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