bzoj 4555

题意：求$\sum_{i=0}^{n}\sum_{j=0}^{i}S(i,j)2^{j}j!$

A look that can not be done ...

But still requires careful analysis of

The most important step is the conversion:

The number of classes is defined a second Stirling: $ S (n, m) $ represents the program sets $ $ $ different articles to the n-m of $

Then consider things inside that summation, the meaning is found different $ $ I $ J $ articles to the sets, each set has two kinds of attributes, then the whole set of these programs are arranged Number

So based on this definition, the reset state $ g (n) = \ sum_ {j = 0} ^ {i} S (i, j) 2 ^ {j} j $, then there recursive formula:! $ G (n) = \ sum_ {j = 1} ^ {n} 2C_ {n} ^ {i} g (ni) $

This recursive enumeration is the origin of the first set of elements obtained

Routine according to expand the number of combinations, to give:

$g(n)=\sum_{i=1}^{n}2\frac{n!}{i!(n-i)!}g(n-i)$

Transposition can be obtained:

$\frac{g(n)}{n!}=\sum_{i=1}^{n}\frac{2}{i!}\frac{g(n-i)}{(n-i)!}$

So the right is clearly a form of convolution

设$F(x)=\sum_{i=1}^{n}\frac{2}{i!}x^{i}$

$G(x)=\sum_{i=0}^{n}\frac{g(i)}{(i)!}x^{i}$

Noting this case the direct convolution $ G (0) = 0 $, but it requires $ G (0) = 1 $ (boundary requirement)

Thus it can be determined $ G (x) = F (x) G (x) + 1 $

Transposition obtain $ G (x) = \ frac {1} {1-F (x)} $

(This method can be used like moisture cure, said FFT)

Etc., to obtain a $ G (x) $ What is the use?

See, for $ G (x) $ I $ by each! $ Obtain $ g (x) $ generating function, can be directly summed

Paste the code:

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#define ll long long
using namespace std;
const ll mode=998244353;
ll F[100005];
ll FF[100005];
ll G[100005];
ll inv[100005];
ll mul[100005];
ll minv[100005];
int to[(1<<20)+5];
int n;
void init()
{
    inv[0]=inv[1]=mul[0]=mul[1]=minv[0]=minv[1]=1;
    for(int i=2;i<=100000;i++)
    {
        inv[i]=(mode-mode/i)*inv[mode%i]%mode;
        minv[i]=minv[i-1]*inv[i]%mode;
        mul[i]=mul[i-1]*i%mode;
    }
}
ll pow_mul(ll x,ll y)
{
    ll ret=1;
    while(y)
    {
        if (y & 1 ) the right = K * x% fashion;
        x=x*x%mode,y>>=1;
    }
    Return the right;
}
void NTT(ll *a,int len,int k)
{
    for(int i=0;i<len;i++)if(i<to[i])swap(a[i],a[to[i]]);
    for(int i=1;i<len;i<<=1)
    {
        ll w0=pow_mul(3,(mode-1)/(i<<1));
        for(int j=0;j<len;j+=(i<<1))
        {
            ll w=1;
            for(int o=0;o<i;o++,w=w*w0%mode)
            {
                ll w1=a[j+o],w2=a[j+o+i]*w;
                a[j+o]=(w1+w2)%mode,a[j+o+i]=((w1-w2)%mode+mode)%mode;
            }
        }
    }
    if(k==-1)
    {
        ll inv=pow_mul(len,mode-2);
        for(int i=1;i<(len>>1);i++)swap(a[i],a[len-i]);
        for(int i=0;i<len;i++)a[i]=a[i]*inv%mode;
    }
}
ll A[(1<<20)+5],B[(1<<20)+5],C[(1<<20)+5];
void get_inv(ll *f,ll *g,int dep)
{
    if(dep==1)
    {
        g[0]=pow_mul(f[0],mode-2);
        return;
    }
    int nxt=(dep+1)>>1;
    get_inv(f,g,nxt);
    int lim=1,l=0;
    while(lim<=2*dep)lim<<=1,l++;
    for(int i=0;i<lim;i++)to[i]=((to[i>>1]>>1)|((i&1)<<(l-1)));
    for(int i=0;i<lim;i++)A[i]=B[i]=0;
    for(int i=0;i<dep;i++)A[i]=f[i];
    for(int i=0;i<nxt;i++)B[i]=g[i];
    NTT(A,lim,1),NTT(B,lim,1);
    for(int i=0;i<lim;i++)C[i]=A[i]*B[i]%mode*B[i]%mode;
    NTT(C,lim,-1);
    for(int i=0;i<dep;i++)g[i]=((2*g[i]-C[i])%mode+mode)%mode;
}
int main ()
{
    scanf("%d",&n);
    init();
    n++;
    for(int i=1;i<n;i++)F[i]=(-2ll*minv[i]%mode+mode)%mode;
    F[0]=1;
    get_inv(F,FF,n);
    ll s=0;
    for(int i=0;i<n;i++)s=(s+FF[i]*mul[i]%mode)%mode;
    printf("%lld\n",s);
    return 0;
}