Taking Pebbles

最新推荐文章于 2018-08-22 21:03:54 发布

原创最新推荐文章于 2018-08-22 21:03:54 发布 · 268 阅读

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本文探讨了一个基于取石子的游戏，通过分析游戏规则和策略，揭示了胜者的关键因素。游戏由Alice和Bob两人进行，轮流取走一定数量的石子，直至取完，胜者为最后取走石子的人。通过对不同初始石子数量的分析，找到了一种规律，能够预测哪一方将最终获胜。

There is a pile of N pebbles initially. Alice and Bob are playing a taking pebbles game as follows:
Alice and Bob moves by turns, Alice moves first. For each move, the player can takes away at least one and at most half of the pebbles remained (“at most half” means you can take way at most k (k >= 1) pebbles if there are 2*k or 2*k+1 pebbles remained). If there is only one pebble remained, the player can also take away this pebble. The player who takes away the last pebble wins.
If Alice and Bob are both clever enough, and they both want to be the winner, who will win?

Input

The first line has one integer T (1 <= T <= 200), means there are T test cases.
For each test case, there is only one line with an integer N (2 <= N <= 109), means the number of pebbles initially.

Output

For each test case, print “Alice” (without quotation marks) in one line if Alice will win. Otherwise print “Bob” (without quotation marks) instead.

Sample Input

Sample Output

Bob
Alice
Alice
Bob
Alice

给你n个石头，以次至少取一个，最多取n/2，n==3 取3/2=1个。Alice先取，
轮流取，谁最后拿完石头就是谁win
刚开始想从奇偶性出发，考虑留给对方多少个时就能赢，但
考虑到每次取的最多的石头数量根本不是固定的，这就没法完了

发现首项为2，只要满足x=2*i+1这个序列的数都市Bob win。
其实从题目就可以看出，最后的结果只能和n本身有关，既然
奇偶性没关系，就试着从n的规律出发

#include<iostream>
#include<cstdio>
#include<bits/stdc++.h>
#define M 100100
using namespace std;
set<int>it;
/*aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa*/
int main()
{
    int st=2;it.insert(2);
    while(st*2+1<=1000000000){
        st=st*2+1;
        it.insert(st);
    }
    int t;cin>>t;
    while(t--){
        int n;cin>>n;
        it.count(n)?printf("Bob\n"):printf("Alice\n");
    }
}