結果
問題 | No.2660 Sweep Cards (Easy) |
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 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | RE | - |
testcase_01 | RE | - |
testcase_02 | RE | - |
testcase_03 | RE | - |
testcase_04 | RE | - |
testcase_05 | RE | - |
testcase_06 | RE | - |
testcase_07 | RE | - |
testcase_08 | RE | - |
testcase_09 | RE | - |
testcase_10 | RE | - |
testcase_11 | RE | - |
testcase_12 | RE | - |
testcase_13 | RE | - |
testcase_14 | RE | - |
testcase_15 | RE | - |
testcase_16 | RE | - |
testcase_17 | RE | - |
testcase_18 | RE | - |
testcase_19 | RE | - |
testcase_20 | RE | - |
testcase_21 | RE | - |
testcase_22 | RE | - |
testcase_23 | RE | - |
testcase_24 | RE | - |
testcase_25 | RE | - |
testcase_26 | RE | - |
testcase_27 | RE | - |
testcase_28 | RE | - |
testcase_29 | RE | - |
testcase_30 | RE | - |
testcase_31 | RE | - |
testcase_32 | RE | - |
testcase_33 | RE | - |
testcase_34 | RE | - |
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")