Codeforces Round #201 (Div. 2) 347C Alice and Bob(脑洞)

C. Alice and Bob

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., an (1 ≤ ai ≤ 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 test(s)

input

22 3

output

Alice

input

25 3

output

Alice

input

35 6 7

output

Bob

Note

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.

Alice or Bob每次选两个不同的数字，如果自己的集合不包含|x - y|，就将其加入自己的集合。Alice先进行，谁无法进行谁输。

用gcd判断即可。

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 105;typedef long long ll;ll gcd(ll a, ll b){return b ? gcd(b, a % b) : a;}int main(int argc, char const *argv[]){int n;ll temp = 0;while(scanf("%d", &n) != EOF) {ll a[maxn];for(int i = 0; i < n; ++i) {scanf("%lld", &a[i]);temp = max(temp, a[i]);}int ans = temp;for(int i = 0; i < n - 1; ++i)for(int j = i + 1; j < n; ++j) {int k = gcd(a[i], a[j]);temp = gcd(temp, k);}int m = ans / temp - n;puts(m & 1 ? "Alice" : "Bob");}return 0;}