Alice and Bob

Posted 2020-09-04 勿忘初心0924

 Alice and Bob

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

It is so boring in the summer holiday, isn‘t it? So Alice and Bob have invented a new game to play. The rules are as follows. First, they get a set of n distinct integers. And then they take turns to make the following moves. During each move, either Alice or Bob (the player whose turn is the current) can choose two distinct integers x and y from the set, such that the set doesn‘t contain their absolute difference |x - y|. Then this player adds integer |x - y| to the set (so, the size of the set increases by one).

If the current player has no valid move, he (or she) loses the game. The question is who will finally win the game if both players play optimally. Remember that Alice always moves first.

Input

The first line contains an integer n (2 ≤ n ≤ 100) — the initial number of elements in the set. The second line contains n distinct space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the elements of the set.

Output

Print a single line with the winner‘s name. If Alice wins print "Alice", otherwise print "Bob" (without quotes).

Sample Input

Input

2
2 3

Output

Alice

Input

2
5 3

Output

Alice

Input

3
5 6 7

Output

Bob

Hint

Consider the first test sample. Alice moves first, and the only move she can do is to choose 2 and 3, then to add 1 to the set. Next Bob moves, there is no valid move anymore, so the winner is Alice.

/*
题意：给出你n个数，两个人轮流想容器中加数，加数的规则是：取x,y x!=y |x-y|不在容器内，添加|x-y|到容器中，如果某人不能添加了就输了

初步思路：先找出所有从1到max的没出现的数的数量cur，然后用cur mod 所有出现数的最大公因子，具体原理：GCD的原理完美解决这个问题，两
    个数辗转相除求最大公因子的时候，实际上__GCD(a,b) 一直都是 a-k*b。。。。。。实际上就是a b互相减，能构造出来的最小数，cur膜上
    最大公因子，得出的就是能添加的数
*/

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int n,a[105];
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&n);
    int res=0;
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    sort(a,a+n);
    int minn=INF;
    for(int i=0;i<n-1;i++){
        if(a[i+1]==a[i]) continue;
        minn=min(minn,__gcd(a[i+1],a[i]));
    }
    res+=(a[0]-1)/minn;
    for(int i=1;i<n;i++){
        res+=(a[i]-a[i-1]-1)/minn;
    }

    if(res%2==1) puts("Alice");
    else puts("Bob");

    return 0;
}