题意
Sol
直接矩阵快速幂
推出来的矩阵应该长这样
\begin{equation*}
\begin{bmatrix}
1&1&1&1&1&1\\
1 & 0&0&0&0&0\\
0 & 0&1&3&3&1\\
0 & 0&0&1&2&1\\
0 & 0&0&0&1&1\\
0 & 0&0&0&0&1\\
\end{bmatrix}^{i - 1}*
\begin{bmatrix}
F_{1}\\
F_0\\
1\\
1\\
1\\
1
\end{bmatrix}=
\begin{bmatrix}
1&1&1&1&1&1\\
1 & 0&0&0&0&0\\
0 & 0&1&3&3&1\\
0 & 0&0&1&2&1\\
0 & 0&0&0&1&1\\
0 & 0&0&0&0&1\\
\end{bmatrix}*
\begin{bmatrix}
F_{i - 1}\\
F_{i - 2}\\
i^3\\
i^2\\
i\\
1
\end{bmatrix}=
\begin{bmatrix}
F_{i}\\
F_{i - 1}\\
(i + 1)^3\\
(i + 1)^2\\
i + 1\\
1
\end{bmatrix}
\end{equation*}
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
//#define int long long
using namespace std;
const int mod = 1e9 + ;
inline LL read() {
char c = getchar(); LL x = , f = ;
while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
while(c >= '' && c <= '') x = x * + c - '', c = getchar();
return x * f;
}
int T;
LL N;
struct Matrix {
LL a[][], N;
Matrix() {
N = ;
memset(a, , sizeof(a));
}
Matrix operator * (const Matrix &rhs) const {
Matrix ans;
for(int k = ; k <= N; k++)
for(int i = ; i <= N; i++)
for(int j = ; j <= N; j++)
(ans.a[i][j] += (1ll * a[i][k] * rhs.a[k][j]) % mod) %= mod;
return ans;
}
};
Matrix fp(Matrix a, LL p) {
Matrix base;
// printf("%d", base.a[0][1]);
for(int i = ; i <= ; i++) base.a[i][i] = ;
while(p) {
if(p & ) base = base * a;
a = a * a; p >>= ;
}
return base;
}
const LL GG[][] = {
{, , , , , , },
{, , , , , , },
{, , , , , , },
{, , , , , , },
{, , , , , , },
{, , , , , , },
{, , , , , , }
};
int main() {
T = read();
while(T--) {
N = read();
if(N == ) {puts(""); continue;}
if(N == ) {puts(""); continue;}
Matrix M;
memcpy(M.a, GG, sizeof(M.a));
Matrix ans = fp(M, N - );
LL out = ;
(out += ans.a[][] * ) %= mod;
(out += ans.a[][] * ) %= mod;
(out += ans.a[][] * ) %= mod;
(out += ans.a[][] * ) %= mod;
(out += ans.a[][] * ) %= mod;
(out += ans.a[][]) %= mod;
printf("%lld\n", out % mod);
}
return ;
}
/*
5
4
1
2
3
100
*/