結果
問題 |
No.1621 Sequence Inversions
|
ユーザー |
![]() |
提出日時 | 2021-07-22 23:16:40 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 1,872 bytes |
コンパイル時間 | 182 ms |
コンパイル使用メモリ | 82,040 KB |
実行使用メモリ | 107,332 KB |
最終ジャッジ日時 | 2024-07-17 20:04:32 |
合計ジャッジ時間 | 8,683 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 6 TLE * 1 -- * 19 |
ソースコード
def way10(ball, box): """ball: False / box: False / constraints: None ans = P(ball, box) Pは分割数 / 計算量 O(ball * box) """ # P[i][j] := 整数iを順序を区別せずに「j以下の自然数」の和に分ける場合の数 P = [[0] * (ball + 1) for _ in range(ball + 1)] for j in range(ball + 1): P[0][j] = 1 for i in range(ball): for j in range(ball): if i - j >= 0: P[i + 1][j + 1] = (P[i + 1][j] + P[i - j][j + 1]) % MOD else: P[i + 1][j + 1] = P[i + 1][j] return P[ball][min(box, ball)] def encoding(s): n = len(s) begin, cnt = 0, 0 ans = [] if n == 0: return ans for end in range(n + 1): if end == n or s[begin] != s[end]: ans.append((s[begin], cnt)) begin, cnt = end, 1 else: cnt += 1 return ans n, k = map(int, input().split()) a = list(map(int, input().split())) MOD = 998244353 P = way10(1000, 1000) a = sorted(a) comp = encoding(a) cnts = [cnt for _, cnt in comp] dp = [0] * (k + 1) dp[0] = 1 ru_cnt = 0 for cnt in cnts: dq = [0] * (k + 1) coeff = [[0] * (cnt + 1) for _ in range(k + 1)] coeff[0][0] = 1 for i in range(ru_cnt + 1): tmp = [[0] * (cnt + 1) for _ in range(k + 1)] for use in range(cnt + 1): for used in range(cnt + 1): for val in range(k + 1): if use + used < cnt + 1 and val + i * use < k + 1: tmp[val + i * use][used + use] += coeff[val][used] coeff, tmp = tmp, coeff for count in range(k + 1): for j in range(k + 1): if count + j < k + 1: dq[count + j] += dp[count] * coeff[j][-1] dq[count + j] %= MOD ru_cnt += cnt dp, dq = dq, dp print(dp[-1] % MOD)