[Luogu P4389] 付公主的背包

洛谷传送门

题目描述

这个背包最多可以装 $10^5$ 大小的东西

付公主有 $n$ 种商品，她要准备出摊了

每种商品体积为 $V_i$ ，都有 $10^5$ 件

给定 $m$ ，对于 $s\in [1,m]$ ，请你回答用这些商品恰好装 $s$ 体积的方案数

输入输出格式

输入格式：

第一行 $n,m$

第二行 $V_1\sim V_n$

输出格式：

$m$ 行，第 $i$ 行代表 $s=i$ 时方案数，对 $998244353$ 取模

输入输出样例

输入样例#1：

2 4
1 2

输出样例#1：

说明

对于30%的数据， $n\le 3000,m\le 3000$

对于60%的数据，纯随机生成

对于100%的数据， $n\le 100000,m\le 100000$

对于100%的数据， $V_i\le m$

解题分析

对于一个体积为 $V_i$ 的物品，其生成函数为 $1+x^{V_i}+x^{2V_i}+...+x^{kVi}=\frac{1}{1-x^{V_i}}$ 。那么很显然我们就是要快速求所有生成函数的积，找到对应项的系数即可。

关键是如何快速求这个玩意。有个很套路的做法是先把这玩意求 $ln$ 加上再 $exp$ 回去。

那么开始推式子：设 $f(x)$ 为第 $i$ 个物品的生成函数， $g(x)=ln(f(x))$ ，那么有：
$g'(x)=\frac{f'(x)}{f(x)} \\ =\frac{\sum_{k\ge 1}kV_ix^{kV_i-1}}{\frac{1}{1-x^{V_i}}} \\ =(1-x^{V_i})\sum_{k\ge 1}kV_ix^{kV_i-1} \\ =\sum_{k\ge 1}V_ix^{kV_i-1} \\ g(x)=\sum_{k\ge 1}\frac{1}{k}x^{kV_i}$
所以这玩意记录一下出现次数可以 $O(Nln(N))$ 求。

那么剩下的就是一个 $exp$ 的板子辣！

代码如下：

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <algorithm>
#include <cstring>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define MX 600500
#define MOD 998244353
#define g 3
#define ginv 332748118
template <class T>
IN void in(T &x)
{
	x = 0; R char c = gc;
	for (; !isdigit(c); c = gc);
	for (;  isdigit(c); c = gc)
	x = (x << 1) + (x << 3) + c - 48;
}
IN int fpow(R int base, R int tim)
{
	int ret = 1;
	W (tim)
	{
		if (tim & 1) ret = 1ll * ret * base % MOD;
		base = 1ll * base * base % MOD, tim >>= 1;
	}
	return ret;
}
int a1[MX], a2[MX], b1[MX], b2[MX], b3[MX], b4[MX], c[MX];
int inv[MX], A[MX], rev[MX], res[MX], cnt[MX];
namespace Poly
{
	IN void NTT(int *dat, R int len, R bool typ)
	{
		for (R int i = 1; i < len; ++i) if (rev[i] > i) std::swap(dat[i], dat[rev[i]]);
		R int seg, step, bd, now, cur, base, deal, buf1, buf2;
		for (seg = 1; seg < len; seg <<= 1)
		{
			base = fpow(typ ? g : ginv, (MOD - 1) / (seg << 1)), step = seg << 1;
			for (now = 0; now < len; now += step)
			{
				deal = 1, bd = now + seg;
				for (cur = now; cur < bd; cur++, deal = 1ll * deal * base % MOD)
				{
					buf1 = dat[cur], buf2 = 1ll * dat[cur + seg] * deal % MOD;
					dat[cur] = (buf1 + buf2) % MOD, dat[cur + seg] = (buf1 - buf2 + MOD) % MOD;
				}
			}
		}
		if (typ) return; int inv = fpow(len, MOD - 2);
		for (R int i = 0; i < len; ++i) dat[i] = 1ll * inv * dat[i] % MOD;
	}
	IN void Dr(int *dat, int *dr, R int len) {for (R int i = 1; i < len; ++i) dr[i - 1] = 1ll * dat[i] * i % MOD; dr[len - 1] = 0;}
	IN void Itg(int *dat, int *itg, R int len) {for (R int i = 1; i < len; ++i) itg[i] = 1ll * dat[i - 1] * fpow(i, MOD - 2) % MOD; itg[0] = 0;}
	IN void Getinv(R int up, R int lg, int *ind, int *ans)
	{
		if (up == 1) return ans[0] = fpow(ind[0], MOD - 2), void();
		Getinv(up >> 1, lg - 1, ind, ans); int len = up << 1, half = up >> 1; lg++;
		for (R int i = 0; i < len; ++i) a1[i] = a2[i] = 0;
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < half; ++i) a1[i] = ans[i];
		for (R int i = 0; i < up; ++i) a2[i] = ind[i];
		NTT(a1, len, 1), NTT(a2, len, 1);
		for (R int i = 0; i < len; ++i) a1[i] = (a1[i] * 2 % MOD - 1ll * a1[i] * a1[i] % MOD * a2[i] % MOD + MOD) % MOD;
		NTT(a1, len, 0);
		for (R int i = 0; i < up; ++i) ans[i] = a1[i];
	}
	IN void Getln(R int up, R int lg, int *dat, int *ans)
	{
		Dr(dat, b1, up); Getinv(up, lg, dat, b2);
		int len = up << 1; lg++;
		for (R int i = 0; i < len; ++i) a1[i] = a2[i] = 0;
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < up; ++i) a1[i] = b1[i], a2[i] = b2[i];
		NTT(a1, len, 1), NTT(a2, len, 1);
		for (R int i = 0; i < len; ++i) a1[i] = 1ll * a1[i] * a2[i] % MOD;
		NTT(a1, len, 0); Itg(a1, ans, up);
	}
	IN void Getexp(R int up, R int lg)
	{
		if (up == 1) return res[0] = 1, void();
		Getexp(up >> 1, lg - 1); Getln(up, lg, res, c);
		int len = up << 1, half = up >> 1; lg++;
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < len; ++i) a1[i] = a2[i] = 0;
		for (R int i = 0; i < up; ++i) a1[i] = (A[i] - c[i] + MOD) % MOD;
		for (R int i = 0; i < half; ++i) a2[i] = res[i];
		a1[0] = (a1[0] + 1) % MOD; NTT(a1, len, 1), NTT(a2, len, 1);
		for (R int i = 0; i < len; ++i) a1[i] = 1ll * a1[i] * a2[i] % MOD;
		NTT(a1, len, 0);
		for (R int i = 0; i < up; ++i) res[i] = a1[i];
	}
}
 int main(void)
 {
	int n, m, foo; in(n), in(m); inv[0] = inv[1] = 1;
	for (R int i = 2; i <= m; ++i) inv[i] = 1ll * inv[MOD % i] * (MOD - MOD / i) % MOD;
	for (R int i = 1; i <= n; ++i) in(foo), ++cnt[foo];
	for (R int i = 1; i <= m; ++i)
	{
		if (cnt[i])
		for (R int j = 1; i * j <= m; ++j)
		(A[j * i] += 1ll * inv[j] * cnt[i] % MOD) %= MOD;
	}
	int len = 1, lg = 0;
	for (; len <= m; len <<= 1, lg++);
	Poly::Getexp(len, lg);
	for (R int i = 1; i <= m; ++i) printf("%d\n", res[i]);
 }