启智树提高组Day4T3 2的幂拆分

问把 \(n\) 拆分成2的幂的和的形式的方案数。注意拆分是无序的,即 \(2^0,2^2,2^1,2^0\)\(2^0,2^0,2^1,2^2\) 是一种方案。\(n \le 10^{18}\)

前置知识

一些定义与约定

定义:

\[S_k(n)=\sum_{i=0}^{n-1}i^k \]

\[\]

部分下取整符号可能有遗漏。

题解

首先通过打表发现前几项是:(从0开始)

\[1,1,2,2,4,4,6,6,10,10,14,14,20,20,26,26,... \]

发现去重差分还是原数组,并且发现:

\[f_i = f_{i - 1} ,i \bmod 2=1 \]

\[f_i = f_{i-1}+f_{i/2},i\bmod2=0 \]

不断重复使用这两个式子拆开发现:

\[f_{n} = \sum_{j=0}^{\left \lfloor n/2 \right \rfloor}f_j \]

\(f_j\) 发现:

\[f_n=\sum_{j=0}^{\left \lfloor n/2 \right \rfloor }\sum_{i=0}^{\left \lfloor j/2 \right \rfloor }f_i \]

\[=\sum_{i=0}^{\left \lfloor n/4 \right \rfloor } f_i\sum_{j=2i}^{\left \lfloor n/2 \right \rfloor }1 \]

\[=\sum_{i=0}^{n/4}f_i(S_{0}(\left \lfloor n/2 \right \rfloor +1)-S_0(2i)) \]

\[=S_0(\left \lfloor n/2 \right \rfloor +1)\sum_{i=0}^{n/4}f_i-\sum_{i=0}^{n/4}f_iS_0(2i) \]

\[=S_0(\left \lfloor n/2 \right \rfloor +1)\sum_{i=0}^{n/4}f_i-2\sum_{i=0}^{n/4}f_ii \]

(公式不太好打,就不打那么多了)

发现前面的那部分可以直接递归子问题,我们难以解决的是后边的那个类似 \(\sum_{i=0}^{n}f_ii^k\) 的式子。我们试图求解这个式子:

\[F(n,k)=\sum_{i=0}^nf_ii^k \]

\[=\sum_{i=0}^ni^k\sum_{j=0}^{i/2}f_j \]

\[=\sum_{j=0}^{n/2}f_j \sum_{i=2j}^ni^k \]

\[=\sum_{j=0}^{n/2}f_j(S_k(n+1)-S_k(2j)) \]

\[=\sum_{j=0}^{n/2}f_jS_k(n+1)-\sum_{j=0}^{n/2}f_j\sum_{d=1}^{k+1}\frac{{k + 1 \choose d}}{k+1}B_{k+1-d}(2j)^d \]

\[=\sum_{j=0}^{n/2}f_jS_k(n+1)-\sum_{d=1}^{k+1}\frac{{k + 1 \choose d}}{k+1}B_{k+1-d}2^d \sum_{j=0}^{n/2}f_jj^d \]

\[=F(n/2,0)S_k(n+1)-\sum_{d=1}^{k+1}\frac{{k + 1 \choose d}}{k+1}B_{k+1-d}2^d F(n/2,d) \]

发现左边是子问题,右边每次需要的 \(k\) 会加一。因此 \(k\) 最多是 \(logn\)。状态数 \(O(log^2n)\),转移复杂度 \(O(logn)\),总复杂度:\(O(log^3n)\)。由于 \(k\) 比较小,不需要多项式求逆加速求 \(B_i\),且这并不是复杂度瓶颈。

注意最好手写个哈希表,不然用 STL map 可能会被卡常。

\(Code:\)

#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <ctime>
#include <algorithm>
#define N 76
typedef long long ll;
template <typename T> inline void read(T &x) {
	x = 0; char c = getchar(); bool flag = false;
	while (!isdigit(c)) { if (c == '-')	flag = true; c = getchar(); }
	while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
	if (flag)	x = -x;
}
using namespace std;
const int P = 1e9 + 7;
inline void MAX(int &a, int b) {
	if (b > a)	a = b;
}
ll c[N][N], B[N];
inline ll quickpow(ll x, ll k) {
	ll res = 1;
	while (k) {
		if (k & 1)	res = res * x % P;
		x = x * x % P;
		k >>= 1;
	}
	return res;
}
ll inv[N];
inline ll get_c(int n, int m) {
	return c[n + 1][m + 1];
}
inline void init_B() {
	const int up = 66;
	for (register int i = 1; i <= up; ++i)	inv[i] = quickpow(i, P - 2);
	c[1][1] = 1;
	for (register int i = 2; i <= up; ++i) {
		for (register int j = 1; j <= i; ++j) {
			c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
		}
	}
	B[0] = 1;
	for (register int i = 1; i <= up; ++i) {
		for (register int j = 0; j <= i - 1; ++j) {
			B[i] = (B[i] + get_c(i + 1, j) * B[j]) % P;
		}
		B[i] = (P - inv[i + 1] * B[i]) % P; 
	}
}
const int PP = 13331;
struct Hashtable {//手写了哈希表,也可以用STL map,但是复杂度会多一个 log
	struct edge{
		int nxt;
		ll to;
		int val;
	}e[N << 1];
	int head[PP + 1000], ecnt;
	inline void init() {
		memset(head, 0, sizeof(head));
		ecnt = 0;
	}
	inline void addedge(ll v, int val) {
		int mirr = v % PP + 1;
		e[++ecnt] = (edge){head[mirr], v, val};
		head[mirr] = ecnt;
	}
	inline int find(ll v) {
		int mirr = v % PP + 1;
		for (register int i = head[mirr]; i; i = e[i].nxt) {
			ll to = e[i].to; if (to == v)	return e[i].val;
		}
		return -1;
	}
}Hash;
int mtot;
ll memo[N << 1][N];
ll calc(ll n, int k) {
	int mirr = Hash.find(n);
	if (mirr == -1)	Hash.addedge(n, ++mtot), mirr = mtot;
	if (memo[mirr][k])	return memo[mirr][k];
	if (n <= 1) {
		if (n == 0) {
			return memo[mirr][k] = (k == 0 ? 1 : 0);
		}
		return memo[mirr][k] = (k == 0 ? 2 : 1);
	}
	ll res = 0, tmp = (n + 1) % P;
	for (register int d = 1; d <= k + 1; ++d) {
		res = (res + get_c(k + 1, d) * B[k - d + 1] % P * inv[k + 1] % P * tmp) % P;
		tmp = (n + 1) % P * tmp % P;
	}
	res = res * calc(n >> 1, 0) % P;
	for (register int d = 1; d <= k + 1; ++d) {
		res = (res - (1ll << d) % P * get_c(k + 1, d) % P * inv[k + 1] % P * B[k - d + 1] % P * calc(n >> 1, d) % P) % P;
	}
	return memo[mirr][k] = res;
}
ll n;
inline void work() {
	Hash.init();
	mtot = 0;
	memset(memo, 0, sizeof(memo));
	read(n);
	printf("%lld\n", (calc(n >> 1, 0) % P + P) % P);
}

int main() {
	init_B();
	int _; read(_);
	while (_--) {
		work();
	}
	return 0;
}
posted @ 2020-08-23 20:03  JiaZP  阅读(187)  评论(0编辑  收藏  举报