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).

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.

起初一眼以为是一道博弈问题,后来发现就是一个GCD();

求出GCD();后即可做倍数。

注意最后要-n

附AC代码:

 #include<bits/stdc++.h>
using namespace std; int gcd(int a, int b)
{
return (b>)?gcd(b,a%b):a;
} int main()
{
int n, a[], i;
cin>>n;
int Max = ;
for(i = ; i < n; i++)
{
cin>>a[i];
if(a[i] > Max)
Max = a[i];
}
int p = a[];
for(i = ; i < n; i++)
p = gcd(p, a[i]);
int ans = Max/p - n;
printf(ans& ? "Alice\n" : "Bob\n");
return ;
}

E - Alice and Bob的更多相关文章

  1. 2016中国大学生程序设计竞赛 - 网络选拔赛 J. Alice and Bob

    Alice and Bob Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  2. bzoj4730: Alice和Bob又在玩游戏

    Description Alice和Bob在玩游戏.有n个节点,m条边(0<=m<=n-1),构成若干棵有根树,每棵树的根节点是该连通块内编号最 小的点.Alice和Bob轮流操作,每回合 ...

  3. Alice and Bob(2013年山东省第四届ACM大学生程序设计竞赛)

    Alice and Bob Time Limit: 1000ms   Memory limit: 65536K 题目描述 Alice and Bob like playing games very m ...

  4. sdutoj 2608 Alice and Bob

    http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2608 Alice and Bob Time L ...

  5. hdu 4268 Alice and Bob

    Alice and Bob Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Tota ...

  6. 2014 Super Training #6 A Alice and Bob --SG函数

    原题: ZOJ 3666 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3666 博弈问题. 题意:给你1~N个位置,N是最 ...

  7. ACdream 1112 Alice and Bob(素筛+博弈SG函数)

    Alice and Bob Time Limit:3000MS     Memory Limit:128000KB     64bit IO Format:%lld & %llu Submit ...

  8. 位运算 2013年山东省赛 F Alice and Bob

    题目传送门 /* 题意: 求(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1) 式子中,x的p次方的系数 二进制位运算:p ...

  9. SDUT 2608:Alice and Bob

    Alice and Bob Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述 Alice and Bob like playing ...

  10. Alice and Bob(贪心HDU 4268)

    Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...

随机推荐

  1. spring security原理图及其解释

    用户发出订单修改页面的请求,Access Decision Manager进行拦截,然后对比用户的授权和次页面需要的授权是不是有重合的部分,如果有重合的部分,那面页面就授权成功,如果失败就通知用户. ...

  2. Windows下Python安装pyecharts

    都说pyechart用来可视化好,可是安装的时候各种坑 正常情况是 pip install pyecharts 然后各种报错,找到一种可行的方式 在https://pypi.org/project/p ...

  3. PPT中的图像失真

    现象:Office PowerPoint 保存出来的PPT文件在WPS下播放的时候会出现图像失真的显现. 解决方法:Office PowerPoint打开PPT将里面的图像另存为BMP格式的图像文件, ...

  4. People seldom do what they believe in. They do what is convenient, then repent.

    People seldom do what they believe in. They do what is convenient, then repent. 人们很少真正实践他们的理想.他们只做比较 ...

  5. 如何给老婆解释什么是RESTful

    如何给老婆解释什么是RESTful Javdroider Hong 知乎专栏<Beautiful Java>的作者,一个热爱足球和健身的上进boy 1,543 人赞了该文章 老婆经常喜欢翻 ...

  6. stream_context_create()模拟POST/GET

    有时候,我们需要在服务器端模拟 POST/GET 等请求,也就是在 PHP 程序中去实现模拟,该怎么做到呢?或者说,在 PHP 程序里,给你一个数组,如何将这个数组 POST/GET 到另外一个地址呢 ...

  7. winfrom桌面程序调用python解释器

    Winfrom桌面程序调用python解释器执行py脚本后台执行完成具体的功能,为什么要这样处理呢?因为我现在的大部分过项目都是后台的脚本处理,界面基本的输入完成之后,将参数按照规则传入到脚本的入口, ...

  8. python -- day 11 考试题

    1. 文件t1.txt里面的内容为:(6分) 1,alex,22,13651054608,IT 2,wusir,23,13304320533,Tearcher 3,taibai,18,13332353 ...

  9. luogu3384 【模板】 树链剖分

    题目大意 已知一棵包含N个结点的树(连通且无环),每个节点上包含一个数值,需要支持以下操作:操作1: 格式: 1 x y z 表示将树从x到y结点最短路径上所有节点的值都加上z操作2: 格式: 2 x ...

  10. oracle long 转varchar2

    函数: /* 其中in_rowid为行id,in_owner为数据库登陆的帐号名,in_table_name为数据库表名,in_column为数据库对应long类型的表字段名称 */ CREATE O ...