bzoj 4555 求和

题目大意：

求$sumlimits_{i=0}^n sumlimits_{j=0}^i S2(i,j) imes 2^j imes j!$

思路：

法1：

首先把斯特林数展开$S2(i,j)=frac{1}{j!} sumlimits_{k=0}^j (-1)^k * inom{j}{k} * (j-k)^i$

然后先交换求和顺序提出只与$j$有关的项，同时把$j$的范围可以扩展到$n$

代入得到：$sumlimits_{j=0}^n 2^j * j! sumlimits_{i=0}^n sumlimits_{k=0}^j frac{(-1)^k}{j!} * inom{j}{k} * (j-k)^i$ 

拆开组合数：$sumlimits_{j=0}^n 2^j * j! sumlimits_{i=0}^n sumlimits_{k=0}^j frac{(-1)^k}{k!} * frac{(j-k)^i}{(j-k)!}$ 

把只于$k$有关的提出来：$sumlimits_{j=0}^n 2^j * j! sumlimits_{k=0}^j frac{(-1)^k}{k!} * frac{sumlimits_{i=0}^n (j-k)^i}{(j-k)!}$ 

设$f(i)=frac {(-1)^i}{i!},g(i)=frac{sumlimits_{j=0}^n i^j}{i!}$

则原式为$sumlimits_{i=0}^n 2^i imes (i!) imes (f*g)(i)$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
#define db double
#define inf 2139062143
#define MAXN 200100
#define MOD 998244353
#define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
#define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
#define ren for(register int i=fst[x];i;i=nxt[i])
#define pb(i,x) vec[i].push_back(x)
#define pls(a,b) ((a+b)%MOD+MOD)%MOD
#define mns(a,b) ((a%MOD-(b)%MOD)%MOD+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,a[MAXN<<1],b[MAXN<<1],fac[MAXN],ifac[MAXN];
int rev[MAXN<<1],lim,lg,pw[30],ipw[30],ans;
int q_pow(int bas,int t,int res=1)
{
    for(;t;t>>=1,bas=mul(bas,bas)) if(t&1) res=mul(res,bas);return res;
}
#define inv(x) q_pow(x,MOD-2)
void ntt(int *a,int n,int f)
{
    rep(i,0,n-1) if(i<rev[i]) swap(a[i],a[rev[i]]);
    for(int i=1,t=1;i<n;i<<=1,++t)
    {
        int wn= f>0?pw[t]:ipw[t];for(int j=0;j<n;j+=i<<1)
        {
            int w=1,x,y;for(int k=0;k<i;++k,w=mul(w,wn))
                x=a[j+k],y=mul(a[j+k+i],w),a[j+k]=pls(x,y),a[j+k+i]=mns(x,y);
        }
    }
    if(f>0) return ;int nv=inv(n);rep(i,0,n-1) a[i]=mul(a[i],nv);
}
int main()
{
    n=read();fac[0]=ifac[0]=1;rep(i,1,n) fac[i]=mul(fac[i-1],i),ifac[i]=inv(fac[i]);
    int tmp=-1;rep(i,0,n) tmp=(MOD-tmp)%MOD,a[i]=mul(tmp,ifac[i]);
    b[0]=1,b[1]=n+1;rep(i,2,n) b[i]=mul(ifac[i],mul(inv(i-1),(q_pow(i,n+1)-1)));
    for(lim=1,lg=1;lim<=(n+1<<1);lim<<=1,lg++)
        pw[lg]=q_pow(3,(MOD-1)/(1<<lg)),ipw[lg]=inv(pw[lg]);
    rep(i,1,lim-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<lg-2);
    ntt(a,lim,1);ntt(b,lim,1);rep(i,0,lim-1) a[i]=mul(a[i],b[i]);ntt(a,lim,-1);
    tmp=1;rep(i,0,n) ans=pls(ans,mul(mul(a[i],tmp),fac[i])),tmp=pls(tmp,tmp);
    printf("%d
",ans);
}

法2：

思考式子的组合意义，考虑$f(i)=sumlimits_{j=0}^i S2(i,j) imes 2^j imes j!$

即表示将$i$个数分成$j$个有序集合且每个集合有两种状态的所有方案

那么我们可以枚举最后一个集合中数的数量即$f(n)=sumlimits_{i=1}^n 2 * inom{n}{i}* f(n-i)$

两边同除以阶乘，然后就可以分治或者求逆了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define ll long long
#define db double
#define inf 2139062143
#define MAXN 200100
#define MOD 998244353
#define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
#define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
#define ren for(register int i=fst[x];i;i=nxt[i])
#define pb(i,x) vec[i].push_back(x)
#define pls(a,b) ((a+b)%MOD+MOD)%MOD
#define mns(a,b) ((a%MOD-(b)%MOD)%MOD+MOD)%MOD
#define mul(a,b) (1LL*(a)*(b))%MOD
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,a[MAXN<<1],b[MAXN<<1],fac[MAXN],ifac[MAXN],iv[MAXN<<1],tmp[MAXN<<1];
int rev[MAXN<<1],lim,lg,pw[30],ipw[30],ans,l2[MAXN<<2];
int q_pow(int bas,int t,int res=1)
{
    for(;t;t>>=1,bas=mul(bas,bas)) if(t&1) res=mul(res,bas);return res;
}
#define inv(x) q_pow(x,MOD-2)
void ntt(int *a,int n,int f)
{
    rep(i,0,n-1) if(i<rev[i]) swap(a[i],a[rev[i]]);
    for(int i=1,t=1;i<n;i<<=1,++t)
    {
        int wn= f>0?pw[t]:ipw[t];for(int j=0;j<n;j+=i<<1)
        {
            int w=1,x,y;for(int k=0;k<i;++k,w=mul(w,wn))
                x=a[j+k],y=mul(a[j+k+i],w),a[j+k]=pls(x,y),a[j+k+i]=mns(x,y);
        }
    }
    if(f>0) return ;int nv=inv(n);rep(i,0,n-1) a[i]=mul(a[i],nv);
}
void solve(int *a,int *b,int lmt)
{
    rep(i,1,lmt-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<l2[lmt]-1);
    ntt(a,lmt,1);ntt(b,lmt,1);
    rep(i,0,lmt-1) a[i]=mul(mns(2,mul(a[i],b[i])),a[i]);ntt(a,lmt,-1);
}
void Inv(int *a,int *f,int lmt)
{
    f[0]=inv(a[0]);for(int t=2;t<=lmt;t<<=1)
        {rep(i,0,t-1)tmp[i]=a[i];solve(f,tmp,t<<1);rep(i,t,(t<<1)-1) f[i]=0;}
}
int main()
{
    n=read();fac[0]=ifac[0]=1;rep(i,1,n) fac[i]=mul(fac[i-1],i),ifac[i]=inv(fac[i]);
    for(lim=1,lg=1;lim<=(n+1)<<1;lim<<=1,lg++)
        l2[1<<lg]=lg,pw[lg]=q_pow(3,(MOD-1)/(1<<lg)),ipw[lg]=inv(pw[lg]);
    lim>>=1;rep(i,1,n) b[i]=mns(MOD,mul(2,ifac[i]));b[0]=pls(b[0],1);
    Inv(b,a,lim);rep(i,0,n) ans=pls(ans,mul(a[i],fac[i]));printf("%d
",ans);
}