[HEOI2016/TJOI2016]求和(第二类斯特林数)

题目

[HEOI2016/TJOI2016]求和

关于斯特林数与反演的更多姿势(Longrightarrow)点这里

做法

[egin{aligned}\ Ans&=sumlimits_{i=0}^n sumlimits_{j=0}^i egin{Bmatrix}i\jend{Bmatrix}2^j×j!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1\ &=sumlimits_{i=0}^n sumlimits_{j=0}^n egin{Bmatrix}i\jend{Bmatrix}2^j×j!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2\ &=sumlimits_{j=0}^n 2^j×j!sumlimits_{i=0}^n egin{Bmatrix}i\jend{Bmatrix}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~3\ &=sumlimits_{j=0}^n 2^j×j!sumlimits_{i=0}^n sumlimits_{k=0}^jfrac{(-1)^k}{k!}cdotfrac{(j-k)^i}{(j-k)!}~~~~~~~~~~~~~~~~~~~~4\ &=sumlimits_{j=0}^n 2^j×j!sumlimits_{k=0}^jfrac{(-1)^k}{k!}cdotfrac{ sumlimits_{i=0}^n (j-k)^i}{(j-k)!}~~~~~~~~~~~~~~~~~~~~~~5\ &=sumlimits_{j=0}^n 2^j×j!sumlimits_{k=0}^jfrac{(-1)^k}{k!}cdot frac{(j-k)^{n+1}-1}{(j-k-1)(j-k)!}~~~~~~~6\ end{aligned}]

• (2：egin{Bmatrix}i\jend{Bmatrix}=0(i>j))
• (3：)移项
• (4：)第二类斯特林的性质
• (5：)移项化卷积
• (6：)等比公式

总结

这题非常有意思，最后一步对于蒟蒻来说还是少见的，推到(4)谁都会，然后就无从下手了
以至于会从头考虑(2^j×j!)的性质，特别容易想偏

Code

#include<bits/stdc++.h>
typedef int LL;
typedef long long L;
const LL maxn=3e5+9,mod=998244353,g=3,_g=332748118;
inline LL Pow(LL base,LL b){
    LL ret(1);
    while(b){
        if(b&1) ret=(L)ret*base%mod; base=(L)base*base%mod; b>>=1;
    }return ret;
}
LL fac[maxn],fav[maxn],r[maxn],F[maxn],G[maxn],W[maxn];
inline void NTT(LL *a,LL n,LL type){
    for(LL i=0;i<n;++i) if(i<r[i]) std::swap(a[i],a[r[i]]);
    for(LL mid=1;mid<n;mid<<=1){
        LL wn(Pow(type?g:_g,(mod-1)/(mid<<1)));
        W[0]=1; for(LL i=1;i<mid;++i) W[i]=(L)W[i-1]*wn%mod;
        for(LL R=mid<<1,j=0;j<n;j+=R)
            for(LL k=0;k<mid;++k){
            	LL x(a[j+k]),y((L)W[k]*a[j+mid+k]%mod);
            	a[j+k]=x+y; if(a[j+k]>=mod) a[j+k]%=mod;
				a[j+mid+k]=x-y; if(a[j+mid+k]<0) a[j+mid+k]+=mod;
            }
    }
}
inline LL Fir(LL n){
    LL limit(1),len(0);
    while(limit<n){
        limit<<=1; ++len;
    }
    for(LL i=0;i<limit;++i) r[i]=(r[i>>1]>>1)|((i&1)<<len-1);
    return limit;
}
inline LL Solve_fg(LL n){
    for(LL i=0;i<=n;++i) F[i]=(L)(i&1?mod-1:1)*fav[i]%mod;
    for(LL i=0;i<=n;++i) G[i]=(L)(Pow(i,n+1)+mod-1)%mod*Pow(i-1<0?i+mod-1:i-1,mod-2)%mod*fav[i]%mod;
    G[1]=n+1;
    LL limit(Fir(n+1<<1));
    NTT(F,limit,1); NTT(G,limit,1);
    for(LL i=0;i<limit;++i) F[i]=(L)F[i]*G[i]%mod;
    NTT(F,limit,0);
    LL ty(Pow(limit,mod-2)); for(LL i=0;i<=n;++i) F[i]=(L)F[i]*ty%mod;
    
    LL ret(0);
    for(LL i=0;i<=n;++i) ret=(L)(ret+(L)Pow(2,i)*fac[i]%mod*F[i]%mod)%mod;
    return ret;
}
LL n;
int main(){
    scanf("%d",&n);
    fac[0]=fac[1]=1;
    for(LL i=2;i<=n;++i) fac[i]=(L)fac[i-1]*i%mod;
    fav[n]=Pow(fac[n],mod-2);
    for(LL i=n;i>=1;--i) fav[i-1]=(L)fav[i]*i%mod;
    
    printf("%d",Solve_fg(n));
    return 0;
}