这种题的关键在于取走最后一个的获胜

因此就会交换先后手, 先手拿大于的堆一定会获胜 根据的奇偶性不同, 先手可以做如下操作

  • 拿个, 将留给后手
  • 直接拿个

如果当前是奇数, 直接拿全部, 如果当前是偶数, 拿, 这样就会保证到最后一堆一定是先手拿

注意特殊处理全是的情况

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()
#define vec1(T, name, n, val) vector<T> name(n, val)
#define vec2(T, name, n, m, val) vector<vector<T>> name(n, vector<T>(m, val))
#define vec3(T, name, n, m, k, val) vector<vector<vector<T>>> name(n, vector<vector<T>>(m, vector<T>(k, val)))
#define vec4(T, name, n, m, k, p, val) vector<vector<vector<vector<T>>>> name((n), vector<vector<vector<T>>>((m), vector<vector<T>>((k), vector<T>((p), (val)))))

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream& operator>>(istream& is, i128& val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char& c : str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream& operator<<(ostream& os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

struct Hash {
    vector<int> h, p;
    int B = 131;
    Hash(const string& s) {
        int n = s.size();
        h.resize(n + 1, 0);
        p.resize(n + 1, 1);
        for (int i = 0; i < n; ++i) {
            p[i + 1] = p[i] * B % MOD;
            h[i + 1] = (h[i] * B + s[i]) % MOD;
        }
    }

    LL get(int l, int r) {
        LL v = h[r] - h[l - 1] * p[r - l + 1] % MOD;
        v = (v % MOD + MOD) % MOD;
        return v;
    }
};

void solve() {
    int n;
    cin >> n;
    vec1(int, a, n, 0);
    for (int i = 0; i < n; ++i) cin >> a[i];
    
    if (n == 1) {
        cout << "Alice" << '\n';
        return;
    }

    if (a[0] > 1) {
        cout << "Alice" << '\n';
        return;
    }

    int idx = -1;
    for (int i = 0; i < n; ++i) {
        if (a[i] > 1) {
            idx = i;
            break;
        }
    }

    if (idx == -1) {
        if (n & 1) {
            cout << "Alice" << '\n';
        }
        else {
            cout << "Bob" << '\n';
        }
        return;
    }

    if (idx & 1) {
        cout << "Bob" << '\n';
    }
    else cout << "Alice" << '\n';

/**/ #ifdef LOCAL
    cout << flush;
/**/ #endif
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T;
    cin >> T;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}