lightoj 1020 - A Childhood Game 【博弈】

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

 

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PROBLEM SETTER: JANE ALAM JAN

题意：有n个石头，每次可以从中选1个或2个。现有两个人Alice 和 Bob，若Alice先手，则最后取完石头者输，若Bob先手，则最后取完石头者赢。给定石头数目和先手者，问谁能赢。 假设两个人走的每一步都是完美的。

思路：拿Alice先手来说

n为1时，后手赢。当n为2或3时，先手者总是可以取走一定数目的石头使得剩余石头数为1，这样先手赢。

n为4时，后手赢。当n为5或6时，先手者总是可以取走一定数目的石头使得剩余石头数为4，这样先手赢。

依次推下去，发现n % 3 == 1时后手赢，反之先手赢。

Bob先手 同样的推导过程

n为1或2时，先手赢。

n为3时，后手赢。当n为4或5时，先手者总是可以取走一定数目的石头使得剩余石头数为3，这样先手赢。

得到规律n % 3 == 0时 后手赢，反之先手赢。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN 500000+10
#define MAXM 50000000
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
using namespace std;
int main()
{
    int t, kcase = 1;
    Ri(t);
    W(t)
    {
        int n; char op[6];
        Ri(n), Rs(op);
        if(strcmp(op, "Alice") == 0)
        {
            if(n % 3 == 1)
                printf("Case %d: %s\n", kcase++, "Bob");
            else
                printf("Case %d: %s\n", kcase++, "Alice");
        }
        else
        {
            if(n % 3 == 0)
                printf("Case %d: %s\n", kcase++, "Alice");
            else
                printf("Case %d: %s\n", kcase++, "Bob");
        }
    }
    return 0;
}