推导或者可以找规律有公式：\(f[n] = f[n-1] + f[n-3]\) 。
构造矩阵乘法：
\[\begin{pmatrix} f_i \\ f_{i-1} \\ f_{i-2} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} f_{i-1} \\ f_{i-2} \\ f_{i-3} \end{pmatrix}
\]
\]
时间复杂度为 \(O(\log n)\) 。
#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int t;
long long n;
struct Matrix{
long long a[5][5];
};
Matrix mul(Matrix M1, Matrix M2)
{
Matrix ret;
memset(ret.a, 0, sizeof(ret.a));
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
for(int k = 0; k < 3; k++){
ret.a[i][j] = (M1.a[i][k] * M2.a[k][j] + ret.a[i][j]) % mod;
}
}
}
return ret;
}
void matrix_pow(long long x)
{
Matrix ret;
memset(ret.a, 0, sizeof(ret.a));
for(int i = 0; i < 3; i++) ret.a[i][i] = 1;
Matrix tmp;
memset(tmp.a, 0, sizeof(tmp.a));
tmp.a[0][0] = tmp.a[0][2] = tmp.a[1][0] = tmp.a[2][1] = 1;
while(x){
if(x & 1LL) ret = mul(ret, tmp);
tmp = mul(tmp, tmp);
x >>= 1LL;
}
long long ans = (ret.a[0][0] * 4 + ret.a[0][2] * 2 + ret.a[0][1] * 3) % mod;
cout << ans << endl;
}
int main()
{
for(scanf("%d", &t); t--; ){
scanf("%lld", &n);
if(n == 1) {puts("2"); continue;}
else if(n == 2) {puts("3"); continue;}
else if(n == 3) {puts("4"); continue;}
else{
matrix_pow(n - 3);
}
}
return 0;
}