2018年湘潭大学程序设计竞赛G又见斐波那契(矩阵快速幂)

题意

题目链接

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
*/