线性递推模板

#include<bits/stdc++.h>
 
using namespace std;
 
#define rep(i,a,n) for(int i=a;i<n;i++)
 
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
 
const ll mod = 1e9+7;
 
ll powmod(ll a,ll b) {
    ll ret = 1;
    a%=mod;
    while(b) {
        if(b&1) ret = ret*a%mod;
        a = a*a%mod;
        b>>=1;
    }
    return ret;
}
int _,n;
 
namespace linear_seq {
const int N = 10010;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;
 
void mul(ll *a,ll *b,int k) {
    for(int i=0; i<k+k; i++)
        _c[i] = 0;
    for(int i=0; i<k; ++i)
        if(a[i])
            for(int j=0; j<k; j++)
                _c[i+j] = (_c[i+j]+a[i]*b[j])%mod;
 
    for(int i=k+k-1; i>=k; i--)
        if(_c[i])
            for(int j=0; j<(int) Md.size(); j++)
                _c[i-k+Md[j]] = (_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
    for(int i=0; i<k; i++)
        a[i] = _c[i];
}
 
int solve(ll n,VI a,VI b) {
    ll ans = 0,pnt = 0;
    int k = SZ(a);
    for(int i=0; i<k; i++)
        _md[k-1-i] = -a[i];
    _md[k] = 1;
    Md.clear();
    for(int i=0; i<k; i++)
        if(_md[i]!=0)
            Md.push_back(i);
    for(int i=0; i<k; i++)
        res[i] = base[i] = 0;
    res[0] = 1;
    while((1ll<<pnt)<=n)
        pnt++;
 
    for(int p = pnt; p>=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;
            for(int j=0; j<(int) Md.size(); ++j)
                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;
    for(int n=0; n<(int)s.size(); ++n) {
        ll d = 0;
        for(int i=0; i<L+1; i++)
            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.push_back(0);
            for(int i=0; i<(int) 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*powmod(b,mod-2)%mod;
            while(SZ(C)<SZ(B)+m)
                C.push_back(0);
            for(int i=0; i<(int)B.size(); i++)
                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());
    for(int i=0; i<(int) c.size(); i++)
        c[i] = (mod-c[i])%mod;
    return (ll) solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
 
}
 
int main() {
    ll t;
    ll nnn;
    VI a;
    a.push_back(3);
    a.push_back(9);
    a.push_back(20);
    a.push_back(46);
    a.push_back(106);
    a.push_back(244);
    a.push_back(560);
    a.push_back(1286);
    a.push_back(2956);
    a.push_back(6794);
    scanf("%lld",&t);
    while(t--) {
        scanf("%lld",&nnn);
        printf("%lld
",linear_seq::gao(a,nnn-1));
    }
}