結果
問題 |
No.2199 lower_bound and upper_bound
|
ユーザー |
![]() |
提出日時 | 2025-06-12 15:24:32 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 3,308 bytes |
コンパイル時間 | 191 ms |
コンパイル使用メモリ | 82,816 KB |
実行使用メモリ | 259,456 KB |
最終ジャッジ日時 | 2025-06-12 15:24:47 |
合計ジャッジ時間 | 4,833 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 9 TLE * 1 -- * 12 |
ソースコード
MOD = 998244353 def main(): import sys input = sys.stdin.read N, L, U = map(int, input().split()) # Transform variables C = [U + i for i in range(1, N+1)] # We need to compute the number of non-decreasing sequences E_1 <= E_2 <= ... <= E_N # where E_i <= C[i-1] (since C is 0-based now), E_0 = 0, and E_N >= L + N # The maximum E_N can be C[-1] max_e = C[-1] target = L + N if target > max_e: print(0) return # The problem reduces to counting the number of non-decreasing sequences E_0=0 <= E_1 <= ... <= E_N <= max_e # with E_i <= C[i-1] for each i, and E_N >= target # The solution is to compute the number of such sequences as the inclusion-exclusion over the constraints # However, for large N, this is computationally intensive, so we need a mathematical approach # The number of valid sequences is the product of (C_i + 1) minus the invalid sequences. # However, this is not directly applicable, so we use a combinatorial approach with stars and bars # The number of valid sequences can be computed using the inclusion-exclusion principle, but for large N, we need a formula # The correct approach involves using dynamic programming with optimized prefix sums # We will use dp[i][j] to represent the number of ways to reach j after i steps # However, given the constraints, we can compute the result using combinatorial mathematics # The number of valid sequences is the product of (C_i + 1) for i in 1..N, minus the sequences where any E_i exceeds C_i # This is not straightforward, so we use the following approach: # The number of valid sequences is the sum over all possible E_N of the number of ways to reach E_N >= target, with E_i <= C_i for all i # To compute this, we can use dynamic programming with prefix sums # Initialize dp array dp = [0]*(max_e + 1) dp[0] = 1 # E_0 = 0 for i in range(N): # Compute the new dp array new_dp = [0]*(max_e + 1) # For each possible E_i, update the new_dp for e in range(0, max_e +1): if dp[e] == 0: continue # The next E_{i+1} can be from e to min(C[i], max_e) # Since E_{i+1} >= e and E_{i+1} <= C[i] # Also, E_{i+1} <= max_e min_next = e max_next = min(C[i], max_e) if min_next > max_next: continue # The number of ways to transition from e is 1 for each possible next_e # So, we add dp[e] to new_dp[min_next..max_next] # To do this efficiently, we can use prefix sums new_dp[min_next] = (new_dp[min_next] + dp[e]) % MOD if max_next + 1 <= max_e: new_dp[max_next + 1] = (new_dp[max_next + 1] - dp[e]) % MOD # Compute the prefix sums to get the new_dp values current = 0 for e in range(max_e +1): current = (current + new_dp[e]) % MOD new_dp[e] = current dp = new_dp # Now, sum dp[target ... max_e] total = 0 for e in range(target, max_e +1): total = (total + dp[e]) % MOD print(total % MOD) if __name__ == '__main__': main()