线性递推规律BM杜教

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <vector>

#include <string>

#include <map>

#include <set>

#include <cassert>

#include<bits/stdc++.h>

using namespace std;

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int 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) ((int)(x).size())

typedef vector<int> VI;

typedef long long ll;

typedef pair<int,int> PII;

const ll mod=;

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

int _,n;

namespace linear_seq {

    const int N=;

    ll res[N],base[N],_c[N],_md[N];

    vector<int> Md;

    void mul(ll *a,ll *b,int 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 (int 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];

    }

    int 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=;

        int 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 (int p=pnt;p>=;p--) {

            mul(res,res,k);

            if ((n>>p)&) {

                for (int 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(,);

        int 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;

    }

    int 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("%d",&n)) {

        vector<int>v;

        v.push_back();

        v.push_back();

        v.push_back();

        v.push_back();

        v.push_back();

        v.push_back();

        //VI{1,2,4,7,13,24}

        printf("%d\n",linear_seq::gao(v,n-));

    }

}

BM