#892948 (PyPy3) No.2472 Tea time in the grand garden

• 問題: No.2472 Tea time in the grand garden
• ユーザー: 👑 SPD_9X2
• 提出日時: 2023-07-19 00:26:43
• 言語: PyPy3 (7.3.15)
• 結果: TLE (最新)
  AC
  (最初)
• 実行時間: -
• コード長: 1,667 bytes
• コンパイル時間: 835 ms
• コンパイル使用メモリ: 82,588 KB
• 実行使用メモリ: 179,916 KB
• 最終ジャッジ日時: 2024-11-07 03:05:11
• 合計ジャッジ時間: 3,894 ms
• ジャッジサーバーID （参考情報）: judge5 / judge3

ファイルパターン	結果
sample	AC * 2 TLE * 1 -- * 1
other	-- * 30

ソースコード

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
"""
Tea Time (5) 想定解
定数倍高速化Ver
注: O(N+K^3) なので、通りません
"""
import sys
from sys import stdin
mod = 998244353
def modfac(n, MOD):
 
    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
    return factorials, invs
def modnCr(n,r):
    if n < 0 or r < 0 or n < r:
        return 0
    return fac[n] * inv[n-r] * inv[r] % mod
fac,inv = modfac(1000,mod)
N,K = map(int,stdin.readline().split())
# 制約チェック
assert 1 <= N <= 400
assert 0 <= K <= 400
if K == 0:
    print (1)
    sys.exit()
# 1つ目の括弧は置いた状態から開始
dp = [[[0],[0]] for l in range(2)]
dp[0][1][0] = 1 #last,s,p の順番
for i in range(2,2*K+1):
    ndp = [[[0] * i for nnum in range(min(K+1,i+1))] for l in range(2)]
    for put in range(2): #置く括弧 0 = ( , 1 = )
        for last in range(2):
            put_neq_last = (1 if last != put else 0)
            for ls in range(len(dp[0])):
                news = ls + (1,-1)[put]
                if 0 <= news < len(ndp[0]):
                    for lp in range(len(dp[0][0])):
                        newp = lp + put_neq_last
                        ndp[put][news][newp] = (ndp[put][news][newp] + dp[last][ls][lp]) % mod
    
    dp = ndp
    #print (dp)
ans = 0
for p in range(len(dp[0][0])):
    now = dp[1][0][p] * modnCr(2*K+N-p,N-p)
    #print (p,dp[0][p][1] , modnCr(2*K+N-p,N-p))
    ans += now
    ans %= mod
print (ans)
 
 
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX