4387 Stone Game

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本文介绍了一款游戏中的策略问题，玩家Alice和Bob在一个1*N格的棋盘上进行游戏，每人各有K颗棋子，分别放置于棋盘两端，双方轮流移动己方棋子到相邻空位，直至一方无法移动为止。文章提供了游戏胜负判断的算法实现。

#这是一道找规律的题。

#问题

Stone Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 943 Accepted Submission(s): 399

Problem Description

　　Alice and Bob are playing a game. It is played in 1*N grids. Each grid can be occupied by one stone. Alice has K white stones on the left (numbered 1 to K from left to right), and Bob has K black stones on the right (numbered 1 to K from right to left). They take turns to move their own stones, and Alice moves first. In each move, the player must choose one stone to move to the nearest empty grid forward (Alice moves to the right, Bob moves to the left). If one player cannot find any stone to move, he wins.
　　Now Alice asks you to find a winning strategy of the game. Can you help him?

Input

　　There are multiple test cases. In each case, there is one line containing two integers N, K.

Technical Specification
　　3 <= N <= 1,000,000, 1 < K*2 < N

Output

　　For each case, print in one line containing the case number (starting with 1) and the winning information. If Alice loses, just print “Bob”, otherwise print “Alice” and the first stone he chooses. If there are multiple stones he can choose, he will choose the rightmost one.

Sample Input

3 1
4 1

Sample Output

Case 1: Bob
Case 2: Alice 1

#AC码

#include<iostream>
using namespace std;
int main()
{
    int n,k,cn=1;
    while(cin >>n>>k)
    {
        printf("Case %d: ",cn++);
        if(k==1)
        {
            if(n%2)
                cout<<"Bob"<<endl;
            else
                cout<<"Alice 1"<<endl;
        }
        else if(n==2*k+1)
            cout<<"Alice "<<k<<endl;
        else
            cout<<"Alice 1"<<endl;
    }
    return 0;
}