結果

問題 No.2199 lower_bound and upper_bound
コンテスト
ユーザー gew1fw
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
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ソースコード

diff # 

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()
0