8.12 第八场 GCD Game

8.12 第八场 GCD Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3640 Accepted Submission(s): 730

Problem Description

Alice and Bob are playing a game.

They take turns to operate. There are n numbers, a1 , a2 , … , an. Every time, the player plays in 3 steps.

1. Arbitrarily chooses one number ai.

2. Arbitrarily chooses another number x(1≤x<ai).

3. Replace the number ai with gcd(ai,x). Here, gcd(u,v) refers to the Greatest Common Divisor of u and v.

When a player can not make a single move he/she loses the game. Alice moves the first and she asks you to tell her who will win the game if both player play optimally.

Input

The first line contains a number T(1≤T≤100), the number of testcases.

For each testcase, there are two lines.
The first line contains one number n(1≤n≤106).
The second line contains n numbers a1 , a2 , … , an(1≤ai≤107).

It is guaranteed that for all testcases, ∑n≤106.

Output

For each testcase, output the answer in one line. If Alice will win the game, print “Alice”, otherwise print “Bob”.

Sample Input

2
1
1
1
2

Sample Output

Bob
Alice

大概题意：

两个人轮流进行游戏，可以将集合中的数变为与另一个数的最大公约数，先无法行动的人输。

思路：

用线性筛法求约数个数，然后进行 Nim 博弈即可

代码：

//
// Created by Black on 2021/8/12.
//

#include <iostream>
#include <cstdio>

using namespace std;
const int N = 1e7 + 5;
int primes[N], cnt;     // primes[]存储所有素数
bool st[N];         // st[x]存储x是否被筛掉
int d[N];

void get_primes(int n) {
    for (int i = 2; i <= n; i++) {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++) {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
    d[1] = 0;
    for (int i = 2; i <= n; i++) {
        if (!st[i]) d[i] = 1;
        for (int j = 0; primes[j] <= n / i; j++) {
            d[primes[j] * i] = d[i] + 1;
            if (i % primes[j] == 0) break;
        }
    }
}
//
//int play(int n) {
//    int res = 0;
//    for (int i = 2; i <= n / i; ++i) {
//        if (n % i == 0) {
//            while (n % i == 0) {
//                n /= i;
//                res++;
//            }
//        }
//    }
//    if (n > 1)res++;
//    return res;
//}

//int getY(int x) {
//    if (x == 1)return 0;
//    int res = 1, i;
//    for (i = 2; i * i <= x; i++) {
//        int r = 0;
//        while (x % i == 0) {
//            r++;
//            x /= i;
//        }
//        if (r > 0) {
//            r++;
//            res *= r;
//        }
//    }
//    if (x > 1)
//        res *= 2;
//    return res == 0 ? 1 : res - 1;
//}

int main() {
    get_primes(N);
    int t;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        int res = 0;
        while (n--) {
            int x;
            scanf("%d", &x);
            //            int h = play(x);
            int h = d[x];
            //            if (x == 1)h = 0;
            //            else h = d[x] == 0 ? 1 : d[x] - 1;
            //            cout << h << endl;
            res ^= h;
        }
        if (res) printf("Alice\n");
        else printf("Bob\n");
    }
    return 0;
}