奇（qi）谋（ji）巧（yin）计（qiao）

一、打表法

0.http://oeis.org/

1.差分序列：https://blog.csdn.net/wu_tongtong/article/details/79115921

对于一个多项式产生的序列，可以多次求差分序列，直到差分序列均为0，这时原序列的表达式就可以表示为：

其中，c0……cp为差分表的第0条对角线。

同时有求和公式

确定有公式后除去可以确定的项

#include <bits/stdc++.h>

using namespace std;
#define rep(i, a, n) for (ll i=a;i<n;i++)
#define SZ(x) ((ll)(x).size())
typedef long long ll;
const ll mod = 1000000007;

ll powmod(ll a, ll b) {
    ll res = 1;
    a %= mod;
    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 = 100010;
    ll res[N], base[N], _c[N], _md[N];

    vector<ll> Md;

    void mul(ll *a, const 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, vector<ll> a, vector<ll> b) {
        ll ans = 0, pnt = 0;
        ll k = SZ(a);
        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;
    }

    vector<ll> BM(vector<ll> s) {
        vector<ll> 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) {
                vector<ll> T = C;
                ll c = mod - d * powmod(b, mod - 2) % mod;
                while (SZ(C) < SZ(B) + m) C.push_back(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.push_back(0);
                rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
                ++m;
            }
        }
        return C;
    }

    vector<ll> temp;

    void init(vector<ll> a) {
        temp = BM(a);
        temp.erase(temp.begin());
        rep(i, 0, SZ(temp))temp[i] = (mod - temp[i]) % mod;
    }

    ll gao(vector<ll> a, ll n) {
        return solve(n, temp, vector<ll>(a.begin(), a.begin() + SZ(temp)));
    }
};
using namespace linear_seq;

int main() {

    vector<ll> v;
    v.push_back(3);
    v.push_back(9);
    v.push_back(20);
    v.push_back(46);
    v.push_back(106);
    v.push_back(244);
    v.push_back(560);
    v.push_back(1286);
    v.push_back(2956);
    v.push_back(6794);
    init(v);
    ll T;
    scanf("%lld", &T);
    while (T--) {
        ll n;
        scanf("%lld", &n);
        printf("%lld
", gao(v, n - 1));
    }
    return 0;
}