lightoj 1020 水博弈

最新推荐文章于 2018-04-07 22:54:54 发布

原创最新推荐文章于 2018-04-07 22:54:54 发布 · 380 阅读

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本文介绍了一个简单的博弈论游戏，玩家轮流取走1至2颗弹珠，初始弹珠数和先手玩家已知。文章提供了游戏胜负的判断逻辑及代码实现，并分析了几种特殊情况。

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
3
1 Alice
2 Alice
3 Bob
Case 1: Bob
Case 2: Alice
Case 3: Alice

只举一个bob先去的例子就够了
1.2的时候是必胜态
3的时候只能从1.2这两个状态上转移显然都是输
4的时候能从2.3这两个状态转移
只要有一个能赢他就能赢…

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<queue>
#include<string>
using namespace std;
int main()
{
    int T;
    cin>>T;
    int u=0;
    while(T--)
    {
        int n;
        string q;
        cin>>n>>q;
        printf("Case %d: ",++u);
        if(q=="Bob")
        {
            if(n%3==0)cout<<"Alice"<<endl;
            else cout<<"Bob"<<endl;
        }
        else
        {
            if(n%3==1)cout<<"Bob"<<endl;
            else cout<<"Alice"<<endl;
        }
    }
    return 0;
}