luoguP4389 付公主的背包 多项式exp

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%%%dkw

话说这是个论文题来着...

考虑生成函数\(OGF\)

对于价值为\(v\)的物品，由于有\(10^5\)的件数，可以看做无限个

那么，其生成函数为\(x^0 + x^{v} + x^{2v} + ... = \frac{1}{1 - x^v}\)

我们所需的答案即\([x^n] \prod \frac{1}{1 - x^{v_i}}\)

只需考虑求出\(A = \prod \frac{1}{1 - x^{v_i}}\)

自然地想到取对数

\(In(A) = \sum In(\frac{1}{1 - x^{v_i}})\)

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不难发现

\(In(\frac{1}{1 - x^v}) = - In(1 - x^v)\)

考虑用麦克劳林级数来模拟，那么

由于\(In^{(n)}(1 - x) = - \frac{1}{(1 - x)^n} * (n - 1)!\)

\(-In(1 - x^v) = \sum \frac{x^{vi}}{i}\)

于是，我们可以直接枚举倍数，在\(O(m \log m)\)的时间内完成计算

最后只要\(O(m \log m)\)的\(exp\)一下即可

---

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

#define gc getchar
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
    while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
    return p * w;
}

const int sid = 500050;
const int mod = 998244353;

int n, m;
int V[sid], F[sid], inv[sid], rev[sid], ans[sid];

inline int Inc(int a, int b) { return (a + b >= mod) ? a + b - mod : a + b; }
inline int Dec(int a, int b) { return (a - b < 0) ? a - b + mod : a - b; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline int fp(int a, int k) {
	int ret = 1;
	for( ; k; k >>= 1, a = mul(a, a))
		if(k & 1) ret = mul(ret, a);
	return ret;
}
	
inline void init(int Maxn, int &n, int &lg) {
	n = 1; lg = 0;
	while(n < Maxn) n <<= 1, lg ++;
}
	
inline void NTT(int *a, int n, int opt) {
	for(ri i = 0; i < n; i ++) if(i < rev[i]) swap(a[i], a[rev[i]]);
	for(ri i = 1; i < n; i <<= 1)
	for(ri j = 0, g = fp(3, (mod - 1) / (i << 1)); j < n; j += (i << 1))
	for(ri k = j, G = 1; k < i + j; k ++, G = mul(G, g)) {
		int x = a[k], y = mul(G, a[i + k]);
		a[k] = (x + y >= mod) ? x + y - mod : x + y;
		a[i + k] = (x - y < 0) ? x - y + mod : x - y;
	}
	if(opt == -1) {
		int ivn = fp(n, mod - 2);
		reverse(a + 1, a + n);
		rep(i, 0, n) a[i] = mul(a[i], ivn);
	}
}

int ia[sid], ib[sid];
inline void Inv(int *a, int *b, int n) {
	if(n == 1) { b[0] = fp(a[0], mod - 2); return; }
	Inv(a, b, n >> 1);
	
	int N = 1, lg = 0; init(n + n, N, lg);
	for(ri i = 0; i < N; i ++)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
	
	for(ri i = 0; i < N; i ++) ia[i] = ib[i] = 0;
	for(ri i = 0; i < n; i ++) ia[i] = a[i], ib[i] = b[i];
	
	NTT(ia, N, 1); NTT(ib, N, 1);
	for(ri i = 0; i < N; i ++)
		ia[i] = Dec((ib[i] << 1) % mod, mul(ia[i], mul(ib[i], ib[i])));
	NTT(ia, N, -1);
	
	for(ri i = 0; i < n; i ++) b[i] = ia[i];
}

inline void Inv_init(int n) {
	inv[0] = inv[1] = 1;
	rep(i, 2, n) inv[i] = mul(inv[mod % i], mod - mod / i);
}

inline void wf(int *a, int *b, int n) { for(ri i = 1; i < n; i ++) b[i - 1] = mul(a[i], i); }
inline void jf(int *a, int *b, int n) { for(ri i = 1; i < n; i ++) b[i] = mul(a[i - 1], inv[i]); }
	
int iv[sid], dx[sid];
inline void In(int *a, int *b, int n) {
	for(ri i = 0; i < n + n; i ++) iv[i] = dx[i] = 0;
	Inv(a, iv, n); wf(a, dx, n);
	
	int N = 1, lg = 0; init(n + n, N, lg);
	for(ri i = 0; i < N; i ++)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
	
	NTT(iv, N, 1); NTT(dx, N, 1);
	for(ri i = 0; i < N; i ++) iv[i] = mul(iv[i], dx[i]);
	NTT(iv, N, -1); jf(iv, b, n);
}


int inb[sid], fb[sid];
inline void Exp(int *a, int *b, int n) {
	if(n == 1) { b[0] = 1; return; }
	Exp(a, b, n >> 1); 
	
	
	for(ri i = 0; i < n + n; i ++) inb[i] = fb[i] = 0;
	In(b, inb, n);
	
	int N = 1, lg = 0; init(n + n, N, lg);
	for(ri i = 0; i < N; i ++) 
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
	
	for(ri i = 0; i < n; i ++) fb[i] = Dec(a[i], inb[i]); fb[0] ++;
	for(ri i = 0; i < n; i ++) inb[i] = b[i];
	
	NTT(inb, N, 1); NTT(fb, N, 1);
	for(ri i = 0; i < N; i ++) fb[i] = mul(fb[i], inb[i]);
	NTT(fb, N, -1);
	
	for(ri i = 0; i < n; i ++) b[i] = fb[i], b[i + n] = 0;
}
	
inline void calc() {
	int N = 1, lg = 0;
	init(m + 5, N, lg); Inv_init(N);
	for(ri i = 1; i <= m; i ++)
		for(ri j = i; j <= m; j += i)
			F[j] = Inc(F[j], mul(V[i], inv[j / i]));
	Exp(F, ans, N);
	rep(i, 1, m) printf("%d\n", ans[i]);
}
	
int main() {
	n = read(); m = read();
	rep(i, 1, n) V[read()] ++;
	calc();
	return 0;
}