【LOJ #3120】「CTS2019 | CTSC2019」珍珠（生成函数+NTT）

传送门

奇数的$EGF$为$\frac{e^x-e^{-x}}{2}$，偶数为$\frac{e^x+e^{-x}}{2}$

考虑枚举奇数的个数 $$ans=n!\sum _{i=0}^{n-2m}(\frac{e^x-e^{-x}}{2}y+\frac{e^x+e^{-x}}{2}{)}^D[x^n][y^i]$$ $$=\frac{n!}{2^D}\sum _{i=0}^{n-2m}((1+y)e^x+(1-y)e^{-x}{)}^D[x^n][y^i]$$ $$=\frac{n!}{2^D}\sum _{i=0}^{n-2m}\sum _{j=0}^D\left(\genfrac{}{}{0ex}{}{D}{j}\right)e^{(2j-D)x}(1+y{)}^i(1-y{)}^{D-i}$$

$$=\frac{n!}{2^D}\sum _{j=0}^D\left(\genfrac{}{}{0ex}{}{D}{j}\right)(2j-D{)}^n\sum _{i=0}^{n-2m}(1+y{)}^D(1-y{)}^{D-i}[y^i]$$

$$=\frac{n!}{2^D}\sum _{j=0}^D\left(\genfrac{}{}{0ex}{}{D}{j}\right)(2j-D{)}^n(1+y{)}^D(1-y{)}^{D-j}(1+y+y^2...)[y^{n-2m}]$$

$$=\frac{n!}{2^D}\sum _{j=0}^D\left(\genfrac{}{}{0ex}{}{D}{j}\right)(2j-D{)}^n(1+y{)}^D(1-y{)}^{D-j}\frac{1}{1-y}[y^{n-2m}]$$

对于$j=D$可以单独算 否则暴力二项式展开后是一个卷积的形式 直接$NTT$算即可

#include<bits/stdc++.h> using namespace std; #define re register #define ll long long #define pb push_back #define cs const #define bg begin #define pii pair<int,int> #define fi first #define se second #define poly vector<int> cs int RLEN=1<<20|1; inline char gc(){ static char ibuf[RLEN],*ib,*ob; (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin)); return (ib==ob)?EOF:*ib++; } inline int read(){ char ch=gc(); int res=0,f=1; while(!isdigit(ch))f^=ch=='-',ch=gc(); while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc(); return f?res:-res; } inline ll readl(){ char ch=gc(); ll res=0;bool f=1; while(!isdigit(ch))f^=ch=='-',ch=gc(); while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc(); return f?res:-res; } template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;} template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;} cs int mod=998244353,G=3; inline int add(int a,int b){a+=b-mod;return a+(a>>31&mod);} inline void Add(int &a,int b){a+=b-mod,a+=a>>31&mod;} inline int dec(int a,int b){a-=b;return a+(a>>31&mod);} inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;} inline int mul(int a,int b){static ll r;r=1ll*a*b;return r>=mod?r%mod:r;} inline void Mul(int &a,int b){static ll r;r=1ll*a*b,a=r>=mod?r%mod:r;} inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;} inline int Inv(int x){return ksm(x,mod-2);} cs int N=100005; poly w[19]; int rev[(1<<18)|5]; inline void init_w(){ cs int C=18; for(int i=1;i<=C;i++)w[i].resize(1<<(i-1)); int wn=ksm(G,(mod-1)/(1<<C)); w[C][0]=1; for(int i=1;i<(1<<(C-1));i++)w[C][i]=mul(w[C][i-1],wn); for(int i=C-1;i;i--) for(int j=0;j<(1<<(i-1));j++)w[i][j]=w[i+1][j<<1]; } inline void init_rev(int lim){ for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1)); } inline void ntt(poly &f,int lim,int kd){ for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]); for(int mid=1,l=1,a0,a1;mid<lim;mid<<=1,l++) for(int i=0;i<lim;i+=(mid<<1)) for(int j=0;j<mid;j++) a0=f[i+j],a1=mul(f[i+j+mid],w[l][j]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1); if(kd==-1){ reverse(f.bg()+1,f.bg()+lim); for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv); } } inline poly operator *(poly a,poly b){ int deg=a.size()+b.size()-1,lim=1; while(lim<deg)lim<<=1; init_rev(lim); a.resize(lim),ntt(a,lim,1); b.resize(lim),ntt(b,lim,1); for(int i=0;i<lim;i++)Mul(a[i],b[i]); ntt(a,lim,-1),a.resize(deg);return a; } int fac[N],ifac[N]; inline void init_inv(){ fac[0]=ifac[0]=1; for(int i=1;i<N;i++)fac[i]=mul(fac[i-1],i); ifac[N-1]=Inv(fac[N-1]); for(int i=N-2;i;i--)ifac[i]=mul(ifac[i+1],i+1); } inline int C(int n,int m){ return n<m?0:mul(fac[n],mul(ifac[m],ifac[n-m])); } int d,n,m; int main(){ #ifdef Stargazer freopen("lx.cpp","r",stdin); #endif init_w(),init_inv(); d=read(),n=read(),m=read(); int lim=n-2*m; poly f(d+1,0),g(d+1,0); for(int i=0;i<=d;i++){ if(i<=lim&&lim-i<=d){ f[i]=mul(ifac[i],ifac[lim-i]); if((lim-i)&1)f[i]=dec(0,f[i]); } if(d>=1+lim+i)g[i]=mul(ifac[i],ifac[d-1-lim-i]); } f=f*g; int res=0; for(int i=0;i<d;i++)Add(res,mul(mul(C(d,i),ksm(((2*i-d)%mod+mod)%mod,n)),mul(mul(fac[i],fac[d-1-i]),f[i]))); for(int i=0;i<=min(lim,d);i++)Add(res,mul(C(d,i),ksm(d,n))); cout<<mul(Inv(ksm(2,d)),res); }