates (+, x) convolution in mod 998244353.
Given two arrays a[0], a[1], ..., a[n - 1] and b[0], b[1], ..., b[m - 1],
it calculates the array c of length n + m - 1, defined by
> c[i] = sum(a[j] * b[i - j] for j in range(i + 1)) % 998244353.
It returns an empty list if at least one of a and b are empty.
Constraints
-----------
> len(a) + len(b) <= 8388609
Complexity
----------
> O(n log n), where n = len(a) + len(b).
"""
n = len(a)
m = len(b)
if n == 0 or m == 0:
return []
if min(n, m) <= 0:
return _convolution_naive(a, b)
if a is b:
return _convolution_square(a)
return _convolution_fft(a, b)
def taylor_shift(f,a):
g = [f[i]*g1[i]%mod for i in range(len(f))][::-1]
e = [g2[i] for i in range(len(f))]
t = 1
for i in range(1,len(f)):
t = t * a % mod
e[i] = e[i] * t % mod
res = convolution(g,e)[:len(f)]
return [res[len(f)-1-i]*g2[i]%mod for i in range(len(f))]
def cmb(n,r,mod):
if r < 0 or n < r:
return 0
return g1[n] * (g2[r] * g2[n-r] % mod) % mod
"""
n項をまとめる方法のfpsを考える
まとめる過程はマージ過程を表す木+左右どちらにまとめたかで表すことができる
例えば(1,2,...,n)の表し方は複数あるが、左にまとめた山を左でまとめるのを禁止すると一意になり、特にマージ過程を表す木+最後のまとめが左右どちらかになる
結局葉がn個あり、子が存在するならば2つ以上存在する順序木の数え上げができればよく、このfpsをfとすると
f = x + f^2/(1-f)
が成り立つ。このfに対してk=1,2,...,nに対して [x^n](2f-x)^k を求めればいい
h = 2f-x と置くと -h^2+h = (1+h)x が成り立ち、g = x(1-x)/(1+x) とすると g(h) = x
（一般の）ラグランジュの反転公式から
[x^n]h^k = k/n[x^(-k)]1/g^n = k/n[x^(n-k)] (1+x)^n/(1-x)^n
となるので、(1+x/(1-x))^n = (-1+2/(1-x))^n のn項目までの列挙
これはmaspyさんのやつに載っていてpolynomial-taylor-shiftを利用してO(NlogN)でできる
"""
N = int(input())
f = [0] * (N+1)
for i in range(N+1):
if (N-i) & 1:
f[i] = -pow(2,i,mod) * cmb(N,i,mod) % mod
else:
f[i] = pow(2,i,mod) * cmb(N,i,mod) % mod
f = f[::-1]
f = taylor_shift(f,1)
for i in range(N+1):
if i & 1:
f[i] = -f[i] % mod
g = [cmb(i+N-1,N-1,mod) for i in range(N+1)]
h = convolution(f,g)
res = [h[N-i]*i*inv[N] % mod for i in range(N+1)]
print(*res[1:],sep="\n")