結果

問題 No.1100 Boxes
ユーザー convexineq
提出日時 2021-05-11 18:21:38
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 270 ms / 2,000 ms
コード長 2,223 bytes
コンパイル時間 204 ms
コンパイル使用メモリ 82,304 KB
実行使用メモリ 87,988 KB
最終ジャッジ日時 2024-09-22 01:05:43
合計ジャッジ時間 6,437 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
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ファイルパターン 結果
sample AC * 4
other AC * 36
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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

SIZE= 2*10**5+1
MOD = 998244353
ROOT = 3
roots = [pow(ROOT,(MOD-1)>>i,MOD) for i in range(24)] # 1 2^i
iroots = [pow(x,MOD-2,MOD) for x in roots] # 1 2^i
def untt(a,n):
for i in range(n):
m = 1<<(n-i-1)
for s in range(1<<i):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m])%MOD, (a[s+p]-a[s+p+m])*w_N%MOD
w_N = w_N*roots[n-i]%MOD
def iuntt(a,n):
for i in range(n):
m = 1<<i
for s in range(1<<(n-i-1)):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m]*w_N)%MOD, (a[s+p]-a[s+p+m]*w_N)%MOD
w_N = w_N*iroots[i+1]%MOD
inv = pow((MOD+1)//2,n,MOD)
for i in range(1<<n):
a[i] = a[i]*inv%MOD
def convolution(a,b):
la = len(a)
lb = len(b)
if min(la, lb) <= 50:
if la < lb:
la,lb = lb,la
a,b = b,a
res = [0]*(la+lb-1)
for i in range(la):
for j in range(lb):
res[i+j] += a[i]*b[j]
res[i+j] %= MOD
return res
deg = la+lb-2
n = deg.bit_length()
N = 1<<n
a += [0]*(N-len(a))
b += [0]*(N-len(b))
untt(a,n)
untt(b,n)
for i in range(N):
a[i] = a[i]*b[i]%MOD
iuntt(a,n)
return a[:deg+1]
#inv = [0]*SIZE # inv[j] = j^{-1} mod MOD
fac = [0]*SIZE # fac[j] = j! mod MOD
finv = [0]*SIZE # finv[j] = (j!)^{-1} mod MOD
fac[0] = fac[1] = 1
finv[0] = finv[1] = 1
for i in range(2,SIZE):
fac[i] = fac[i-1]*i%MOD
finv[-1] = pow(fac[-1],MOD-2,MOD)
for i in range(SIZE-1,0,-1):
finv[i-1] = finv[i]*i%MOD
#inv[i] = finv[i]*fac[i-1]%MOD
def choose(n,r): # nCk mod MOD
if 0 <= r <= n:
return (fac[n]*finv[r]%MOD)*finv[n-r]%MOD
else:
return 0
###############################################################
n,k = map(int,input().split())
a = [pow(i,n,MOD)*finv[i]%MOD for i in range(k+1)]
b = finv[:k+1]
for i in range(1,k+1,2): b[i] = -b[i]
p = convolution(a,b)[:k+1]
for i in range(k+1):
p[i] = p[i]*fac[i]%MOD*choose(k,i)%MOD
print(sum(p[(k-1)%2::2])%MOD)
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