题目链接：这里写链接内容
先求出式子的前几项
然后这个式子和h的式子是非常像的
得到递推式后就可以用矩阵快速幂了
详细看代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<iomanip>
#include<cctype>
using namespace std;
#define ll long long
#define edl putchar('\n')
#define sscc ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define FORLL(i,a,b) for(ll i=a;i<=b;i++)
#define ROFLL(i,a,b) for(ll i=a;i>=b;i--)
#define mst(a) memset(a,0,ssizeof(a))
#define mstn(a,n) memset(a,n,ssizeof(a))
#define zero(x)(((x)>0?(x):-(x))<eps)
const int ssize=3;
const int mod=1000000007;
struct Matrix {ll a[ssize][ssize];Matrix() {memset(a,0,sizeof(a));}void init() {for(int i=0; i<ssize; i++)for(int j=0; j<ssize; j++)a[i][j]=(i==j);}Matrix operator + (const Matrix &B)const {Matrix C;for(int i=0; i<ssize; i++)for(int j=0; j<ssize; j++)C.a[i][j]=(a[i][j]+B.a[i][j])%mod;return C;}Matrix operator * (const Matrix &B)const {Matrix C;for(int i=0; i<ssize; i++)for(int k=0; k<ssize; k++)for(int j=0; j<ssize; j++)C.a[i][j]=(C.a[i][j]%mod+1LL*a[i][k]*B.a[k][j]%mod+mod)%mod;return C;}Matrix operator ^ (const ll &t)const {Matrix A=(*this),res;res.init();ll p=t;while(p) {if(p&1)res=res*A;A=A*A;p>>=1;}return res;}
};int main() {int t;scanf("%d",&t);Matrix a,b;Matrix ans;a.a[0][0]=4,a.a[0][1]=17,a.a[0][2]=-12;a.a[1][0]=1,a.a[1][1]=0,a.a[1][2]=0;a.a[2][0]=0,a.a[2][1]=1,a.a[2][2]=0;b.a[0][0]=1255,b.a[1][0]=197,b.a[2][0]=31;while(t--){ll n;scanf("%lld",&n);if (n == 2){printf("31\n");continue;}else if (n == 3){printf("197\n");continue;}else if (n == 4){printf("1255\n");continue;}else{ans=a^(n-4);ll res=(ans.a[0][0]*1255%mod+ans.a[0][1]*197%mod+ans.a[0][2]*31%mod+mod)%mod;printf("%lld\n",res);}}
}