BZOJ 5093[Lydsy1711月赛]图的价值 线性做法

博主曾更过一篇复杂度为$O( k· log k)$的多项式做法在这里

惊闻本题有$ O(k)$的神仙做法，说起神仙我就想起了于是就去学习了一波

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幂与第二类斯特林数

推导看这里

$$ x^k=sum_{j=0}^kj!inom{x}{j}egin{Bmatrix}k\jend{Bmatrix}$$

$$ egin{Bmatrix}k\jend{Bmatrix}=frac{1}{j!}sum_{i=0}^ji^kinom{j}{i}(-1)^{j-i}$$

以上是两个非常实用的公式

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推式子

现在开始推式子

原博文已经推出了我们真正需要求的是$ f(n,k)=sumlimits_{i=0}^ninom{n}{i}i^k$

根据上面的公式可以推得

$$
egin{aligned}
f(n,k) & =sum_{i=0}^ninom{n}{i}i^k=sum_{j=0}egin{Bmatrix}k\jend{Bmatrix}frac{n!}{(n-j)!}2^{n-j}\
& =sum_{j=0}^kfrac{n!}{(n-j)!}2^{n-j}frac{1}{j!}sum_{i=0}^j(-1)^{j-i}inom{j}{i}i^k\
&=sum_{j=0}^kinom{n}{j}2^{n-j}sum_{i=0}^j(-1)^{j-i}inom{j}{i}i^k\
&=sum_{i=0}^kinom{n}{i}i^ksum_{j=i}^k2^{n-j}(-1)^{j-i}inom{n-i}{j-i}\
&=sum_{i=0}^kinom{n}{i}i^k2^{n-i}sum_{j=0}^{k-i}inom{n-i}{j}(-frac{1}{2})^j
end{aligned}
$$

我们需要快速递推出$A(i)=displaystylesum_{j=0}^{k-i}inom{n-i}{j}(-frac{1}{2})^j$

再推波式子得

$$
egin{aligned}
sum_{j=0}^{k-i}inom{n-i}{j}(-frac{1}{2})^j&=sum_{j=0}^{k-i}left(inom{n-i-1}{j}+inom{n-i-1}{j-1} ight)(-frac{1}{2})^j\
&=sum_{j=1}^{k-i}(-frac{1}{2})^jinom{n-i-1}{j-1}+sum_{j=0}^{k-i}(-frac{1}{2})^jinom{n-i-1}{j}\
&=-frac{1}{2}sum_{j=0}^{k-i-1}(-frac{1}{2})^jinom{n-i-1}{j}+sum_{j=0}^{k-i}(-frac{1}{2})^jinom{n-i-1}{j}\
&=frac{1}{2}sum_{j=0}^{k-i-1}(-frac{1}{2})^jinom{n-i-1}{j}+(-frac{1}{2})^{k-i}inom{n-i-1}{k-i}
end{aligned}
$$

因此$A(i)=frac{1}{2}A(i+1)+(-frac{1}{2})^{k-i}inom{n-i-1}{k-i}$

就假装推完了

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大常数代码

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#define p 998244353
#define inv2 499122177
#define rt register int
#define ll long long
using namespace std;
inline ll read(){
    ll x = 0; char zf = 1; char ch = getchar();
    while (ch != '-' && !isdigit(ch)) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}
void write(ll y){if(y<0)putchar('-'),y=-y;if(y>9)write(y/10);putchar(y%10+48);}
void writeln(const ll y){write(y);putchar('
');}
int i,j,k,m,n,x,y,z,cnt;
int ksm(int x,int y=p-2){
    int ans=1;
    for(rt i=y;i;i>>=1,x=1ll*x*x%p)if(i&1)ans=1ll*ans*x%p;
    return ans;
}
int inv[500010],A[500010];
int v[500010],ss[500010];bool b[500010];
int main(){
    n=read()-1;k=read();
    inv[0]=inv[1]=1;
    v[1]=1;v[0]=(k==0);
    for(rt i=2;i<=k;i++){
        if(!b[i])ss[++cnt]=i,v[i]=ksm(i,k);
        for(rt j=1;i*ss[j]<=k&&j<=cnt;j++){
            b[i*ss[j]]=1;v[i*ss[j]]=1ll*v[i]*v[ss[j]]%p;
            if(i%ss[j]==0)break;
        }
    }
    for(rt i=2;i<=k;i++)inv[i]=1ll*inv[p%i]*(p-p/i)%p;
    int ans=0;
    if(n<=k){
        for(rt i=0,c=1;i<=n;c=1ll*(n-i)*inv[i+1]%p,i++)
        ans+=1ll*c*v[i]%p;
        cout<<(1ll*ans*(n+1)%p*ksm(2,(ll)n*(n-1)/2%(p-1))%p+p)%p;
        return 0;
    }
    
    A[k]=1;
    for(rt i=k-1,y=-inv2,c=n-i-1;i>=0;i--,y=1ll*y*-inv2%p){
        A[i]=(1ll*A[i+1]*inv2%p+1ll*c*y%p)%p;
        c=1ll*c*(n-i)%p*inv[k-i+1]%p;
    }

    for(rt i=0,d=ksm(2,n),c=1;i<=k&&i<=n;c=1ll*c*(n-i)%p*inv[i+1]%p,i++,d=1ll*d*inv2%p)
    (ans+=1ll*c*v[i]%p*d%p*A[i]%p)%=p;
    cout<<(1ll*ans*(n+1)%p*ksm(2,(ll)n*(n-1)/2%(p-1))%p+p)%p;
    return 0;
}