Gym 100952 G. The jar of divisors

最新推荐文章于 2018-06-27 14:56:23 发布

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最新推荐文章于 2018-06-27 14:56:23 发布

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原文链接：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7199270.html

本文解析了一道关于博弈论的游戏题，题目描述了Alice和Bob通过移除数及其因数的方式进行游戏，最终确定胜者。文章给出了简洁的代码实现，并得出结论：当输入数n模6等于1时，Bob获胜；否则，Alice获胜。

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Alice and Bob play the following game. They choose a number N to play with. The rules are as follows:

- They write each number from 1 to N on a paper and put all these papers in a jar.

- Alice plays first, and the two players alternate.

- In his/her turn, a player can select any available number M and remove its divisors including M.

- The person who cannot make a move in his/her turn wins the game.

Assuming both players play optimally, you are asked the following question: who wins the game?

Input

The first line contains the number of test cases T (1 ≤ T ≤ 20). Each of the next T lines contains an integer (1 ≤ N ≤ 1,000,000,000).

Output

Output T lines, one for each test case, containing Alice if Alice wins the game, or Bob otherwise.

Examples

Input

2
5
1

Output

Alice
Bob
这道题我自己都有点迷茫，是个博弈题，结论就是n%6==1?"Bob":"Alice"

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int main()
{
    ll t,n;
    scanf("%lld",&t);
    while(t--)
    {
        cin>>n;
        if(n%6==1) puts("Bob");
        else puts("Alice");
    }
    return 0;
}