题目链接：https://www.luogu.com.cn/problem/P1962
解题思路：
因为
\[\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} F_{i} \\ F_{i+1} \end{bmatrix} = \begin{bmatrix} F_{i+1} \\ F_{i+2} \end{bmatrix}
\]
所以
\[\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^{n-2} \times \begin{bmatrix} F_{1} \\ F_{2} \end{bmatrix} = \begin{bmatrix} F_{n-1} \\ F_{n} \end{bmatrix}
\]
示例程序：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5;
const long long mod = 1e9 + 7;struct Matrix {int n, m;long long a[maxn][maxn];void init(int _n, int _m) {n = _n;m = _m;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)a[i][j] = 0;}Matrix operator * (const Matrix &b) const {Matrix res;res.init(n, b.m);for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)for (int k = 1; k <= b.m; k++)res.a[i][k] += a[i][j] * b.a[j][k],res.a[i][k] %= mod;return res;}Matrix operator ^ (long long k) const {Matrix res, t;res.init(n, n);for (int i = 1; i <= n; i++)res.a[i][i] = 1;t.init(n, n);memcpy(t.a, a, sizeof a);for (; k; k >>= 1, t = t * t)if (k & 1ll)res = res * t;return res;}
} a, b;
long long n;int main() {cin >> n;if (n <= 2) {cout << 1 << endl;return 0;}a.init(2, 2);a.a[1][2] = 1;a.a[2][1] = a.a[2][2] = 1;a = a ^ (n-2);b.init(2, 1);b.a[1][1] = b.a[2][1] = 1;a = a * b;cout << a.a[2][1] << endl;return 0;
}