洛谷 P4389 付公主的背包

付公主有一个可爱的背包qwq

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 \(v_i\)，都有无限件

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

考虑对每个物品写出OGF，以体积作为\(x\)的次数

\[F_i(x)=\sum_{j\ge0}x^{jv_i}=\frac{1}{1-x^{v_i}} \]

那么最后的答案为

\[G(x)=\prod_{i=1}^nF_i(x)=\prod_{i=1}^n\frac{1}{1-x^{v_i}} \]

而这个东西是不好求的，因为有\(\prod\)，而我们擅长求\(\sum\)，所以我们可以先用ln把这个变为加法，再exp回去

\[\begin{aligned} explnG(x)&=exp\sum_{i=1}^nln\frac{1}{1-x^{v_i}}\\ &=exp\sum_{i=1}^n\sum_{j\ge1}\frac{x^{jv_i}}{j}\\ \end{aligned}\]

我们可以把体积相同的物品合成一个，假设体积为\(i\)的物品有\(b_i\)个，于是得到

\[exp\sum_{k\ge1}b_k\sum_{j\ge1}^{jk\le m}\frac{x^{jk}}{j} \]

而我们知道对于给定的\(m\)，只有前\(m\)项是有用的，所以\(jk\le m\)，于是这么直接枚举\(k\)和\(j\)的复杂度是\(O(mlogm)\)的，最后再exp回去即可

Code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
const int N = 4e5;
const int p = 998244353;
using namespace std;
int n,m,v[N + 5],b[N + 5],a[N + 5],maxn,lg,rev[N + 5],G[N + 5][2],ans[N + 5];
int mypow(int a,int x){int s = 1;for (;x;x & 1 ? s = 1ll * s * a % p : 0,a = 1ll * a * a % p,x >>= 1);return s;}
void pre(int n)
{
    maxn = 1,lg = 0;
    while (maxn <= n)
        maxn <<= 1,lg++;
    for (int i = 0;i < maxn;i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
}
void clear(int *a,int n,int l = 0)
{
    for (int i = l;i < n;i++)
        a[i] = 0;
}
void ntt(int *a,int typ)
{
    for (int i = 0;i < maxn;i++)
        if (i < rev[i])
            swap(a[i],a[rev[i]]);
    for (int i = 1;i < maxn;i <<= 1)
        for (int j = 0;j < maxn;j += i << 1)
            for (int k = 0;k < i;k++)
            {
                int x = a[j + k],t = 1ll * G[k + i][typ] * a[j + k + i] % p;
                a[j + k] = (x + t) % p;
                a[j + k + i] = (x - t) % p;
            }
    if (!typ)
    {
        int inv = mypow(maxn,p - 2);
        for (int i = 0;i < maxn;i++)
            a[i] = 1ll * a[i] * inv % p;
    }
}
int INVa[N + 5];
void INV(int *a,int *ans,int n)
{
    if (n == 1)
    {
        ans[0] = mypow(a[0],p - 2);
        return;
    }
    INV(a,ans,n + 1 >> 1);
    pre(n * 2);
    for (int i = 0;i < n;i++)
        INVa[i] = a[i];
    clear(INVa,maxn,n);
    ntt(INVa,1);
    ntt(ans,1);
    for (int i = 0;i < maxn;i++)
        ans[i] = (2ll * ans[i] % p - 1ll * INVa[i] * ans[i] % p * ans[i] % p) % p;
    ntt(ans,0);
    clear(ans,maxn,n);
}
int Lna[N + 5],Lnb[N + 5];
void DOV(int *a,int *f,int n)
{
    for (int i = 1;i < n;i++)
        f[i - 1] = 1ll * i * a[i] % p;
    f[n - 1] = 0;
}
void DOVINV(int *a,int *f,int n)
{
    f[0] = 0;
    for (int i = 1;i < n;i++)
        f[i] = 1ll * mypow(i,p - 2) * a[i - 1] % p;
}
void Ln(int *a,int *ans,int n)
{
    DOV(a,Lna,n);
    pre(n * 2);
    clear(Lnb,maxn);
    INV(a,Lnb,n);
    pre(n * 2);
    clear(Lna,maxn,n);
    ntt(Lna,1);
    ntt(Lnb,1);
    for (int i = 0;i < maxn;i++)
        Lna[i] = 1ll * Lna[i] * Lnb[i] % p;
    ntt(Lna,0);
    DOVINV(Lna,ans,n);
    clear(ans,maxn,n);
}
int expa[N + 5],expb[N + 5];
void exp(int *a,int *ans,int n)
{
    if (n == 1)
    {
        ans[0] = 1;
        return;
    }
    exp(a,ans,n + 1 >> 1);
    Ln(ans,expa,n);
    pre(n * 2);
    for (int i = 0;i < n;i++)
        expb[i] = a[i];
    clear(expb,maxn,n);
    ntt(ans,1);
    ntt(expa,1);
    ntt(expb,1);
    for (int i = 0;i < maxn;i++)
        ans[i] = 1ll * ans[i] * ((1 - expa[i] + expb[i]) % p) % p;
    ntt(ans,0);
    clear(ans,maxn,n);
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i = 1;i <= n;i++)
        scanf("%d",&v[i]),b[v[i]]++;
    pre(m * 2 + 2);
    for (int i = 1;i < maxn;i <<= 1)
    {
        int g1 = mypow(3,(p - 1) / (i << 1)),g2 = mypow(mypow(3,p - 2),(p - 1) / (i << 1));
        G[i][0] = G[i][1] = 1;
        for (int j = 1;j < i;j++)
            G[i + j][1] = 1ll * G[i + j - 1][1] * g1 % p,G[i + j][0] = 1ll * G[i + j - 1][0] * g2 % p;
    }
    for (int k = 1;k <= m;k++)
        if (b[k])
            for (int j = 1;j * k <= m;j++)
                a[j * k] += 1ll * b[k] * mypow(j,p - 2) % p,a[j * k] %= p;
    exp(a,ans,m + 1);
    for (int i = 1;i <= m;i++)
        printf("%d\n",(ans[i] + p) % p);
    return 0;
}