BM算法

 不知道原理..

听说扔一定的前n项,然后就可以得到答案了.

#include<bits/stdc++.h>

using namespace std;


typedef long long ll;
const ll mod=1000000007;
ll pwr(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 ll N=10010;
    ll res[N],base[N],_c[N],_md[N];

    vector<int> Md;
    void mul(ll *a,ll *b,int k) {
        int i, j;
        for (i=0; i<k+k; ++i) _c[i]=0;
        for (i=0; i<k; ++i) 
            if (a[i]) 
                for (j=0; j<k; ++j)  
                    _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (i=k+k-1;i>=k;i--) if (_c[i])
            for(j=0; j<Md.size(); ++j) 
                _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        for(i=0; i<k; ++i)  a[i]=_c[i];
    }
    int solve(ll n,vector<ll> a,vector<ll> b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        printf("%d
",SZ(b));
        ll ans=0,pnt=0;
        int p, i, j, k = a.size();
        assert(a.size()==b.size());
        for(i=0; i<k; ++i) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        for(i=0; i<k; ++i) if (_md[i]!=0) Md.push_back(i);
        for(i=0; i<k; ++i) res[i]=base[i]=0;
        res[0] = 1;
        while ( (1ll<<pnt) <= n) pnt++;
        for (p=pnt;p>=0;p--) {
            mul(res,res,k);
            if ((n>>p)&1) {
                for (i=k-1;i>=0;i--) 
                    res[i+1]=res[i];res[0]=0;
                for (j=0; j<Md.size(); ++j) 
                    res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        for(i=0; i<k; ++i) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    vector<ll> BM(vector<ll> s) {
        vector <ll> C(1,1),B(1,1);
        int L = 0, m = 1, b = 1;
        int i, n;
        for (n=0; n<s.size(); ++n) {
            ll d=0;
            for(i=0; i<L+1; ++i)  d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                vector<ll> T=C;
                ll c=mod-d*pwr(b,mod-2)%mod;
                while (C.size()<B.size()+m) C.push_back(0);
                for(i=0; i<B.size(); ++i) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*pwr(b,mod-2)%mod;
                while (C.size()<B.size()+m) C.push_back(0);
                for(i=0; i<B.size(); ++i) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    ll gao(vector<ll> a,ll n) {
        vector<ll> c=BM(a);
        c.erase(c.begin());
        int i;
        for(i=0; i<c.size(); ++i) c[i]=(mod-c[i])%mod;
        return solve(n,c,vector<ll>(a.begin(),a.begin()+c.size()));
    }
};

int main() {
    while (~scanf("%lld",&n)) {
        vector<ll>v;
        v.push_back(1);
        v.push_back(1);
        v.push_back(2);
        v.push_back(3);
        v.push_back(5);
        v.push_back(8);
        //VI{1,2,4,7,13,24}
        printf("%lld
",linear_seq::gao(v,n-1));
    }
}