[LOJ3120][CTS2019|CTSC2019]珍珠:生成函数+NTT

分析

容易发现\(D \leq n - 2m\)时,任意数列都满足要求,直接判掉,下文所讨论的均为\(D > n - 2m\)的情况。

考虑把两个数列合并,显然可以认为是两个带标号对象的合并,可以使用EGF相乘。

我们可以枚举有\(k\)个数出现了奇数次,答案即为:

\[ \begin{aligned} ans=&n!\sum_{k=0}^{n-2m}(EVEN(x)+yODD(x))^D[x^n][y^k]\\ =&n!\sum_{k=0}^{n-2m}(\frac{e^x+e^{-x}}{2}+y\frac{e^x-e^{-x}}{2})^D[x^n][y^k]\\ =&n!\frac{1}{2^D}\sum_{k=0}^{n-2m}((e^x+e^{-x})+y(e^x-e^{-x}))^D[x^n][y^k]\\ =&n!\frac{1}{2^D}\sum_{k=0}^{n-2m}(e^x(1+y)+e^{-x}(1-y))^D[x^n][y^k]\\ =&n!\frac{1}{2^D}\sum_{k=0}^{n-2m}\sum_{i=0}^{D}\binom{D}{i}e^{(2i-D)x}(1+y)^i(1-y)^{D-i}[x^n][y^k]\\ =&\frac{1}{2^D}\sum_{k=0}^{n-2m}\sum_{i=0}^{D}\binom{D}{i}(2i-D)^n(1+y)^i(1-y)^{D-i}[y^k]\\ =&\frac{1}{2^D}\sum_{i=0}^{D}\binom{D}{i}(2i-D)^n\sum_{k=0}^{n-2m}(1+y)^i(1-y)^{D-i}[y^k]\\ \end{aligned} \]

(以下部分参考了bestfy的博客。)

现在我们需要对\(\forall i \in [0,D]\),求出\(f(i)=\sum_{k=0}^{n-2m}(1+y)^i(1-y)^{D-i}[y^k]\)

\(i < D\)时,

\[ \begin{aligned} f(i)=&\sum_{k=0}^{n-2m}(1+y)^i(1-y)^{D-i}[y^k]\\ =&(1+y)^i(1-y)^{D-i}(1+y+y^2+...)[y^{n-2m}]\\ =&(1+y)^i(1-y)^{D-i-1}[y^{n-2m}]\\ =&\sum_{j=0}^{n-2m}\binom{i}{j}\binom{D-i-1}{n-2m-j}(-1)^{n-2m-j}\\ \end{aligned} \]

上式可以化成卷积的形式然后NTT。

\(i = D\)时,

\[f(D)=\sum_{k=0}^{n-2m}(1+y)^D[y^k]\]

因为\(D>n-2m\),所以这个东西可以直接暴力算。

代码

#include <bits/stdc++.h>

#define rin(i,a,b) for(int i=(a);i<=(b);++i)
#define irin(i,a,b) for(int i=(a);i>=(b);--i)
#define trav(i,a) for(int i=head[a];i;i=e[i].nxt)
#define Size(a) (int)a.size()
#define pb push_back
#define mkpr std::make_pair
#define fi first
#define se second
#define lowbit(a) ((a)&(-(a)))
typedef long long LL;

using std::cerr;
using std::endl;

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int MAXN=100005;
const int MOD=998244353;
const int NTT=1048576;
const int G=3;
const int INVG=332748118;

int D,N,M,n,m,len,rev[MAXN<<2];
int fac[MAXN],invf[MAXN];
int w[NTT+5],iw[NTT+5];
int A[MAXN<<2],B[MAXN<<2];

inline int qpow(int x,int y){
    int ret=1,tt=x%MOD;
    while(y){
        if(y&1)ret=1ll*ret*tt%MOD;
        tt=1ll*tt*tt%MOD;
        y>>=1;
    }
    return ret;
}

inline int binom(int n,int m){
    if(n<0||m<0||n<m)return 0;
    return 1ll*fac[n]*invf[n-m]%MOD*invf[m]%MOD;
}

void ntt(int *c,int dft){
    rin(i,0,n-1)if(i<rev[i])std::swap(c[i],c[rev[i]]);
    for(int mid=1;mid<n;mid<<=1){
        int r=(mid<<1),u=NTT/r;
        for(int l=0;l<n;l+=r){
            int v=0;
            for(int i=0;i<mid;++i,v+=u){
                int x=c[l+i],y=1ll*c[l+mid+i]*(dft>0?w[v]:iw[v])%MOD;
                c[l+i]=x+y<MOD?x+y:x+y-MOD;
                c[l+mid+i]=x-y>=0?x-y:x-y+MOD;
            }
        }
    }
    if(dft<0){
        int invn=qpow(n,MOD-2);
        rin(i,0,n-1)c[i]=1ll*c[i]*invn%MOD;
    }
}

void prepare(){
    for(n=1,len=0;n<=m;n<<=1,++len);
    rin(i,1,n-1)rev[i]=((rev[i>>1]>>1)|((i&1)<<(len-1)));
}

void init(int N){
    fac[0]=1;
    rin(i,1,N)fac[i]=1ll*fac[i-1]*i%MOD;
    invf[N]=qpow(fac[N],MOD-2);
    irin(i,N-1,0)invf[i]=1ll*invf[i+1]*(i+1)%MOD;
    w[0]=iw[0]=1;
    w[1]=qpow(G,(MOD-1)/NTT),iw[1]=qpow(INVG,(MOD-1)/NTT);
    rin(i,2,NTT-1){
        w[i]=1ll*w[i-1]*w[1]%MOD;
        iw[i]=1ll*iw[i-1]*iw[1]%MOD;
    } 
}

int main(){
    D=read(),N=read(),M=read();int S=N-2*M;
    if(M>N/2){printf("0\n");return 0;}
    if(D<=S){printf("%d\n",qpow(D,N));return 0;}
    init(D);
    rin(i,0,S)A[i]=1ll*(((S-i)&1)?MOD-1:1)*invf[i]%MOD*invf[S-i]%MOD;
    rin(i,0,D)if(D-1-S-i>=0)B[i]=1ll*invf[i]*invf[D-1-S-i]%MOD;
    m=S+D;prepare();
    ntt(A,1);ntt(B,1);
    rin(i,0,n-1)A[i]=1ll*A[i]*B[i]%MOD;
    ntt(A,-1);
    int ans=0;
    rin(i,0,D){
        if(D-1-i>=0){
            A[i]=1ll*A[i]*fac[i]%MOD*fac[D-i-1]%MOD;
            ans=(ans+1ll*binom(D,i)*qpow(2*i-D,N)%MOD*A[i])%MOD; 
        }
        else{
            int temp=0;
            rin(j,0,S)temp=(temp+binom(D,j))%MOD;
            ans=(ans+1ll*binom(D,i)*qpow(2*i-D,N)%MOD*temp)%MOD;
        }
    }
    printf("%lld\n",(1ll*ans*qpow(qpow(2,D),MOD-2)%MOD+MOD)%MOD);
    return 0;
}

posted on 2019-05-22 08:52 ErkkiErkko 阅读(...) 评论(...) 编辑 收藏

统计