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 + 7;
inline ll read() {
    char c = getchar(); ll x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int t;
ll n;
struct matrix {
    ll a[10][10], n;
    matrix() {
        n = 6;
        memset(a, 0, sizeof(a));
    }
    matrix operator * (const matrix &rhs) const {
        matrix ans;
        for(int k = 1; k <= n; k++) 
            for(int i = 1; i <= n; i++)
                for(int j = 1; 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 = 1; i <= 6; i++) base.a[i][i] = 1;
    while(p) {
        if(p & 1) base = base * a;
        a = a * a; p >>= 1;
    }
    return base;
}
const ll gg[10][10] = {
       {0, 0, 0, 0, 0, 0, 0},
       {0, 1, 1, 1, 1, 1, 1},
       {0, 1, 0, 0, 0, 0, 0},
       {0, 0, 0, 1, 3, 3, 1},
       {0, 0, 0, 0, 1, 2, 1},
       {0, 0, 0, 0, 0, 1, 1},
       {0, 0, 0, 0, 0, 0, 1}
};
int main() {
    t = read();
    while(t--) {
        n = read();
        if(n == 1) {puts("1"); continue;}
        if(n == 2) {puts("16"); continue;}
        matrix m;
        memcpy(m.a, gg, sizeof(m.a));
        matrix ans = fp(m, n - 2);
        ll out = 0;
        (out += ans.a[1][1] * 16) %= mod;
        (out += ans.a[1][2] * 1) %= mod;
        (out += ans.a[1][3] * 27) %= mod;
        (out += ans.a[1][4] * 9) %= mod;
        (out += ans.a[1][5] * 3) %= mod;
        (out += ans.a[1][6]) %= mod;
        printf("%lld\n", out % mod);
    }
    return 0;
}
/*
5
4
1
2
3
100
*/