BM递推杜教版

#include <bits/stdc++.h>

using namespace std;
#define rep(i,a,n) for (long long i=a;i<n;i++)
#define per(i,a,n) for (long long i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((long long)(x).size())
typedef vector<long long> VI;
typedef long long ll;
typedef pair<long long,long long> PII;
const ll mod=1e9+;
ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
// head

long long _,n;
namespace linear_seq
{
    const long long N=;
    ll res[N],base[N],_c[N],_md[N];

    vector<long long> Md;
    void mul(ll *a,ll *b,long long k)
    {
        rep(i,,k+k) _c[i]=;
        rep(i,,k) if (a[i]) rep(j,,k)
            _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (long long i=k+k-;i>=k;i--) if (_c[i])
            rep(j,,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,,k) a[i]=_c[i];
    }
    long long solve(ll n,VI a,VI b)
    { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d\n",SZ(b));
        ll ans=,pnt=;
        long long k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,,k) _md[k--i]=-a[i];_md[k]=;
        Md.clear();
        rep(i,,k) if (_md[i]!=) Md.push_back(i);
        rep(i,,k) res[i]=base[i]=;
        res[]=;
        while ((1ll<<pnt)<=n) pnt++;
        for (long long p=pnt;p>=;p--)
        {
            mul(res,res,k);
            if ((n>>p)&)
            {
                for (long long i=k-;i>=;i--) res[i+]=res[i];res[]=;
                rep(j,,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<) ans+=mod;
        return ans;
    }
    VI BM(VI s)
    {
        VI C(,),B(,);
        long long L=,m=,b=;
        rep(n,,SZ(s))
        {
            ll d=;
            rep(i,,L+) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==) ++m;
            else if (*L<=n)
            {
                VI T=C;
                ll c=mod-d*powmod(b,mod-)%mod;
                while (SZ(C)<SZ(B)+m) C.pb();
                rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+-L; B=T; b=d; m=;
            }
            else
            {
                ll c=mod-d*powmod(b,mod-)%mod;
                while (SZ(C)<SZ(B)+m) C.pb();
                rep(i,,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    long long gao(VI a,ll n)
    {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};

int main()
{
    while(~scanf("%I64d", &n))
    {
        /*求第n项*/
        printf("%lld\n",linear_seq::gao(VI{,,,,,,,,,, },n-));

        /*输出系数*/
        /*前k项递推，需要2*k项能确定*/
        VI res = linear_seq::BM(VI{,,,,,,,,,, });
        for(int i = ;i < res.size();i++)  printf("%lld\n", (mod-res[i]) % mod);

        //f(n) = f(n-1) + 5*f(n-2) + f(n-3) - f(n-4)
    }
}

几个测试板子的数据：

Input 1
1 2 4 9 20 40 90

Output 1

0 10 0

Input 2
2 4 8 16 32 64 128 256 512 2 4 8 16 32 64 128 256 512

Output 2
0 0 0 0 0 0 0 1

Code From:

https://blog.csdn.net/qq_36876305/article/details/80275708

https://blog.csdn.net/running_acmer/article/details/82722111

Data From:

https://www.cnblogs.com/zhouzhendong/p/Berlekamp-Massey.html