Codeforces 1194D 【博弈】

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原文链接：https://blog.youkuaiyun.com/weixin_43895356/article/details/99645117

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本文深入探讨了Codeforces 1194D.1-2-KGame的博弈策略，通过分析不同k值对游戏结果的影响，揭示了如何根据纸条长度和k值预测胜者。

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D. 1-2-K Game

Alice and Bob play a game. There is a paper strip which is divided into n + 1 cells numbered from left to right starting from 0. There is a chip placed in the n-th cell (the last one).

Players take turns, Alice is first. Each player during his or her turn has to move the chip 1, 2 or k cells to the left (so, if the chip is currently in the cell i, the player can move it into cell i - 1, i - 2 or i - k). The chip should not leave the borders of the paper strip: it is impossible, for example, to move it k cells to the left if the current cell has number i < k. The player who can’t make a move loses the game.

Who wins if both participants play optimally?

Alice and Bob would like to play several games, so you should determine the winner in each game.

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of games. Next T lines contain one game per line. All games are independent.

Each of the next T lines contains two integers n and k (0 ≤ n ≤ 109, 3 ≤ k ≤ 109) — the length of the strip and the constant denoting the third move, respectively.

Output

For each game, print Alice if Alice wins this game and Bob otherwise.

Example

input
4
0 3
3 3
3 4
4 4
output
Bob
Alice
Bob
Alice

思路

题目意思是有0-n的共n+1个格子，然后每次可以往左移动1、2、k格，当某个人到达0时无法移动则为输。之前校赛做过一道打牌题，但是那个可以移动1-n中任何格数，多了个k就有点难想了。首先我们需要知道一件事，1+2=3，可以借助两人一共移动格数为3来控制当前节奏，并且当Alice准备移动时，场上格数n % 3 == 0时，Alice必输。这道题需要根据k的情况来判断。（以下m代表正整数，3m代表3的倍数）
第一种情况：k % 3 != 0，这种时候，其实k == 3m+1或者k ==3m+2（如4和5则是1+3和2+3）。有无k对题目影响不大。
第二种情况：k % 3 == 0，这种情况比较复杂。如果n < k，实则是只有1和2，那么就是最简单的情。如果n > k，k+1的时候Alice也必输，如果Alice走1步，则此时格数为k，答案显而易见；如果Alice走2步，则此时格数为3m+2，Bob走2步，此时格数为3m，Alice必输；如果Alice走k步，则此时步数为1，答案显而易见。而大于k+1的时候，所有k+1都相当于重复走了m个0-k的过程。虽然不一定能走到k+1但是一定能走到必输的地方。然后再控制节奏就ok了。

代码

#include<iostream>
using namespace std;

int t,n,k;

int main() {
	cin >> t;
	while(t--) {
		cin >> n >> k;
		if(k % 3 == 0) {
			n %= k + 1;
		}
		if(n % 3 == 0 && n != k) {
			puts("Bob");
		}
		else {
			puts("Alice");
		}
	}
}