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  • HDU 2147

    kiki's game

    Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others)
    Total Submission(s): 3808 Accepted Submission(s): 2210


    Problem Description
    Recently kiki has nothing to do. While she is bored, an idea appears in his mind, she just playes the checkerboard game.The size of the chesserboard is n*m.First of all, a coin is placed in the top right corner(1,m). Each time one people can move the coin into the left, the underneath or the left-underneath blank space.The person who can't make a move will lose the game. kiki plays it with ZZ.The game always starts with kiki. If both play perfectly, who will win the game?
     
    Input
    Input contains multiple test cases. Each line contains two integer n, m (0<n,m<=2000). The input is terminated when n=0 and m=0.

     
    Output
    If kiki wins the game printf "Wonderful!", else "What a pity!".
     
    Sample Input
    5 3 5 4 6 6 0 0
     
    Sample Output
    What a pity! Wonderful! Wonderful!
     
     
     1 //一切博弈都是找规律
     2 //题目的意思是从棋盘的最右上角到左下角,其中只可以走三个方向, 左边, 下边,右边,不能移动着失败,问先手是否胜利
     3 //根据博弈论的理论,先从左下角开始分析
     4 /*
     5 * 博弈论:组合博弈
     6 * 必败点(P点) :前一个选手(Previous player)将取胜的位置称为必败点。
     7 * 必胜点(N点) :下一个选手(Next player)将取胜的位置称为必胜点。
     8 * 必败(必胜)点的属性:
     9 * (1) 所有终结点是必败点(P点);
    10 * (2) 从任何必胜点(N点)操作,至少有一种方法可以进入必败点(P点);
    11 * (3)无论如何操作, 从必败点(P点)都只能进入必胜点(N点).
    12 * 由上面的属性得到该题的算法:
    13 * 步骤1:将所有终结位置标记为必败点(P点);
    14 * 步骤2: 将所有一步操作能进入必败点(P点)的位置标记为必胜点(N点)
    15 * 步骤3:如果从某个点开始的所有一步操作都只能进入必胜点(N点) ,则将该点标记为必败点(P点) ;
    16 * 步骤4: 如果在步骤3未能找到新的必败(P点),则算法终止;否则,返回到步骤2。
    17 * 由上面的算法计算一个例子:
    18 * 我们可以把问题转换成从(1,1)走到(n,m) (方便等下得出结论)
    19 * 但n=8,m=9的情况
    20 *  NNNNNNNNN
    21 *  PNPNPNPNP
    22 *  NNNNNNNNN
    23 *  PNPNPNPNP
    24 *  NNNNNNNNN
    25 *  PNPNPNPNP
    26 *  NNNNNNNNN
    27 *  PNPNPNPNP
    28 *初始点(1,1)为N所以输出Wonderful!
    29 *从这里例子就可以很清楚得看出当n和m都为奇数时,初始点(1,1)才会是P。
    30 *因此该题只需判断n,m是否同时为奇数即可。
    31 */
    32 #include<stdio.h>
    33 int main()
    34 {
    35   int n,m;
    36   while(scanf("%d%d",&n,&m)!=EOF&&n!=0&&m!=0)
    37   {
    38      if(m&1&&n&1)
    39           printf("What a pity!\n");
    40      else
    41           printf("Wonderful!\n");//行数或者列数为偶数就赢 
    42   }
    43   return 0;
    44 }
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  • 原文地址:https://www.cnblogs.com/hxsyl/p/2667888.html
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