結果
| 問題 | No.1621 Sequence Inversions | 
| コンテスト | |
| ユーザー |  mkawa2 | 
| 提出日時 | 2021-07-22 23:35:11 | 
| 言語 | PyPy3 (7.3.15) | 
| 結果 | 
                                MLE
                                 
                             | 
| 実行時間 | - | 
| コード長 | 1,593 bytes | 
| コンパイル時間 | 253 ms | 
| コンパイル使用メモリ | 82,472 KB | 
| 実行使用メモリ | 849,308 KB | 
| 最終ジャッジ日時 | 2024-07-17 20:47:36 | 
| 合計ジャッジ時間 | 6,091 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 | MLE * 1 -- * 2 | 
| other | -- * 26 | 
ソースコード
import sys
sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = 10**16
md = 998244353
# md = 10**9+7
mx = 100
mx2=mx*(mx+1)//2+5
dd = [[[0]*(mx+1) for _ in range(mx+1)] for _ in range(mx2)]
for s in range(mx2):
    for j in range(1, mx+1):
        for lim in range(mx+1):
            if j == 1:
                if s <= lim:
                    dd[s][j][lim] = 1
                else:
                    dd[s][j][lim] = 0
            elif s == 0:
                dd[s][j][lim] = 1
            elif s > lim*j:
                dd[s][j][lim] = 0
            else:
                dd[s][j][lim] = dd[s-j][j][lim-1]+dd[s][j-1][lim]
from collections import Counter
n, k = LI()
aa = LI()
aa.sort()
ac = Counter(aa)
dp = [0]*(k+1)
dp[0] = 1
pa = -1
for i, a in enumerate(aa):
    if a == pa: continue
    dp1 = dp
    dp = [0]*(k+1)
    c = ac[a]
    for j in range(k+1):
        pre = dp1[j]
        if pre == 0: continue
        for nj in range(j, k+1):
            diff = nj-j
            dp[nj] += pre*dd[diff][c][i]%md
            dp[nj] %= md
    pa = a
# p2D(dp)
print(dp[k])
            
            
            
        