CTS2019 珍珠

link

$$
\begin{align}
ans&=\frac{\sum_{i=0}^k \binom{d}{i}(e^x+e^{-x})^{d-i}(e^x-e^{-x})^i}{2^d}[x^n]\\
&=\frac{e^{-dx}\sum_{i=0}^k \binom{d}{i}(e^{2x}+1)^{d-i}(e^{2x}-1)^i}{2^d}[x^n]
\end{align}\\
y=e^{2x}\\
F(y)=\sum_{i=0}^k\binom{d}{i}(y+1)^{d-i}(y-1)^i\\
ans=\sum_{i}(2i-d)^nF(y)[y^i]\\
F(y)=(y+1)^d\sum_{i=0}^k\binom{d}{i}(\frac{y-1}{y+1})^i\\
G(x)=\sum_{i=0}^k\binom{d}{i}x^i\\
F(y)=(y+1)^dG(1-\frac{2}{y+1})\\
H(x)=G(1-2x)\\
h_i=\sum_{j}\binom{j}{i}\binom{d}{j}=\sum_j\frac{j!d!}{i!(j-i)!j!(d-j)!}=\frac{d!}{i!}\sum_j\frac{1}{(j-i)!(d-j)!}=\frac{d!}{i!}\sum_x\\
F(y)=(y+1)^dH(\frac{1}{y+1})\\
\frac{1}{y+1}=\sum_{i}(-x)^i\\
(\frac{1}{y+1})^k=\sum_{i}\binom{i+k-1}{k-1}(-y)^i\\
$$
$$
\begin{align}
H(\frac{1}{y+1})&=\sum_{j}(-y)^j\sum_i\binom{i+j-1}{i-1}h_i\\
&=\sum_{j}\frac{(-y)^j}{j!}\sum_i(i+j-1)!\frac{h_i}{(i-1)!}\\
\end{align}
$$
然后可以NTT

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/*
Author: QAQ Automaton
Lang: C++
Prog: pearl.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}

inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}

inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int inf;
struct _init_{
_init_(){
memset(&inf,0x3f,sizeof(inf));
}
};
const int p=998244353,yg=3,ig=332748118;
const int N=1<<18,M=N|5;
int fac[200005],inv[200005],invf[200005];
int fpm(int a,int b){
if(a<0)a+=p;
int c=1;
for(;b;b>>=1,a=a*a%p)if(b&1)c=c*a%p;
return c;
}
int rev[M],f[M],g[M],h[M],s[M],t[M];
void NTT(int *a,int n,int flg){
for(int i=1;i<n;++i){
rev[i]=(rev[i>>1]>>1)|((i&1)*(n>>1));
if(i<rev[i])swap(a[i],a[rev[i]]);
}
for(int i=1;i<n;i<<=1){
int w=fpm(flg==1?yg:ig,(p-1)/(i<<1)),ww;
for(int j=0;j<n;j+=i+i){
ww=1;
for(int k=0;k<i;++k){
int u=a[j+k],v=a[j+k+i];
a[j+k]=(u+ww*v)%p;
a[j+k+i]=((u-ww*v)%p+p)%p;
ww=ww*w%p;
}
}
}
if(flg==-1){
int w=fpm(n,p-2)%p;
for(int i=0;i<n;++i)a[i]=a[i]*w%p;
}
}
int C(int a,int b){return fac[a]*invf[b]%p*invf[a-b]%p;}
signed main(){
#ifdef QAQAutoMaton
freopen("pearl.in","r",stdin);
freopen("pearl.out","w",stdout);
#endif
int d,n,m;
read(d,n,m);
if(n-m*2<0)return write("0\n");
int XD=d*2+5;
fac[0]=fac[1]=inv[1]=invf[0]=invf[1]=1;
for(int i=2;i<=XD;++i){
fac[i]=fac[i-1]*i%p;
inv[i]=(p-p/i)*inv[p%i]%p;
invf[i]=inv[i]*invf[i-1]%p;
}
int w=min(n-m*2,d);
for(int i=0;i<=w;++i){
h[d-i]=invf[d-i];
}
for(int i=0;i<=d;++i)g[i]=invf[i];
NTT(h,N,1);
NTT(g,N,1);
for(int i=0;i<N;++i){g[i]=g[i]*h[i]%p;h[i]=0;}
NTT(g,N,-1);
int x=1;
for(int i=0;i<=w;++i){
h[i]=g[d-i]*fac[d]%p*invf[i]%p*x%p;
x=x*(p-2)%p;
}
for(int i=0;i<N;++i)g[i]=0;
int i2=fpm(2,p-2)%p;
for(int i=0;i<=w;++i)s[0]=(s[0]+h[i])%p;
h[0]=0;
for(int i=1;i<=w+1;++i){h[i]=h[i]*invf[i-1]%p;}
for(int i=1;i<=XD;++i)g[XD-i]=fac[i-1];
NTT(h,N,1);
NTT(g,N,1);
for(int i=0;i<N;++i)f[i]=h[i]*g[i]%p;
NTT(f,N,-1);

int ans=0;
t[0]=1;
for(int i=1;i<=d;++i){
s[i]=f[XD-i]%p*invf[i]%p*(i&1?p-1:1)%p;
t[i]=C(d,i);
}
NTT(s,N,1);
NTT(t,N,1);
for(int i=0;i<N;++i)f[i]=s[i]*t[i]%p;
NTT(f,N,-1);
for(int i=0;i<=d;++i){
ans=(ans+fpm(i+i-d,n)*f[i])%p;
}
ans=ans*fpm(i2,d)%p;
write(ans,'\n');
return 0;
}

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