Codeforces Round #201 (Div. 1) / 346A Alice and Bob (数论&想法题)

A. Alice and Bob

http://codeforces.com/problemset/problem/346/A

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, ..., 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 test(s)

input

2
2 3

output

Alice

input

2
5 3

output

Alice

input

3
5 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.

思路：

两个观察：

1. 数组6,8,10的结果和数组3,4,5的结果相同——求n个数的最大公约数

2. 数组3,5,7(最大公约数为1)的最终结果为1,2,3,4,5,6,7——移动次数 = 最大数 - n

综上得：移动次数 = 最大数/gcd - n

完整代码：

/*30ms,0KB*/

#include
#include
using namespace std;

int num[105];

int gcd(int a, int b)
{
	return b ? gcd(b, a % b) : a;
}

int main(void)
{
	int n, temp = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d", &num[i]);
		temp = max(temp, num[i]);
	}
	int maxn = temp;
	for (int i = 0; i < n - 1; ++i)
		for (int j = i + 1; j < n; ++j)
		{
			int k = gcd(num[i], num[j]);
			temp = gcd(temp, k); ///求n个数的gcd
		}
	int movetimes = maxn / temp - n; ///移动次数 = 最大数/gcd - 数的个数
	puts(movetimes & 1 ? "Alice" : "Bob");
	return 0;
}