CTS 2019 Pearl

CTS2019 Pearl

每种颜色有奇偶两种情况，只有偶数和只有奇数个，其对应的的指数型生成函数分别为 \(\frac{e^x+e^{-x}}2,\frac{e^x-e^{-x}}2\)

考虑选出现奇数次的颜色受一定限制，所以本质上就是要求这个东西

\[\frac{n!}{2^D}\sum_{i=0}^{n-2m} [(e^x+e^{-x})+y(e^x-e^{-x})]^D[x^ny^i] \\
=\frac{n!}{2^D}\sum_{i=0}^{n-2m} [(1+y)e^x+(1-y)e^{-x}]^D[x^ny^i] \\
=\frac{n!}{2^D}\sum_{i=0}^{n-2m}\sum_{k=0}^D \binom{D}{k}(1+y)^k(1-y)^{D-k}e^{(2k-D)x} [x^ny^i] \\
=\frac1{2^D}\sum_{i=0}^{n-2m}\sum_{k=0}^D (2k-D)^D\binom{D}{k}(1+y)^k(1-y)^{D-k} [y^i] \\
=\frac1{2^D}\sum_{k=0}^D (2k-D)^D\binom{D}{k} \sum_{i=0}^{n-2m} (1+y)^k(1-y)^{D-k} [y^i] \\
------------------------- \\
F(k,D)=\sum_{i=0}^{n-2m} (1+y)^k(1-y)^{D-k} [y^i] \\
=\sum_{i=0}^{n-2m} (2-(1-y))(1+y)^{k-1}(1-y)^{D-k} [y^i] \\
=\sum_{i=0}^{n-2m} 2(1+y)^{k-1}(1-y)^{D-k} [y^i] -\sum_{i=0}^{n-2m}(1+y)^{k-1}(1-y)^{D-k+1} [y^i] \\
=2F(k-1,D-1)-F(k-1,D)\\
------------------------- \\
G_i=F(0,i)\ = \sum_{j=0}^{n-2m} (1-y)^i[y^j]\ = \sum_{j=0}^{n-2m} (-1)^j\binom{i}{j}\\
= \sum_{j=0}^{n-2m} (-1)^j(\binom{i-1}{j-1}+\binom{i-1}{j})\\
= \sum_{j=0}^{n-2m} (-1)^j\binom{i-1}{j}-\sum_{j=0}^{n-2m-1}(-1)^j\binom{i-1}{j}\\
= (-1)^{n-2m}\binom{i-1}{n-2m} \\
------------------------- \\
F(k,D)=\sum_{i=0}^D F(0,i)2^{D-i}(-1)^{k-(D-i)}\binom{k}{D-i} \\
=k!\sum_{i=0}^D \frac{G_i(-1)^{k-(D-i)}2^{D-i}}{(D-i)!(k-(D-i))!} \\
=k!\sum_{i=0}^D \frac{G_{D-i}(-1)^{k-i}2^i}{i!(k-i)!} \\
------------------------- \\
T_i=\frac{G_{D-i}2^i}{i!} \ ,\ H_i=\frac{(-1)^i}{i!} \ ,\ F_i=(H*T)_i\times i!\\
Ans=\frac1{2^D}\sum_{k=0}^D (2k-D)^D\binom{D}{k}F_k \\
\]

一次卷积即可，注意要定义 \(G_0=1\)

式子没有仔细检查，有错见谅qwq

#include<bits/stdc++.h>
#define LL long long
using namespace std;
int read(){
	int nm=0,fh=1; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
#define pii pair<int,int>
#define mp make_pair
#define mod 998244353
#define M 600010
#define inv3 332748118
#define inv2 499122177
namespace CALC{
	inline int add(int x,int y){x+=y;return (x>=mod)?(x-mod):x;}
	inline int mns(int x,int y){x-=y;return (x<0)?(x+mod):x;}
	inline int mul(LL x,LL y){return (LL)x*(LL)y%mod;}
	inline void upd(int &x,int y){x+=y;(x>=mod)?(x-=mod):0;}
	inline void dec(int &x,int y){x-=y;(x<0)?(x+=mod):0;}
	inline int qpow(int x,int sq){
		int res=1;
		for(;sq;sq>>=1,x=mul(x,x)) if(sq&1) res=mul(res,x);
		return res;
	}
}using namespace CALC;
int n,m,D,od[M],fac[M+2],ifac[M+2],G[M],H[M],T[M],F[M];
inline int C(int tot,int tk){return (tk<0||tot<tk)?0:mul(fac[tot],mul(ifac[tk],ifac[tot-tk]));}
inline void NTT(int *x,int len,int kd){
	for(int i=1;i<len;i++) if(i<od[i]) swap(x[i],x[od[i]]);
	for(int tt=1;tt<len;tt<<=1){
		int wn=qpow(kd>0?3:inv3,(mod-1)/(tt<<1));
		for(int st=0,now=1;st<len;st+=(tt<<1),now=1){
			for(int pos=st;pos<st+tt;pos++,now=mul(now,wn)){
				int t1=x[pos],t2=mul(now,x[pos+tt]);
				x[pos]=add(t1,t2),x[pos+tt]=mns(t1,t2);
			}
		}
	} if(kd>0) return; int rv=qpow(len,mod-2);
	for(int i=0;i<len;i++) x[i]=mul(x[i],rv);
}
int main(){
	D=read(),n=read(),m=read();
	if(n-m-m>=D){printf("%d\n",qpow(D,n)); return 0;}
	if(n<m+m){puts("0"); return 0;} fac[0]=1;
	for(int i=1;i<=M;i++) fac[i]=mul(fac[i-1],i); ifac[M]=qpow(fac[M],mod-2);
	for(int i=M;i>=1;i--) ifac[i-1]=mul(ifac[i],i); int len=1,nw=-1,ans=0;
	while(len<=D+D) len<<=1,nw++;
	for(int i=1;i<len;i++) od[i]=(od[i>>1]>>1)|((i&1)<<nw);
	for(int i=1;i<=D;i++) G[i]=C(i-1,n-m-m); G[0]=1;
	if(n&1){for(int i=1;i<=D;i++) G[i]=mns(0,G[i]);}
	for(int i=0;i<=D;i++) H[i]=(i&1)?mns(0,ifac[i]):ifac[i];
	for(int i=0;i<=D;i++) T[i]=mul(G[D-i],mul(ifac[i],qpow(2,i)));
	NTT(T,len,1),NTT(H,len,1);
	for(int i=0;i<len;i++) F[i]=mul(T[i],H[i]); NTT(F,len,-1);
	for(int i=0;i<=D;i++) F[i]=mul(F[i],fac[i]);
	for(int i=0;i<=D;i++) upd(ans,mul(qpow(mns(i+i,D),n),mul(C(D,i),F[i])));
	printf("%d\n",mul(qpow(inv2,D),ans)); return 0;
}