zoukankan      html  css  js  c++  java
  • HDU 1564 Play a game (找规律博弈)

    Play a game

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1030    Accepted Submission(s): 838


    Problem Description
    New Year is Coming! 
    ailyanlu is very happy today! and he is playing a chessboard game with 8600. 
    The size of the chessboard is n*n. A stone is placed in a corner square. They play alternatively with 8600 having the first move. Each time, player is allowed to move the stone to an unvisited neighbor square horizontally or vertically. The one who can't make a move will lose the game. If both play perfectly, who will win the game?
     
    Input
    The input is a sequence of positive integers each in a separate line. 
    The integers are between 1 and 10000, inclusive,(means 1 <= n <= 10000) indicating the size of the chessboard. The end of the input is indicated by a zero.
     
    Output
    Output the winner ("8600" or "ailyanlu") for each input line except the last zero. 
    No other characters should be inserted in the output.
     
    Sample Input
    2 0
     
    Sample Output
    8600
     
    Author
    ailyanlu
     
    Source
     
    Recommend
    8600

    从一个n*n的角落出发,每次移动到相邻的,而且没有经过的格子上。谁不能操作了谁输。

    结论就是n为偶数,先手赢,奇数,后手赢。

    可以看下图

    S表示起点。

    如果n为偶数,那么所有格子可以被2*1的砖块覆盖掉。

    这样先手每次都移动到当前1*2的另外一块。先手必赢。

    如果n为奇数。出了起始那个店,其余点都可以被覆盖。

    所有后手赢

    #include <string.h>
    #include <stdio.h>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    int main()
    {
        int n;
        while(scanf("%d",&n)==1 &&n)
        {
            if(n%2==0)printf("8600
    ");
            else printf("ailyanlu
    ");
        }
        return 0;
    }
  • 相关阅读:
    面试题21 包含min函数的栈
    面试题20 顺时针打印矩阵
    基于熵的方法计算query与docs相似度
    使用信息检索和深度学习方法的智能对话
    常用的激活函数
    spark实现smote近邻采样
    wide&deep用于ltr排序
    deepfm用于ltr排序
    lightgbm用于排序
    静态工厂方法+服务提供者框架模板
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3204654.html
Copyright © 2011-2022 走看看