题解 P4091 【[HEOI2016/TJOI2016]求和】
y2823774827y
2019-04-15 13:47:25
## 题目
[[HEOI2016/TJOI2016]求和](https://www.luogu.org/problemnew/show/P4091)
关于斯特林数与反演的更多姿势$\Longrightarrow$[点这里](https://www.cnblogs.com/y2823774827y/p/10700231.html)
## 做法
$$
\begin{aligned}
Ans&=\sum_{i=0}^n \sum_{j=0}^i \begin{Bmatrix}i\\j\end{Bmatrix}2^j×j!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1\\
&=\sum_{i=0}^n \sum_{j=0}^n \begin{Bmatrix}i\\j\end{Bmatrix}2^j×j!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2\\
&=\sum_{j=0}^n 2^j×j!\sum_{i=0}^n \begin{Bmatrix}i\\j\end{Bmatrix}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~3\\
&=\sum_{j=0}^n 2^j×j!\sum_{i=0}^n \sum_{k=0}^j\frac{(-1)^k}{k!}\cdot\frac{(j-k)^i}{(j-k)!}~~~~~~~~~~~~~~~4\\
&=\sum_{j=0}^n 2^j×j!\sum_{k=0}^j\frac{(-1)^k}{k!}\cdot\frac{ \sum_{i=0}^n (j-k)^i}{(j-k)!}~~~~~~~~~~~~~5\\
&=\sum_{j=0}^n 2^j×j!\sum_{k=0}^j\frac{(-1)^k}{k!}\cdot \frac{(j-k)^{n+1}-1}{(j-k-1)(j-k)!}~~~~6\\
\end{aligned}
$$
- $2$：$\displaystyle \begin{Bmatrix}i\\j\end{Bmatrix}=0(i>j)$
- $3$：移项
- $4$：[第二类斯特林的性质](https://www.cnblogs.com/y2823774827y/p/10700231.html)
- $5$：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
```cpp
#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;
}
```