UVA 12293 Box Game（博弈入门）

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#UVA #入门 #博弈 #思维

编程题——博弈专栏收录该内容

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本文针对BoxGame问题进行了详细的解析，通过模拟发现必输状态的规律，并给出了对应的代码实现。问题中两个玩家轮流操作，目标是让对手无法进行有效操作。

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题目链接：Box Game

题面：

12293 Box Game

There are two identical boxes. One of them contains n balls, while the other box contains one ball.
Alice and Bob invented a game with the boxes and balls, which is played as follows:
Alice and Bob moves alternatively, Alice moves first. For each move, the player finds out the box
having fewer number of balls inside, and empties that box (the balls inside will be removed forever),
and redistribute the balls in the other box. After the redistribution, each box should contain at least
one ball. If a player cannot perform a valid move, he loses. A typical game is shown below:
When both boxes contain only one ball, Bob cannot do anything more, so Alice wins.
Question: if Alice and Bob are both clever enough, who will win? Suppose both of them are very
smart and always follows a perfect strategy.

Input
There will be at most 300 test cases. Each test case contains an integer n (2 ≤ n ≤ 109) in a single
line. The input terminates by n = 0.

Output
For each test case, print a single line, the name of the winner.

Sample Input
2
3
4
0

Sample Output
Alice
Bob
Alice

解题：

博弈基本不会，这道题很好推，先模拟一下发现1,3,7为必输态，只要后面的情况能够让分解为之前的必输态给Bob那么都是Alice赢，很容易推出下一必输态为15。因为此时最小只能得到8了。这道题也可以找规律，1,3,7,15,31...总是2^n-1。

代码：

#include <iostream>
using namespace std;
int main()
{
	int n;
	while(cin>>n&&n)
	{
	  bool flag=true;
	  while(n)
	  {
		  if((n-1)%2)
		  {
			  flag=false;
			  break;
		  }
		  n=(n-1)/2;
	  }
	  if(flag)cout<<"Bob\n";
	  else cout<<"Alice\n";
	}
}