lightoj1020 - A Childhood Game【简单博弈】

最新推荐文章于 2018-07-07 18:56:53 发布

最新推荐文章于 2018-07-07 18:56:53 发布 · 621 阅读

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#lightoj1020 - A Chil

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本文介绍了一个简单的回合制弹珠游戏，玩家轮流取走一到两颗弹珠，最终目标是确定谁会赢得游戏。文章详细解释了游戏规则、输入输出格式及示例，并提供了一段C++代码实现。

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1020 - A Childhood Game

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

Alice and Bob are playing a game with marbles; you may have played this game in childhood. The game is playing by alternating turns. In each turn a player can take exactly one or two marbles.

Both Alice and Bob know the number of marbles initially. Now the game can be started by any one. But the winning condition depends on the player who starts it. If Alice starts first, then the player who takes the last marble looses the game. If Bob starts first, then the player who takes the last marble wins the game.

Now you are given the initial number of marbles and the name of the player who starts first. Then you have to find the winner of the game if both of them play optimally.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer n (1 ≤ n < 2³¹) and the name of the player who starts first.

Output

For each case, print the case number and the name of the winning player.

Sample Input

Output for Sample Input

1 Alice

2 Alice

3 Bob

Case 1: Bob

Case 2: Alice

Case 3: Alice

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
#include<vector>
using namespace std;
int main()
{
	int t,i,j,k=1,n;
	char str[10];
	scanf("%d",&t);
	while(t--){
		scanf("%d%s",&n,str);
		printf("Case %d: ",k++);
		if(strcmp(str,"Alice")==0){
			if(n%3==1){
				printf("Bob\n");
			}
			else {
				printf("Alice\n");
			}
		}
		else {
			if(n%3==0){
				printf("Alice\n");
			}
			else {
				printf("Bob\n");
			}
		}
	}
	return 0;
}