137 lines
3.0 KiB
Markdown
137 lines
3.0 KiB
Markdown
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##[$AcWing$ $1303$. 斐波那契前 $n$ 项和](https://www.acwing.com/problem/content/1305/)
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### 一、题目描述
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大家都知道 $Fibonacci$ 数列吧，$f_1=1,f_2=1,f_3=2,f_4=3,…,f_n=f_{n−1}+f_{n−2}$。
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现在问题很简单，输入 $n$ 和 $m$，求 $f_n$ 的前 $n$ 项和 $S_n \ mod \ m$。
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**输入格式**
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共一行，包含两个整数 $n$ 和 $m$。
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**输出格式**
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输出前 $n$ 项和 $S_n\ mod\ m$ 的值。
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**数据范围**
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$1≤n≤2000000000,1≤m≤1000000010$
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**输入样例**：
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$5$ $1000$
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**输出样例**：
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$12$
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### 二、解题思路
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普通的迭代法，是线性的，$n$的上限是$2e9$,循环的话会超时，需要想其它办法。
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**矩阵快速幂**
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$$\large \begin{bmatrix} s_n & f_{n+1}& f_n \end{bmatrix} \times
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\begin{bmatrix}
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a&b &c \\
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d&e & f \\
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h&i & j \\
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\end{bmatrix} = \begin{bmatrix} s_{n+1} & f_{n+2}& f_{n+1} \end{bmatrix}$$
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$\therefore \\
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s_{n+1}= s_n \times a+f_{n+1} \times d+f_n \times h \\
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f_{n+2}= s_n \times b + f_{n+1} \times e + f_n \times i \\
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f_{n+1}= s_n \times c + f_{n+1} \times f + f_n \times j
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$
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根据题目的现实含义：
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$s_{n+1}=s_n+f_{n+1}$
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$f_{n+2}=f_n+f_{n+1}$
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推导出:
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$a=1,d=1,h=0$
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$b=0,e=1,i=1$
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$c=0,f=1,j=0$
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整理出矩阵就是：
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$m=\begin{bmatrix}
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1 & 0 & 0 \\
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1 & 1 & 1 \\
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0 & 1 & 0
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\end{bmatrix}$
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已知 $s_1=1，f_2=1，f_1=1$，使用矩阵快速幂可以求解 $s_n$。
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即$b=\begin{bmatrix}
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1 & 1& 1
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\end{bmatrix}$
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$\begin{bmatrix}s_1 & f_2 & f_1 \end{bmatrix} \times
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\begin{bmatrix}
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1 & 0 & 0 \\
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1 & 1 & 1 \\
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0 & 1 & 0
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\end{bmatrix} =\begin{bmatrix}s_2 & f_3 & f_2 \end{bmatrix}
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$ ①
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$\begin{bmatrix}s_2 & f_3 & f_2 \end{bmatrix} \times
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\begin{bmatrix}
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1 & 0 & 0 \\
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1 & 1 & 1 \\
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0 & 1 & 0
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\end{bmatrix} =\begin{bmatrix}s_3 & f_4 & f_3 \end{bmatrix}
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$ ②
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将 ① 代入 ② 得到：
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$\begin{bmatrix}s_1 & f_2 & f_1 \end{bmatrix} \times
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\begin{bmatrix}
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1 & 0 & 0 \\
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1 & 1 & 1 \\
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0 & 1 & 0
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\end{bmatrix}^2 =\begin{bmatrix}s_3 & f_4 & f_3 \end{bmatrix}
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$
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通过数学归纳法，得:
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$\begin{bmatrix}s_1 & f_2 & f_1 \end{bmatrix} \times
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\begin{bmatrix}
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1 & 0 & 0 \\
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1 & 1 & 1 \\
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0 & 1 & 0
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\end{bmatrix}^{n-1} =\begin{bmatrix}s_n & f_{n+1} & f_n \end{bmatrix}
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$
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### 三、实现代码
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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#define int long long
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#define endl "\n"
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const int N = 3;
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int n, mod;
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// 矩阵乘法
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void mul(int a[][N], int b[][N], int c[][N]) {
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int t[N][N] = {0};
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < N; j++)
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for (int k = 0; k < N; k++)
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t[i][j] = (t[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
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}
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memcpy(c, t, sizeof t);
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}
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signed main() {
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int b[N][N] = {1, 1, 1};
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int m[N][N] = {
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{1, 0, 0},
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{1, 1, 1},
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{0, 1, 0}};
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cin >> n >> mod;
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for (int i = n - 1; i; i >>= 1) {
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if (i & 1) mul(b, m, b);
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mul(m, m, m);
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}
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printf("%d\n", b[0][0]);
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}
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```
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