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CF392C Yet Another Number Sequence

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LuoGu: CF392C Yet Another Number Sequence

CF: C. Yet Another Number Sequence

如果一般的斐波那契数列前缀和可以直接矩阵加速。考虑增加了 $i ^ k$ 如何实现。

$$
\begin {aligned}
& F _ {i + 1} (i + 1) ^ k \\
= & (F _ {i - 1} + F _ i) (i + 1) ^ k \\
= & F _ {i - 1} ((i - 1) + 2) ^ k + F _ i (i + 1) ^ k \\
= & \sum _ {j = 0} ^ k \binom k j 2 ^ {k - j} F _ {i - 1} (i - 1) ^ j + \sum _ {j = 0} ^ k \binom k j F _ i i ^ j
\end {aligned}
$$

这样，维护 $F _ i, F _ {i - 1}$ 和 $i ^ 0, i ^ 1, \ldots , i ^ k$ 乘积即可转移。

查看代码 
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#include <cstdio>
using namespace std;
template <class Type>
void read(Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 90, mod = 1e9 + 7;
struct Matrix
{
    int n, m, w[N][N];
    Matrix (int _n, int _m) { n = _n, m = _m; }
    void clear ()
    {
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                w[i][j] = 0;
    }
    void init ()
    {
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                w[i][j] = i == j;
    }
    Matrix operator * (Matrix _) const
    {
        Matrix res(n, _.m);
        res.clear();
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < _.m; ++j)
                for (int k = 0; k < m; ++k)
                    res.w[i][j] = (res.w[i][j] + (LL)w[i][k] * _.w[k][j]) % mod;
        return res;
    }
};
LL n;
int m, p[N], C[N][N];
void init ()
{
    p[0] = 1;
    for (int i = 1; i <= m; ++i)
        p[i] = p[i - 1] * 2 % mod;
    for (int i = 0; i <= m; ++i) C[i][0] = 1;
    for (int i = 1; i <= m; ++i)
        for (int j = 1; j <= m; ++j)
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
}
Matrix binpow (Matrix b, LL k)
{
    Matrix res(b.n, b.m);
    res.init();
    for (; k; k >>= 1, b = b * b)
        if (k & 1) res = res * b;
    return res;
}
int main ()
{
    read(n, m);
    init();
    static Matrix A(1, 2 * m + 3), B(2 * m + 3, 2 * m + 3);
    A.clear(), B.clear();
    B.w[2 * m + 1][2 * m + 2] = B.w[2 * m + 2][2 * m + 2] = 1;
    for (int i = 0; i <= m; ++i)
    {
        B.w[i + m + 1][i] = 1;
        for (int j = 0; j <= i; ++j)
        {
            B.w[j + m + 1][i + m + 1] = C[i][j];
            B.w[j][i + m + 1] = (LL)C[i][j] * p[i - j] % mod;
        }
    }
    A.w[0][2 * m + 2] = 1;
    for (int i = 0; i <= m; ++i)
    {
        A.w[0][i] = 1;
        A.w[0][i + m + 1] = p[i] * 2 % mod;
    }
    write(((A * binpow(B, n - 1)).w[0][2 * m + 2] + mod) % mod);
    return 0;
}
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