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  • Educational Codeforces Round 68 (Rated for Div. 2)-D. 1-2-K Game

    output
    standard output

    Alice and Bob play a game. There is a paper strip which is divided into n + 1 cells numbered from left to right starting from 0. There is a chip placed in the n-th cell (the last one).

    Players take turns, Alice is first. Each player during his or her turn has to move the chip 1, 2 or k cells to the left (so, if the chip is currently in the cell i, the player can move it into cell i - 1, i - 2 or i - k). The chip should not leave the borders of the paper strip: it is impossible, for example, to move it k cells to the left if the current cell has number i < k. The player who can't make a move loses the game.

    Who wins if both participants play optimally?

    Alice and Bob would like to play several games, so you should determine the winner in each game.

    Input

    The first line contains the single integer T (1 ≤ T ≤ 100) — the number of games. Next T lines contain one game per line. All games are independent.

    Each of the next T lines contains two integers n and k (0 ≤ n ≤ 109, 3 ≤ k ≤ 109) — the length of the strip and the constant denoting the third move, respectively.

    Output

    For each game, print Alice if Alice wins this game and Bob otherwise.

    Example
    input
    Copy
    4
    0 3
    3 3
    3 4
    4 4
    output
    Copy
    Bob
    Alice
    Bob
    Alice

     代码:

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<stack>
    #include<set>
    #include<vector>
    #include<map>
    #include<cmath>
    const int maxn=1e5+5;
    typedef long long ll;
    using namespace std;
     
    int main()
    {
        int T;
        scanf("%d", &T);
        while (T--)
        {
            int n, k;
            scanf("%d%d", &n, &k);
            if (k == 3)
            {
                if(n%4!=0)
                {
                    puts("Alice
    ");
                }
                else
                {
                    puts("Bob");
                }
     
            }
            else if (k % 3 == 0)
            {
                n %= (k + 1);
                if (n % 3 == 0 && n != k)
                    puts("Bob");
                else
                    puts("Alice");
            }
            else
            {
                if(n%3!=0)
                {
                    puts("Alice
    ");
                }
                else
                {
                    puts("Bob");
                }       
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/11190395.html
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