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+## [$P1939$ 【模板】矩阵加速（数列）](https://www.luogu.com.cn/problem/P1939)
+
+
+### 一、题目描述
+![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/01/25d5dfdbf4b65a4f4b6ae43abb1c3604.png)
+
+
+### 二、解题思路
+从题目上来看，知道需要递推求公式，但$n<=2e9$,我们知道简单递推肯定要挂掉。
+
+所以想到需要一个$O(NlogN)$的算法,递推求式子，线性的还不行，联想到矩阵快速幂。
+
+$$\large \begin{bmatrix}f_{x-3} & f_{x-2} & f_{x-1} \end{bmatrix} \times 
+\begin{bmatrix}
+a&b &c \\ 
+d&e &f \\ 
+g&h &i 
+\end{bmatrix} = \begin{bmatrix}f_{x-2} & f_{x-1} & f_{x} \end{bmatrix} 
+$$
+
+$\therefore$
+$\large f_{x-3} \times a+f_{x-2}\times d + f_{x-1}\times g =f_{x-2}$
+$\large f_{x-3} \times b+f_{x-2}\times e + f_{x-1}\times h =f_{x-1}$
+$\large f_{x-3} \times c+f_{x-2}\times f + f_{x-1}\times i =f_{x}$
+
+对比观察得到:
+$a=0,d=1,g=0$
+$b=0,e=0,h=1$
+$c=1,f=0,i=1$
+
+得到
+$$m=\begin{bmatrix}
+0& 0 & 1 \\ 
+1& 0 & 0 \\ 
+0& 1 & 1 
+\end{bmatrix}$$
+
+而初始化矩阵$b=\begin{bmatrix} f_{1}& f_{2} & f_3 \end{bmatrix}=\begin{bmatrix} 1&1&1 \end{bmatrix}$
+
+**递推式**：
+$\large b=\begin{bmatrix}f_{x-2}& f_{x-1} & f_x  \end{bmatrix}=
+\begin{bmatrix}f_{1}& f_{2} & f_3  \end{bmatrix} \times 
+\begin{bmatrix}
+0& 0 & 1 \\ 
+1& 0 & 0 \\ 
+0& 1 & 1 
+\end{bmatrix}^{n-3}
+$
+
+**答案**：
+$\large b[0][2]$
+
+$Code$
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+#define int long long
+#define endl "\n"
+const int MOD = 1e9 + 7;
+const int N = 4;
+int n;
+
+// 矩阵乘法
+void mul(int a[][N], int b[][N], int c[][N]) {
+    int t[N][N] = {0};
+    for (int i = 0; i < N; i++)
+        for (int j = 0; j < N; j++)
+            for (int k = 0; k < N; k++)
+                t[i][j] = (t[i][j] + (a[i][k] * b[k][j] % MOD)) % MOD;
+    memcpy(c, t, sizeof t);
+}
+
+void solve() {
+    int b[N][N] = {1, 1, 1}, m[N][N] = {0};
+    m[0][2] = m[1][0] = m[2][1] = m[2][2] = 1;
+
+    for (int i = n - 3; i; i >>= 1) {
+        if (i & 1) mul(b, m, b);
+        mul(m, m, m);
+    }
+
+    printf("%lld\n", b[0][2]);
+}
+
+signed main() {
+    int T;
+    cin >> T;
+    while (T--) {
+        cin >> n;
+        if (n <= 3) {
+            printf("1\n");
+            continue;
+        }
+        solve();
+    }
+}
+
+```