矩阵快速幂和普通快速幂道理一样，主要用来加速数列的递推，关键在于构造矩阵。

这几个博客的板子挺好

https://www.cnblogs.com/fzl194/p/8877276.html

题目 落谷模板题p1939

https://www.luogu.com.cn/problem/P1939

贴个代码吧，没用重载

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 100;
const int mod = 1e9 + 7;
int n;
struct Mat
{
    LL a[maxn][maxn]; //一定要用long long存矩阵，否则在过程中会爆掉
    Mat()
    {
        memset(a, 0, sizeof a);
    }
    inline void build()
    {
        for (int i = 1; i <= 3; ++i)
            a[i][i] = 1;
    }
};
Mat mul(Mat x, Mat y)
{
    Mat z;
    for (int i = 1; i <= 3; ++i)
        for (int k = 1; k <= 3; ++k)
            for (int j = 1; j <= 3; ++j)
                z.a[i][j] = (z.a[i][j] + x.a[i][k] * y.a[k][j] % mod) % mod;
    return z;
}
Mat Mat_qpow(Mat a, LL p)
{
    Mat res;
    res.build();
    while (p)
    {
        if (p & 1)
            res = mul(res, a);
        a = mul(a, a);
        p = p >> 1;
    }
    return res;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    Mat t;
    t.a[1][1] = 1;
    t.a[1][3] = 1;
    t.a[2][1] = 1;
    t.a[3][2] = 1;
    while (T--)
    {
        cin >> n;
        Mat ans = Mat_qpow(t, n - 1);
        cout << ans.a[1][1] << endl;
    }
    return 0;
}