题解 Yet Another Number Sequence

题目传送门

Description

给出 \(n,k\) ，求出：

\[\sum_{i=1}^{n} f_i·i^k \]

其中 \(f_i\) 表示斐波拉契第 \(i\) 项。\(n\le 10^{18},k\le 40\) 。

Solution

哎，被这个 \(k\le 40\) 误导了，以为是斯特林数。/kk

发现假如我们用矩阵表示 \(f_i\)，那么我们就只需要求出：

\[\sum_{i=1}^{n} G^ii^k \]

其中 \(G=\begin{bmatrix}1 & 1\\ 1 & 0\end{bmatrix}\) 。

考虑倍增，设 \(f(n,m)=\sum_{i=1}^{n} G^ii^m\)，则有：

\[f(2n,m)=f(n,m)+G^n\sum_{h=0}^{m} n^h\binom{m}{h}f(n,m-h) \]

然后我们就可以做到 \(\Theta(k^2\log n)\) 了，不过似乎分治的时候用 \(\text{fft}\) 的话可以做到 \(\Theta(k\log k\log n)\)，不过这个题中 \(k\) 不是很大，所以可以接受。

（矩阵的 \(\text{fwt}\) 可以对于每一项搞）

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define mod 1000000007
#define int long long
#define MAXN 45

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> inline void chkmax (T &a,T b){a = max (a,b);}
template <typename T> inline void chkmin (T &a,T b){a = min (a,b);}

int k,C[MAXN][MAXN];
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
void Add (int &a,int b){a = add (a,b);}
void Sub (int &a,int b){a = dec (a,b);}

struct matrix{
	int mat[2][2];
	matrix(){memset (mat,0,sizeof (mat));}
	int * operator [] (const int key){return mat[key];}
	matrix operator * (const matrix &p)const{
		matrix now;
		for (Int i = 0;i < 2;++ i)
			for (Int j = 0;j < 2;++ j)
				for (Int k = 0;k < 2;++ k)
					Add (now.mat[i][k],mul (mat[i][j],p.mat[j][k]));
		return now;
	}
	matrix operator * (const int &p)const{
		matrix now;
		for (Int i = 0;i < 2;++ i)
			for (Int j = 0;j < 2;++ j)
				now.mat[i][j] = mul (mat[i][j],p);
		return now;
	}
	matrix operator + (const matrix &p)const{
		matrix now;
		for (Int i = 0;i < 2;++ i)
			for (Int j = 0;j < 2;++ j)
				now.mat[i][j] = add (mat[i][j],p.mat[i][j]);
		return now;
	}
}one,ret;

struct node{
	matrix gn,f[MAXN];
};

node getit (int n){
	node now;
	if (n == 1){
		now.gn = ret;
		for (Int i = 0;i <= k;++ i) now.f[i] = ret;
		return now;
	}
	if (n & 1){
		node lst = getit (n - 1);
		now.gn = lst.gn * ret;
		for (Int i = 0,res = 1;i <= k;++ i,res = mul (res,n % mod)) now.f[i] = lst.f[i] + now.gn * res;
	}
	else{
		int mid = n >> 1;
		node lst = getit (mid);now.gn = lst.gn * lst.gn;
		for (Int m = 0;m <= k;++ m){
			now.f[m] = lst.f[m];
			for (Int h = 0,res = 1;h <= m;++ h,res = mul (res,mid % mod)) now.f[m] = now.f[m] + (lst.gn * lst.f[m - h]) * mul (res,C[m][h]);
		} 
	}
	return now;
}

int n;

signed main(){
	read (n,k),one[0][0] = one[1][1] = 1,ret[0][0] = ret[0][1] = ret[1][0] = C[0][0] = 1;
	for (Int i = 1;i <= k;++ i){
		C[i][0] = 1;
		for (Int j = 1;j <= i;++ j) C[i][j] = add (C[i - 1][j - 1],C[i - 1][j]);
	}
	node now = getit (n);
	write (now.f[k][0][0]),putchar ('\n');
	return 0;
}