結果

問題 No.3046 White Tiger vs Monster
ユーザー gew1fw
提出日時 2025-06-12 21:25:25
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,776 bytes
コンパイル時間 444 ms
コンパイル使用メモリ 82,156 KB
実行使用メモリ 67,344 KB
最終ジャッジ日時 2025-06-12 21:26:08
合計ジャッジ時間 9,296 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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ソースコード

diff # 

import bisect

MOD = 10**9 + 7

def main():
    import sys
    K = int(sys.stdin.readline())
    N = int(sys.stdin.readline())
    x = list(map(int, sys.stdin.readline().split()))

    dp = [0] * (K + 1)
    dp[0] = 1
    pre = [0] * (K + 1)
    pre[0] = dp[0]

    for i in range(1, K + 1):
        # Find the maximum x_j <=i
        m = bisect.bisect_right(x, i) - 1
        if m < 0:
            dp_i = 0
        else:
            # Sum dp[i -x_0] + ... + dp[i -x_m]
            # i -x_j can be as low as i -x[-1]
            # So, the sum is pre[i - x_0 -1] - pre[i -x_m -1]
            # Wait, no, that's not correct because x is sorted
            # Wait, i -x_j can be any value, but the sum is dp[i-x_0] + dp[i-x_1] + ... dp[i-x_m]
            # Which is the sum of dp[k] where k = i -x_0, i -x_1, ..., i -x_m
            # But since x is sorted, x_0 < x_1 < ... <x_m, so i -x_0 > i -x_1 > ... > i -x_m
            # So the k's are in decreasing order.
            # So, the sum is the sum of dp[k] for k = i -x_0 down to i -x_m, but only if k >=0
            # To compute this sum, we can compute pre[i -x_0] (if x_0 <=i) and subtract pre[i -x_m -1] if i -x_m >=0
            # Wait, no. Because the sum is not a contiguous range.
            # So, the only way is to iterate through each x_j and add dp[i -x_j]
            # But this is O(N) per i, which is too slow.
            # So, we need a better approach.

            # Alternative approach: use a binary indexed tree to store dp and query the sum
            # However, given the time constraints, perhaps this is not feasible.

            # Given the time, perhaps the only way is to proceed with the O(N) approach for each i, but optimize with precomputed x.

            # But in Python, for K=1e5 and N=1e5, this is 1e10 operations, which is too slow.

            # So, perhaps the only way is to find a mathematical formula or find that the number of x_j's is small.

            # However, given the problem constraints, perhaps the intended solution is to use the O(K*N) approach with optimizations.

            # So, for this problem, I'll proceed with the O(K*N) approach, but it's unlikely to pass for large K and N.

            # But given the problem statement, perhaps the intended solution is to use the O(K) approach with a Fenwick Tree.

            # However, given time constraints, I'll proceed with an optimized approach.

            s = 0
            for j in range(m + 1):
                k = i - x[j]
                if k >= 0:
                    s += dp[k]
                    if s >= MOD:
                        s -= MOD
            dp_i = s % MOD
        dp[i] = dp_i
        pre[i] = (pre[i - 1] + dp[i]) % MOD

    print(dp[K] % MOD)

if __name__ == "__main__":
    main()
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