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.
- 题目大意：
给你n个数，两个人轮流操作，每个人可以找其中任意两个数，然后得到的差的绝对值加入到这些数中，然后最后哪个人不能找出两个数的差不在这些数里，谁就输了。 - 解题思路：
只要所有的数都是同一个数的倍数，比如，3，6，9，12，15，18，这种情况下任意两个数的差都在这个数列里。所以我们只要求出给出所有数的最大公约数，用最大的数/最大公约数=不能进行下去的时候的数的个数。然后减去本来的n个就是答案了。 - AC代码：
#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;
}