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

    Play a game

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1430    Accepted Submission(s): 1168


    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   |   We have carefully selected several similar problems for you:  1404 1536 1517 1524 1729 
     
    规律就是如果是奇数的话就后手赢,偶数的话时先手赢
    一开始还以为要搜索= =
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<stdlib.h>
     4 #include<algorithm>
     5 using namespace std;
     6 int main()
     7 {
     8     int n;
     9     while(scanf("%d",&n)&&n)
    10     {
    11         if(n%2==0)
    12             printf("8600
    ");
    13         else
    14             printf("ailyanlu
    ");
    15     }
    16     return 0;
    17 }
    View Code
  • 相关阅读:
    request、bs4爬虫
    1031 查验身份证
    1029 旧键盘
    1028 人口普查
    1027 打印沙漏
    1026 程序运行时间
    1025 反转链表
    1024 科学计数法
    1022 D进制的A+B
    1021 个位数统计
  • 原文地址:https://www.cnblogs.com/clliff/p/3928377.html
Copyright © 2011-2022 走看看