算法竞赛模板 BM算法（找线性规律万能模板）

(1) n是指要找该数列的第n项。

(2) 往vec中放入该数列前几项的值，越多越精确。

#include<set>
#include<cmath>
#include<vector>
#include<string>
#include<cstdio>
#include<cassert>
#include<cstring>
#include<algorithm>
using namespace std;
#define rep(i,a,n) for (ll i=a;i<n;i++)
#define per(i,a,n) for (ll i=n-1;i>=a;i--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define SZ(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
typedef long long ll;
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod=1000000007;
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 ll N=10010;
    ll res[N],base[N],_c[N],_md[N];
    vector<ll>Md;
    void mul(ll*a,ll*b,ll 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(ll 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];
    }
    ll solve(ll n, VI a, VI b) 
    { 
        ll ans=0,pnt=0;
        ll 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<<pnt)<=n) pnt++;
        for(ll p=pnt;p>=0;p--) 
        {
            mul(res,res,k);
            if((n>>p)&1) 
            {
                for(ll 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);
        ll 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)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) 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*powmod(b,mod-2)%mod;
                while(SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    ll gao(VI a,ll n)
    {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};
int main() 
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld", &n);
        VI vec;
        vec.push_back(1);
        vec.push_back(1);
        vec.push_back(2);
        vec.push_back(3);
        vec.push_back(5);
        vec.push_back(8); 
        vec.push_back(13);
        printf("%lld
", linear_seq::gao(vec, n - 1)%mod);
    }
}