BM-线性递推板子

//杜教BM
#include
using namespace std;
#define rep(i,a,n) for (int i=a;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) ((int)(x).size())
typedef vector VI;
typedef long long ll;
typedef pair PII;
const ll mod=1e9+7;
ll powmod(ll a,ll b)
{
    ll res=1;
    a%=mod;
    assert(b>=0);
    for(; b; b>>=1)
    {
        if(b&1)res=res*a%mod;
        a=a*a%mod;
    }
    return res;
}
ll _,n;
namespace linear_seq
{
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];
    vector Md;
    void mul(ll *a,ll *b,int k)
    {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1; i>=k; i--) if (_c[i])
                rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    int solve(ll n,VI a,VI b)
    {
        ll ans=0,pnt=0;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];
        _md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<=0; p--)
        {
            mul(res,res,k);
            if ((n>>p)&1)
            {
                for (int i=k-1; i>=0; i--) res[i+1]=res[i];
                res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s)
    {
        VI C(1,1),B(1,1);
        int L=0,m=1,b=1;
        rep(n,0,SZ(s))
        {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n)
            {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)v;
    n++;
    for(int i=1;i<=200;i++)
    v.push_back(f[i]);  
    printf("%lld\n",linear_seq::gao(v,n-1)%mod);
}