結果

問題 No.2807 Have Another Go (Easy)
コンテスト
ユーザー qwewe
提出日時 2025-05-14 13:21:11
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2,328 ms / 3,000 ms
コード長 4,906 bytes
コンパイル時間 1,041 ms
コンパイル使用メモリ 86,088 KB
実行使用メモリ 11,200 KB
最終ジャッジ日時 2025-05-14 13:23:49
合計ジャッジ時間 61,507 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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ファイルパターン 結果
other AC * 46
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ソースコード

diff # 

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>

using namespace std;

long long MOD = 998244353;

const int D = 6; // Dimension of recurrence

// Matrix multiplication for DxD matrices
vector<vector<long long>> multiply(const vector<vector<long long>>& A, const vector<vector<long long>>& B) {
    vector<vector<long long>> C(D + 1, vector<long long>(D + 1, 0));
    for (int i = 0; i <= D; ++i) {
        for (int j = 0; j <= D; ++j) {
            for (int k = 0; k <= D; ++k) {
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
            }
        }
    }
    return C;
}

// Matrix power using binary exponentiation
vector<vector<long long>> matrix_pow(vector<vector<long long>> A, long long p) {
    vector<vector<long long>> res(D + 1, vector<long long>(D + 1, 0));
    for (int i = 0; i <= D; ++i) res[i][i] = 1;
    A[D][D] = 1; // Ensure the (D,D) element for affine transform is 1 initially

    while (p > 0) {
        if (p & 1) res = multiply(res, A);
        A = multiply(A, A);
        p >>= 1;
    }
    return res;
}

// Apply matrix to vector V = M * U
vector<long long> apply_matrix_to_vector(const vector<vector<long long>>& M, const vector<long long>& U) {
    vector<long long> V(D + 1, 0);
    for (int i = 0; i <= D; ++i) {
        for (int j = 0; j <= D; ++j) {
            V[i] = (V[i] + M[i][j] * U[j]) % MOD;
        }
    }
    return V;
}


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int N_val, M_val, k_queries;
    cin >> N_val >> M_val >> k_queries; // M_val is always 2

    int limit_sum = 2 * N_val;

    vector<long long> f(limit_sum + D, 0);
    f[0] = 1;
    for (int i = 1; i < limit_sum + D; ++i) {
        for (int d = 1; d <= D; ++d) {
            if (i - d >= 0) {
                f[i] = (f[i] + f[i - d]) % MOD;
            }
        }
    }

    vector<vector<long long>> T_std(D + 1, vector<long long>(D + 1, 0));
    for (int j = 0; j < D; ++j) {
        T_std[0][j] = 1;
    }
    for (int i = 1; i < D; ++i) {
        T_std[i][i - 1] = 1;
    }
    T_std[D][D] = 1;

    vector<vector<long long>> T_reset_proto(D + 1, vector<long long>(D + 1, 0));
    // First row of M_zr is all zeros, so T_reset_proto[0][j] for j<D are 0.
    for (int i = 1; i < D; ++i) {
        T_reset_proto[i][i - 1] = 1;
    }
    T_reset_proto[D][D] = 1;
    
    vector<int> C_values(k_queries);
    for (int i = 0; i < k_queries; ++i) {
        cin >> C_values[i];
    }

    for (int qi = 0; qi < k_queries; ++qi) {
        int C = C_values[qi];

        vector<long long> U(D + 1, 0); // Represents (g[x-1]...g[x-6], 1)^T
        U[D] = 1; // for affine transformation
        // Initial state corresponds to g[-1]...g[-6] all 0.
        
        long long current_x = -1;
        vector<long long> critical_points;
        critical_points.push_back(C);
        if (C + N_val < limit_sum) {
            critical_points.push_back(C + N_val);
        }

        for (long long p : critical_points) {
            long long segment_len = p - 1 - current_x;
            if (segment_len > 0) {
                vector<vector<long long>> T_std_pow_L = matrix_pow(T_std, segment_len);
                U = apply_matrix_to_vector(T_std_pow_L, U);
            }
            current_x = p - 1;

            // Reset at point p
            vector<vector<long long>> T_reset_concrete = T_reset_proto;
            T_reset_concrete[0][D] = f[p]; // B_x vector's first component is f[p]
            U = apply_matrix_to_vector(T_reset_concrete, U);
            current_x = p;
        }

        long long final_segment_len = (limit_sum - 1) - current_x;
        if (final_segment_len > 0) {
            vector<vector<long long>> T_std_pow_L = matrix_pow(T_std, final_segment_len);
            U = apply_matrix_to_vector(T_std_pow_L, U);
        }
        // Now U contains (g[2N-1], g[2N-2], ..., g[2N-6], 1)^T

        vector<long long> g_values(D);
        for(int i=0; i<D; ++i) {
            g_values[i] = U[i];
        }

        long long ans_C = 0;
        for (int s_idx = 0; s_idx < D; ++s_idx) {
            long long s = limit_sum - 1 - s_idx; // s goes from 2N-1 down to 2N-6
            if (s < 0) continue;

            long long term_rolls_count = 0;
            // Number of d in [1,6] such that s+d >= limit_sum
            // d >= limit_sum - s
            // Smallest d is max(1, limit_sum - s)
            // Largest d is 6
            // If max(1, limit_sum - s) > 6, count is 0
            // Else count = 6 - max(1, limit_sum - s) + 1
            
            long long min_d_to_terminate = limit_sum - s;
            if (min_d_to_terminate <= D) { // D is 6
                 term_rolls_count = D - max(1LL, min_d_to_terminate) + 1;
            }
            
            ans_C = (ans_C + g_values[s_idx] * term_rolls_count) % MOD;
        }
        cout << ans_C << "\n";
    }

    return 0;
}
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