loj2058 「TJOI / HEOI2016」求和 NTT
链接
思路
[S(i,j)=frac{1}{j!}sumlimits_{k=0}^{j}(-1)^{k}C_{j}^{k}(j-k)^{i}
]
[sumlimits_{i=0}^{n}sumlimits_{j=0}^{i}S(i,j)·2^j·j!
]
[sumlimits_{i=0}^{n}sumlimits_{j=0}^{n}S(i,j)·2^j·j!
]
[sumlimits_{j=0}^{n}2^j·j!sumlimits_{i=0}^{n}S(i,j)
]
先看后边
[sumlimits_{i=0}^{n}frac{1}{j!}sumlimits_{k=0}^{j}(-1)^{k}C_{j}^{k}(j-k)^{i}
]
[frac{1}{j!}sumlimits_{k=0}^{j}(-1)^{k}C_{j}^{k}sumlimits_{i=0}^{n}(j-k)^{i}
]
[sumlimits_{k=0}^{j}(-1)^{k}frac{1}{k!(j-k)!}sumlimits_{i=0}^{n}(j-k)^{i}
]
(f(j-k)=sumlimits_{i=0}^{n}(j-k)^{i})等比数列求和。
[sumlimits_{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。
[sumlimits_{j=0}^{n}2^j·j!sumlimits_{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
```cpp
#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
",ans);
return 0;
}
```]