Project Euler 435 Polynomials of Fibonacci numbers (矩阵快速幂)

题目链接：

https://projecteuler.net/problem=435

题意：

The Fibonacci numbers $ {f_n, n ≥ 0}$ are defined recursively as \(f_n = f_{n-1} + f_{n-2}\) with base cases \(f_0 = 0\) and \(f_1 = 1\).

Define the polynomials $ {F_n, n ≥ 0} $ as $F_n(x) =\sum_{i=0}^{n} f_i x^i $.

For example, \(F_{7}(x) = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7\), and$ F_7(11) = 268357683$.

Let \(n = 10^{15}\). Find [$\sum_{x=0}^{100} F_{n}(x)] $ mod \(1307674368000 (= 15!)\).

题解：

\[\begin{pmatrix}
f_{n}x^{n} & f_{n+1}x^{n+1} & F_{n}(x)
\end{pmatrix}
\]

\[=
\begin{pmatrix}
f_{n-1}x^{n-1} & f_{n}x^{n} & F_{n-1}(x)
\end{pmatrix}
\begin{pmatrix}
0 & 0 & x^{2} \\
0 & 1 & 1 \\
1 & 0 & i
\end{pmatrix}
\]

\[=
\begin{pmatrix}
f_{0}x^{0} & f_{1}x^{1} & F_{1}(x)
\end{pmatrix}
\begin{pmatrix}
0 & 0 & x^{2} \\
0 & 1 & 1 \\
1 & 0 & i
\end{pmatrix}^{n-1}
\]

\[=
\begin{pmatrix}
0 & x & x
\end{pmatrix}
\begin{pmatrix}
0 & 0 & x^{2} \\
0 & 1 & 1 \\
1 & 0 & i
\end{pmatrix}^{n-1}
\]

然后跑矩阵快速幂就可以得到 \(F_{n}(x)\)了。\(C\)++ 会爆 \(long long\)... 用 \(Python\)吧...

其实用 \(C\)++也行，就是将模数分解再用 \(crt\) 合并。

代码：

#coding: utf-8

from math import sqrt

mod = 1307674368000

def matrix_mult(a, b) :

    n = len(a); m = len(b); h = len(b[0])

    ans = [[0, 0, 0],[0, 0, 0],[0, 0, 0]]

    for i in range(n) :

        for j in range(m) :

            for k in range(h) :

                ans[i][k] += a[i][j] * b[j][k]

                if ans[i][k] >= mod :

                    ans[i][k] %= mod

                ans[i][k] %= mod

        ans[i][j] %= mod

    return ans

def qpower(a, n, i) :

    ans = [[0, i, i],[0, 0, 0],[0, 0, 0]]

    while n > 0 :

        if n & 1 : ans = matrix_mult(ans, a)

        n >>= 1

        a = matrix_mult(a, a)

    return ans[0][2]

if __name__ =="__main__":

    ans = 0

    for i in range(101):

        a = [[0, 0, i ** 2],

             [0, 1, 1],

             [1, 0, i]]

        ans += qpower(a, 10 ** 15 - 1, i)

    print( ans % mod )