P5487 【模板】Berlekamp-Massey算法

链接

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题解

这个题就是堆板子

最短线性递推的那一大坨拿来，然后再放一个 $BM$的板子就行了

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
        return ans;
    }
}em;
#define mod 998244353ll
struct berlekamp_massey
{
    static const ll maxl = 5050;
    ll l, m[maxl], r[maxl<<1][maxl], pos[maxl], d[maxl], cnt;
    pair<ll,ll*> run(ll* a, ll n)
    {
        l=n, cnt=0;
        cl(m), cl(r), cl(pos), cl(d);
        ll i, j;
        rep(i,1,l)
        {
            ll D=a[i];
            rep(j,1,m[cnt])(D-=r[cnt][j]*(a[i-j]%mod))%=mod;
            if(D)
            {
                cnt++;
                if(cnt==1)
                {
                    pos[cnt]=i;
                    m[cnt]=i;
                    d[cnt]=D;
                }
                else
                {
                    ll p=cnt-1;
                    rep(j,1,cnt-1)if(i-pos[j]+m[j-1]<i-pos[p]+m[p-1])p=j;
                    ll t=em.inv(d[p],mod)*D%mod;
                    m[cnt]=max(m[cnt-1],i-pos[p]+m[p-1]);
                    rep(j,1,m[cnt-1])r[cnt][j]=r[cnt-1][j];
                    (r[cnt][i-pos[p]]+=t)%=mod;
                    rep(j,1,m[p-1])(r[cnt][i-pos[p]+j]-=t*r[p-1][j])%=mod;
                    pos[cnt]=i;
                    d[cnt]=D;
                }
            }
        }
        return make_pair(m[cnt],r[cnt]);
    }
}bm;
struct NTT
{
    ll n;
    vector<ll> R;
    void init(ll bound)    //bound是积多项式的最高次幂
    {
        ll L(0);
        for(n=1;n<=bound;n<<=1,L++);
        R.resize(n);
        for(ll i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    }
    void ntt(vector<ll>& a, int opt)
    {
        ll i, j, k, wn, w, x, y, inv(em.fastpow(n,mod-2,mod));
        for(i=0;i<n;i++)if(i>R[i])swap(a[i],a[R[i]]);
        for(i=1;i<n;i<<=1)
        {
            if(opt==1)wn=em.fastpow(3,(mod-1)/(i<<1),mod);
            else wn=em.fastpow(3,(mod-1-(mod-1)/(i<<1)),mod);
            for(j=0;j<n;j+=i<<1)
                for(w=1,k=0;k<i;k++,w=w*wn%mod)
                {
                    x=a[k+j], y=a[k+j+i]*w%mod;
                    a[k+j]=(x+y)%mod, a[k+j+i]=(x-y)%mod;
                }
        }
        if(opt==-1)for(i=0;i<n;i++)(a[i]*=inv)%=mod;
    }
};
struct Poly
{
    vector<ll> v;
    ll n;	//n是最高次项的次数
    Poly(ll N){v.resize(N+1);n=N;}
    Poly(const Poly& p){v=p.v; n=p.n;}
    void resize(ll N){n=N; v.resize(N+1);}
    ll& operator[](ll id){return v[id];}
    void show()
    {
        printf("n=%lld\n",n);
        ll i; rep(i,0,n-1)printf("%lldx^%lld + ",(v[i]+mod)%mod,i);
        printf("%lldx^%lld\n",(v[n]+mod)%mod,n);
    }
};
Poly operator*(Poly A, Poly B)
{
    NTT ntt;
    ll i, n=A.n+B.n;
    ntt.init(A.n+B.n);
    A.resize(ntt.n-1), B.resize(ntt.n-1);
    ntt.ntt(A.v,1), ntt.ntt(B.v,1);
    Poly C(ntt.n-1);
    rep(i,0,C.n)C[i]=(A[i]*B[i])%mod;
    ntt.ntt(C.v,-1);
    C.resize(n);
    return C;
}
Poly operator~(Poly F)
{
    Poly H(0), f(0);
    ll i, tar, j;
    H[0]=em.inv(F[0],mod);
    f.resize(0), f[0]=F[0];
    for(tar=1;tar<=F.n;tar*=2);
    F.resize(tar-1);
    for(i=1;i<=tar;i*=2)
    {
        f.resize(i-1);
        rep(j,i>>1,i-1)f[j]=F[j];
        Poly t=H*H*f; t.resize(i-1);
        H.resize(i-1);
        rep(j,0,i-1)H[j]=(2*H[j]-t[j])%mod;
    }
    H.resize(F.n);
    return H;
}
Poly gg(0);
pair<Poly,Poly> operator/(Poly F, Poly G)
{
	ll n=F.n, m=G.n, i;
	if(m>n)return make_pair(Poly(0),F);
	Poly f=F, g=gg;
	reverse(f.v.begin(),f.v.end());
	f.resize(n-m), g.resize(n-m);
	Poly Q=f*g;
	Q.resize(n-m);
	reverse(Q.v.begin(),Q.v.end());
	Poly t(Q*G), R(m-1);
	rep(i,0,m-1)R[i]=F[i]-t[i];
	return make_pair(Q,R);
}
ll calc(vector<ll> f, vector<ll> a, ll n)
{
    ll k=f.size(), i, j;
    Poly t(1), r(k-1), F(k);
    rep(i,0,k-1)F[i]=-f[k-1-i];
    F[k]=1;
    gg=F;
    reverse(gg.v.begin(),gg.v.end());
    gg=~gg;
    t[1]=1;
    r[0]=1;
    for(ll b=n;b;b>>=1,t=(t*t/F).second)
        if(b&1)r=(r*t/F).second;
    ll ans=0;
    rep(i,0,k-1)(ans+=r[i]*a[i])%=mod;
    return ans;
}
ll a[maxn], n, m;
int main()
{
    ll i;
    n=read(); m=read();
    rep(i,1,n)a[i]=read();
    auto pr = bm.run(a,n);
    ll l=pr.first, *r=pr.second;
    rep(i,1,l)printf("%lld ",(r[i]+mod)%mod); putchar(10);
    vector<ll> p(a+1,a+l+1), f(r+1,r+l+1);
    ll ans=calc(f,p,m);
    printf("%lld",(ans+mod)%mod);
    return 0;
}