[luogu 4389] 付公主的背包

题意：求一个较大的多重背包对于每个i的方案数，答案对998244353取模。

思路：

生成函数：

对于一个\(V\)

设：

\(f(x) = \sum_{i=0}^{oo} x ^ {V * i} = {1 \over {1 - x ^ V}}\)

那么就是求这个生成函数的积。

首先将\(f(x)\)取\(ln\)为\(g(x)\)，最后\(exp\)回去得到答案。

\(g'(x) = {f'(x) \over f(x)} = (1 - x^V)\sum_{i = 1}^{oo}V * i * x ^ {V}\)

最终\(exp\)一遍得到答案。

时间复杂度\(O(mlogm)\)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000010;
const int mod = 998244353;
const int g = 3;
#define ll long long
int A[maxn];
int B[maxn];
int c[maxn],d[maxn];
int e[maxn];
int f[maxn];
int w[maxn][2];
int rev[maxn];
int cnt[maxn];
int v[maxn];
int ans[maxn];
int inv[maxn];
const int vg = (mod + 1) / 3;
int n,m,l;
int len;
inline int read() {
	int q=0,f=1;char ch = getchar();
	while(!isdigit(ch)){
		if(ch=='-')f=-1;ch=getchar();
	}
	while(isdigit(ch)){
		q=q*10+ch-'0';ch=getchar();
	}
	return q*f;
}
inline int pow_mod (int x,int y) {
	int res = 1;
	while(y) {
		if(y & 1) res = (ll) res * x % mod;
		x = (ll) x * x % mod;
		y >>= 1;
	}
	return res;
}
inline void NTT(int *a,int n,int type) {
	for(len = 1,l = 0;len <= n;len <<= 1, ++l);
	for(int i = 0;i < len; ++i) {
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
	}
	for(int i = 0;i < len; ++i) {
		if(i < rev[i]) {
			swap(a[i],a[rev[i]]);
		}
	}
	for(int i = 1;i < len; i <<= 1) {
		int wn = (~type) ? w[i][0]:w[i][1];
		for(int j = 0;j < len;j += (i << 1)) {
			int w = 1;
			for(int k = 0;k < i; ++k,w = (ll)w * wn % mod) {
				int x = (ll)a[i + j + k];
				int y = a[j + k];
				a[j + k] = (x + y) % mod;
				a[i + j + k] = (y + mod - x) % mod;
			}
		}
	}
	if(type == -1) {
		int tmp = pow_mod(len,mod - 2);
		for(int i = 0;i < len; ++i) {
			a[i] = (ll)a[i] * tmp % mod;
		}
	}
}
inline void get_Inv(int *a,int *b,int n) {
	if(n == 1) {
		b[0] = pow_mod(a[0],mod - 2);
		return;
	}
	get_Inv(a,b,n >> 1);
	for(int i = 0;i < n; ++i) {
		A[i] = a[i];
		B[i] = b[i];
	}
	NTT(A,n << 1,1);
	NTT(B,n << 1,1);
	for(int i = 0;i < len; ++i) {
		A[i] = (ll)A[i] * B[i] % mod * B[i] % mod;
	}
	NTT(A,n << 1,-1);
	for(int i = 0;i < n; ++i) {
		b[i] = ((b[i] << 1) % mod - A[i] + mod) % mod;
	}
	for(int i = 0;i < len; ++i) {
		A[i] = B[i] = 0;
	}
}

inline void get_ln(int *a,int *b,int n) {
	get_Inv(a,c,n);
	for(int i = 0;i < n - 1; ++i) {
		d[i] = (ll)(i + 1) * a[i + 1] % mod;
	}
	NTT(c,n << 1,1);
	NTT(d,n << 1,1);
	for(int i = 0;i < len; ++i) {
		c[i] = (ll) c[i] * d[i] % mod;
	}
	NTT(c,n << 1,-1);
	for(int i = 1;i < n; ++i) {
		b[i] = (ll)inv[i] * c[i - 1] % mod;
	}
	for(int i = 0;i < len; ++i) {
		c[i] = d[i] = 0;
	}
}

inline void get_exp(int *a,int *b,int n) {
	if(n == 1) {
		b[0] = 1;
		return;
	}
	get_exp(a,b,n >> 1);
	for(int i = 0;i < n; ++i) {
		e[i] = b[i];
	}
	get_ln(b,f,n);
	for(int i = 0;i < n; ++i) {
		f[i] = (mod - f[i] + a[i]) % mod;
		f[0] = (f[0] + 1)%mod;
	}
	NTT(e,n << 1,1);
	NTT(f,n << 1,1);
	for(int i = 0;i < len; ++i) {
		e[i] = (ll)e[i] * f[i] % mod;
	}
	NTT(e,n << 1,-1);
	for(int i = 0;i < n; ++i) {
		b[i] = e[i];
	}
	for(int i = 0;i < len; ++i) {
		e[i] = f[i] = 0;
	}
}

int main () {
	n = read(),m = read();
	inv[1] = 1;
	for(int i = 1;i <= n; ++i) {
		++cnt[read()];
	}
	int tmp = 1;
	for(;tmp <= m;tmp <<= 1);
	for(int i = 1;i <= (tmp << 1);i <<= 1) {
		w[i][0] = pow_mod(g,(mod - 1)/(i << 1));
		w[i][1] = pow_mod(vg,(mod - 1)/(i<<1));
	}
	for(int i = 2;i <= tmp; ++i) {
		inv[i] = (ll)(mod - mod / i) * inv[mod % i]%mod;
	}
	for(int i = 1;i <= m; ++i) {
		if(cnt[i]) {
			int dl = (ll) i * cnt[i] % mod;
			for(int j = 1;i * j <= m; ++j) {
				v[i * j] = (v[i * j] + dl) % mod;
			}
		}
	}
	for(int i = 1;i <= m; ++i) {
		v[i] = (ll)inv[i] * v[i] % mod;
	}
	get_exp(v,ans,tmp);
	for(int i = 1;i <= m; ++i) {
		printf("%d\n",ans[i] % mod);
	}
	return 0;
}
//原地爆炸