[AT697]フィボナッチ
题目大意:给你$n,k(n\leqslant10^9,k\leqslant10^3)$,求$f_n$。$f$数组满足$f_1=f_2=\cdots=f_k=1$,$f_n=\sum\limits_{i=n-k}^{n-1}f_i$
题解:线性齐次递推:
$$
\left[
\begin{matrix}
f_1&f_2&\cdots&f_k
\end{matrix}
\right]
\left[
\begin{matrix}
0&0&0&\cdots&0&1\\
1&0&0&\cdots&0&1\\
0&1&0&\cdots&0&1\\
0&0&1&\cdots&0&1\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&1&1
\end{matrix}
\right]
=
\left[
\begin{matrix}
f_2&f_3&\cdots&f_{k+1}
\end{matrix}
\right]
$$
特征多项式$G_k(x)$为:
$$
\begin{align*}
G_k(x)&=|\lambda I-A|\\
&=\left|
\left[
\begin{matrix}
\lambda&0&0&\cdots&0&-1\\
-1&\lambda&0&\cdots&0&-1\\
0&-1&\lambda&\cdots&0&-1\\
0&0&-1&\cdots&0&-1\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&-1&\lambda-1
\end{matrix}
\right]\right|
\end{align*}
$$
可以对第一行展开
$$
\begin{align*}
G_k(x)&=(-1)^{1+1}\lambda G_{k-1}(x)+(-1)(-1)^{k-1}(-1)^{k+1}\\
&=\lambda G_{k-1}(x)-1\\
&=\lambda^k-\lambda^{k-1}-\lambda^{k-2}-\cdots-1
\end{align*}
$$
发现模数是$10^9+7$,但是$k$只有$10^3$,所以直接$O(k^2)$卷积和取模,总复杂度$O(k^2\log_2n)$
卡点:无
C++ Code:
#include <algorithm> #include <cstdio> #include <cstring> #define maxn 2010 const int mod = 1e9 + 7; #define mul(x, y) static_cast<long long> (x) * (y) % mod inline void reduce(int &x) { x += x >> 31 & mod; } int n, K; int f[maxn], g[maxn]; void PW(int n) { if (n == 0) { f[0] = 1; return ; } PW(n >> 1); std::memset(g, 0, K << 3); for (int i = 0; i < K; ++i) for (int j = 0; j < K; ++j) reduce(g[i + j + (n & 1)] += mul(f[i], f[j]) - mod); for (int i = K + K - 1 + (n & 1); i >= K; --i) { for (int j = 1; j <= K; ++j) reduce(g[i - j] += g[i] - mod); } std::memcpy(f, g, K << 2); } int main() { scanf("%d%d", &K, &n); PW(n - 1); int ans = 0; for (int i = 0; i < K; ++i) reduce(ans += f[i] - mod); printf("%d\n", ans); return 0; }