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LightOJ1020(博弈)

編輯：C++入門知識

LightOJ1020(博弈)

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

直接找循環節

    /*************************************************************************
        > File Name: LightOJ1020.cpp
        > Author: ALex
        > Mail: [email protected]
        > Created Time: 2015年06月09日 星期二 18時46分32秒
     ************************************************************************/

    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include
    #include 
    #include 
    #include 

    using namespace std;

    const double pi = acos(-1.0);
    const int inf = 0x3f3f3f3f;
    const double eps = 1e-15;
    typedef long long LL;
    typedef pair  PLL;

    int main() {
        int t, icase = 1;
        scanf("%d", &t);
        while (t--) {
            int n;
            string s;
            cin >> n >> s;
            n %= 3;
            cout << "Case " << icase++ << ": ";
            if (s[0] == 'A') {
                if (n == 1) {
                    printf("Bob\n");
                }
                else {
                    printf("Alice\n");
                }
            }
            else {
                if (n == 0) {
                    printf("Alice\n");
                }
                else {
                    printf("Bob\n");
                }
            }
        }
        return 0;
    }