結果
問題 | No.2019 Digits Filling for All Substrings |
ユーザー |
![]() |
提出日時 | 2025-06-12 14:06:40 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,086 bytes |
コンパイル時間 | 246 ms |
コンパイル使用メモリ | 82,176 KB |
実行使用メモリ | 98,048 KB |
最終ジャッジ日時 | 2025-06-12 14:06:48 |
合計ジャッジ時間 | 3,836 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | TLE * 1 -- * 29 |
ソースコード
MOD = 998244353 def main(): import sys input = sys.stdin.read data = input().split() N = int(data[0]) S = data[1] # Precompute prefix sums for sum_mod and number of '?' pre_mod = [0] * (N + 1) pre_q = [0] * (N + 1) for i in range(1, N+1): c = S[i-1] if c == '?': pre_q[i] = pre_q[i-1] + 1 pre_mod[i] = pre_mod[i-1] else: d = int(c) pre_mod[i] = (pre_mod[i-1] + d) % 3 pre_q[i] = pre_q[i-1] # Precompute pow_dp[k][r] where r is 0,1,2 max_k = N a, b, c = 4, 3, 3 # coefficients of f(x) = 4 + 3x +3x² pow_dp = [ [0]*3 for _ in range(max_k+1) ] pow_dp[0][0] = 1 for k in range(1, max_k+1): for r in range(3): # Multiply by f(x): for each term x^r, multiply by 4, 3x, 3x² # and accumulate to new_r = (r + s) mod3 for s=0,1,2 new_r0 = (pow_dp[k-1][r] * a) % MOD new_r1 = (pow_dp[k-1][r] * b) % MOD new_r2 = (pow_dp[k-1][r] * c) % MOD pow_dp[k][(r + 0) %3] = (pow_dp[k][(r +0) %3] + new_r0) % MOD pow_dp[k][(r +1) %3] = (pow_dp[k][(r +1) %3] + new_r1) % MOD pow_dp[k][(r +2) %3] = (pow_dp[k][(r +2) %3] + new_r2) % MOD # Calculate sum_f_part1: sum for k>0 sum_f_part1 = 0 for L in range(1, N+1): for R in range(L, N+1): k = pre_q[R] - pre_q[L-1] if k == 0: continue sum_mod = (pre_mod[R] - pre_mod[L-1]) % 3 t = (-sum_mod) %3 sum_f_part1 = (sum_f_part1 + pow_dp[k][t]) % MOD # Calculate sum_f_part2: sum for k=0 and sum_mod ==0 sum_f_part2 = 0 for L in range(1, N+1): for R in range(L, N+1): k = pre_q[R] - pre_q[L-1] if k !=0: continue sum_mod = (pre_mod[R] - pre_mod[L-1]) %3 if sum_mod ==0: sum_f_part2 +=1 total = (sum_f_part1 + sum_f_part2) % MOD print(total) if __name__ == '__main__': main()