思路
神仙题.jpg
脑子一抽，想把\(f(x)\)表示成下降幂的形式，也就是
\[ f(x)=\sum_{i=0}^m f_ix_{(i)}\\ x_{(i)}=\prod_{k=0}^{i-1}(x-k)=[x\ge i]\frac{x!}{(x-i)!} \]
这样有什么好处呢？回到原来的式子，我们有
\[ \begin{align*} &\sum_{k=0}^n f(k){n\choose k}x^k(1-x)^{n-k}\\ =&\sum_{k=0}^n \sum_{i=0}^m f_ik_{(i)}{n\choose k}x^k(1-x)^{n-k}\\ =&\sum_{i=0}^mf_i \sum_{k=i}^n \frac{n!}{(n-k)!(k-i)!} x^k(1-x)^{n-k}\\ =&\sum_{i=0}^m f_i n_{(i)}\sum_{k=i}^n {n-i\choose k-i} x^{k}(1-x)^{n-k}\\ =&\sum_{i=0}^m f_in_{(i)}x^i \end{align*} \]
可以非常快地求出来。
还有一个问题：\(f_i\)怎么求？
再一次脑子一抽想到这样一个式子：
\[ \begin{align*} &\sum_n f(n)\frac{1}{n!}x^n\\ =&\sum_{i=0}^m f_i\sum_{n\ge i} x^n\frac{1}{(n-i)!}\\ =&(\sum_{i=0}^m f_ix^i)e^x \end{align*} \]
于是\(f(x)\)在\([0,m]\)上的取值卷上一个\(e^x\)就可以得到\(f_i\)，用NTT优化到\(m\log m\)。
脑洞大开，神仙题.jpg
代码
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 100101
#define mod 998244353ll
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char __sr[1<<21],__z[20];int __C=-1,__zz=0;
inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
inline void print(register int x)
{
if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
while(__z[++__zz]=x%10+48,x/=10);
while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
}
void file()
{
#ifdef NTFOrz
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int r[sz],limit;
void NTT_init(int n)
{
limit=1;int l=-1;
while (limit<=n+n) limit<<=1,++l;
rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);
}
void NTT(ll *a,int type)
{
rep(i,0,limit-1) if (i<r[i]) swap(a[i],a[r[i]]);
for (int mid=1;mid<limit;mid<<=1)
{
ll Wn=ksm(3,(mod-1)/mid>>1);if (type==-1) Wn=inv(Wn);
for (int len=mid<<1,j=0;j<limit;j+=len)
{
ll w=1;
for (int k=0;k<mid;k++,w=w*Wn%mod)
{
ll x=a[j+k],y=a[j+k+mid]*w%mod;
a[j+k]=(x+y)%mod;a[j+k+mid]=(x-y+mod)%mod;
}
}
}
if (type==1) return;
ll I=inv(limit);
rep(i,0,limit-1) a[i]=a[i]*I%mod;
}
ll n,X;
int m;
ll tmp1[sz],tmp2[sz],f[sz];
ll _fac[sz];
void init(){_fac[0]=1;rep(i,1,sz-1) _fac[i]=_fac[i-1]*inv(i)%mod;}
int main()
{
file();
init();
read(n,m,X);
rep(i,0,m) read(tmp1[i]),tmp1[i]=tmp1[i]*_fac[i]%mod;
rep(i,0,m) tmp2[i]=((i&1)?mod-1:1ll)*_fac[i]%mod;
NTT_init(m);
NTT(tmp1,1);NTT(tmp2,1);
rep(i,0,limit-1) f[i]=tmp1[i]*tmp2[i]%mod;
NTT(f,-1);
ll cur=1,ans=0;
rep(i,0,m)
{
(ans+=f[i]*cur%mod)%=mod;
cur=cur*(mod+n-i)%mod*X%mod;
}
cout<<ans;
return 0;
}