CodeForces - 347C - Alice and Bob

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).

Examples
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.

• 題目大意：
  給你n個數，兩個人輪流操作，每個人可以找其中任意兩個數，然後得到的差的絕對值加入到這些數中，然後最後哪個人不能找出兩個數的差不在這些數裡，誰就輸了。
• 解題思路：
  只要所有的數都是同一個數的倍數，比如，3，6，9，12，15，18，這種情況下任意兩個數的差都在這個數列裡。所以我們只要求出給出所有數的最大公約數，用最大的數/最大公約數=不能進行下去的時候的數的個數。然後減去本來的n個就是答案了。
• AC程式碼：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
int a[1100];
int gcd(int a,int b)
{
	if(b==0)
		return a;
	else
		return gcd(b,a%b); 
}
int main()
{
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	int g=a[0];
	for(int i=1;i<n;i++)
		g=gcd(a[i],g);
	int	ans;
	ans=a[n-1]/g-n;
	if(ans%2==0)
		cout<<"Bob"<<endl;
	else
		cout<<"Alice"<<endl;
}