Game with Board

题面翻译

Alice 和 Bob 玩游戏，他们有一块黑板。最初，有 $n$ 个整数 $1$。Alice 和 Bob 轮流操作，Alice 先手。

轮到时，玩家必须在棋盘上选择几个（至少两个）相等的整数，擦除它们，然后写一个等于它们总和的新整数。

如果玩家不能移动（棋盘上的所有整数都不同），该玩家将赢得游戏。
如果两名选手都发挥最佳，则确定谁获胜。

题目描述

Alice and Bob play a game. They have a blackboard; initially, there are $ n $ integers written on it, and each integer is equal to $ 1 $ .

Alice and Bob take turns; Alice goes first. On their turn, the player has to choose several (at least two) equal integers on the board, wipe them and write a new integer which is equal to their sum.

For example, if the board currently contains integers $ {1, 1, 2, 2, 2, 3} $ , then the following moves are possible:

choose two integers equal to $ 1 $ , wipe them and write an integer $ 2 $ , then the board becomes $ {2, 2, 2, 2, 3} $ ;
choose two integers equal to $ 2 $ , wipe them and write an integer $ 4 $ , then the board becomes $ {1, 1, 2, 3, 4} $ ;
choose three integers equal to $ 2 $ , wipe them and write an integer $ 6 $ , then the board becomes $ {1, 1, 3, 6} $ .

If a player cannot make a move (all integers on the board are different), that player wins the game.

Determine who wins if both players play optimally.

输入格式

The first line contains one integer $ t $ ( $ 1 \le t \le 99 $ ) — the number of test cases.

Each test case consists of one line containing one integer $ n $ ( $ 2 \le n \le 100 $ ) — the number of integers equal to $ 1 $ on the board.

输出格式

For each test case, print Alice if Alice wins when both players play optimally. Otherwise, print Bob.

样例 #1

样例输入 #1

2
3
6

样例输出 #1

Bob
Alice

分析

因为A先走, A有主动权, 我们看A能不能找到一个策略让自己必赢, 如果有那么A赢, 如果没有那么B赢, 所以这道题的理解很容易出错, 不明白题目所说的 "两人均发挥最佳" 是什么意思

A如果要赢, 就要让B无论如何选A都是赢的, 那么A可以在第一步的时候构造出 1 1 n-2 的情况, 然后B不得不选两个1去合, 合完之后 2 和 n-2 不一样, 肯定是A赢.

要让 2 和 n-2 不一样, n 要大于等于5, 那么也就是说 n 大于等于5的时候, A必赢

当n是2到4的时候, 推一下就知道无论A怎么选, 都是B赢

综上就可以总结出来了

感觉是一个好大的坑啊妈的

代码

感觉这个代码写得好屈辱啊妈的

#include <iostream>
using namespace std;

int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	
	int t; cin >> t;
	while(t--)
	{
		int n; cin >> n;
		if(n >= 5) cout << "Alice\n";
		else cout << "Bob\n";
	}
}

JZTXT

CF1841A