15年山东第六届acm省赛 C题 Game！（博弈）

本文介绍了一种基于博弈论的游戏，玩家轮流从一圈石头中取走一个或两个相邻的石头，探讨了最优策略下胜者归属的问题。通过分析游戏规则，得出结论：当初始石头数量为1或2时，先手玩家获胜；否则后手玩家获胜。

Game!

Time Limit: 1000MS Memory Limit: 65536KB

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Problem Description

One day, zbybr is playing a game with blankcqk, here are the rules of the game:

There is a circle of N stones, zbybr and blankcqk take turns taking the stones.

Each time, one player can choose to take one stone or take two adjacent stones.

You should notice that if there are 4 stones, and zbybr takes the 2nd, the 1st and 3rd stones are still not adjacent.

The winner is the one who takes the last stone.

Now, the game begins and zbybr moves first.

If both of them will play with the best strategy, can you tell me who will win the game?

Input

The first line of input contains an integer T, indicating the number of test cases (T≈100000).

For each case, there is a positive integer N (N ≤ 10^18).

Output

Output the name of the winner.

Example Input

2
1
2

Example Output

zbybr
zbybr

分析：博弈题，难推易实现，成环的两个之间被拿走一个，不是相邻的，所以长度为n的拿走后有两种状态：长度为n-1的链子合或是长度为n-2的链子；

代码如下：

#include<stdio.h>

int main()
{
	int a,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&a);
		if(a==1 ||a==2)
		{
			printf("zbybr\n");
		}
		else
		{
			printf("blankcqk\n");
		}
	}
	return 0;
}