loj2058 「TJOI / HEOI2016」求和 NTT

链接

loj

思路

\[S(i,j)=\frac{1}{j!}\sum\limits_{k=0}^{j}(-1)^{k}C_{j}^{k}(j-k)^{i}\]
\[\sum\limits_{i=0}^{n}\sum\limits_{j=0}^{i}S(i,j)·2^j·j!\]
\[\sum\limits_{i=0}^{n}\sum\limits_{j=0}^{n}S(i,j)·2^j·j!\]
\[\sum\limits_{j=0}^{n}2^j·j!\sum\limits_{i=0}^{n}S(i,j)\]
先看后边
\[\sum\limits_{i=0}^{n}\frac{1}{j!}\sum\limits_{k=0}^{j}(-1)^{k}C_{j}^{k}(j-k)^{i}\]
\[\frac{1}{j!}\sum\limits_{k=0}^{j}(-1)^{k}C_{j}^{k}\sum\limits_{i=0}^{n}(j-k)^{i}\]
\[\sum\limits_{k=0}^{j}(-1)^{k}\frac{1}{k!(j-k)!}\sum\limits_{i=0}^{n}(j-k)^{i}\]
\(f(j-k)=\sum\limits_{i=0}^{n}(j-k)^{i}\)等比数列求和。
\[\sum\limits_{k=0}^{j}\frac{(-1)^{k}}{k!}\frac{f(j-k)}{(j-k)!}\]
nice，这很卷积，用NTT预处理就好了，然后后面的式子就很好求ans了。
$\(\sum\limits_{j=0}^{n}2^j·j!\sum\limits_{k=0}^{j}\frac{(-1)^{k}}{k!}\frac{f(j-k)}{(j-k)!}\)

坑点

f(x)要特判q=1和q=0。因为公式本来就不能做
其他的照的推出来的式子做就可以辣

代码

#include <bits/stdc++.h>
using namespace std;
const int N=4e5+7,mod=998244353;
int n,ans,jc[N],a[N],b[N],limit=1,p,r[N];
int q_pow(int a,int b) {int ans=1;while(b) {if(b&1) ans=1LL*a*ans%mod;a=1LL*a*a%mod;b>>=1;}return ans;
}
int inv(int a) {return q_pow(a,mod-2);}
int ntt(int *a,int type) {for(int i=0;i<limit;++i)if(i<r[i]) swap(a[i],a[r[i]]);for(int mid=1;mid<limit;mid<<=1) {int Wn=q_pow(3,(mod-1)/(mid<<1));for(int i=0;i<limit;i+=(mid<<1)) {for(int j=0,w=1;j<mid;j++,w=1LL*w*Wn%mod) {int x=a[i+j],y=1LL*w*a[i+j+mid]%mod;a[i+j]=(x+y)%mod;a[i+j+mid]=(x-y+mod)%mod;}}}if(type==-1) {reverse(&a[1],&a[limit]);int inv=q_pow(limit,mod-2);for(int i=0;i<limit;++i) a[i]=1LL*a[i]*inv%mod;}
}
int f(int x) {if(!x) return 1;if(x==1) return n+1;return 1LL*(q_pow(x,n+1)-1)*inv(x-1)%mod;
}
int main() {scanf("%d",&n);jc[0]=1;for(int i=1;i<=n;++i) jc[i]=1LL*jc[i-1]*i%mod;for(int i=0;i<=n;++i) a[i]=1LL*(i&1?(mod-1):1)*inv(jc[i])%mod;for(int i=0;i<=n;++i) b[i]=1LL*f(i)*inv(jc[i])%mod;while(limit<n+n) limit<<=1,p++;for(int i=0;i<limit;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(p-1));ntt(a,1),ntt(b,1);for(int i=0;i<limit;++i) a[i]=1LL*a[i]*b[i]%mod;ntt(a,-1);for(int i=0;i<=n;++i) ans=(ans+1LL*q_pow(2,i)*jc[i]%mod*a[i]%mod)%mod;printf("%d\n",ans);return 0;
}