• 1404-Digital Deletions


    Digital deletions is a two-player game. The rule of the game is as following.

    Begin by writing down a string of digits (numbers) that's as long or as short as you like. The digits can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and appear in any combinations that you like. You don't have to use them all. Here is an example:

    1404-Digital <wbr>Deletions


    On a turn a player may either:
    Change any one of the digits to a value less than the number that it is. (No negative numbers are allowed.) For example, you could change a 5 into a 4, 3, 2, 1, or 0.
    Erase a zero and all the digits to the right of it.


    The player who removes the last digit wins.


    The game that begins with the string of numbers above could proceed like this:

    1404-Digital <wbr>Deletions


    Now, given a initial string, try to determine can the first player win if the two players play optimally both.
     

    Input
    The input consists of several test cases. For each case, there is a string in one line.

    The length of string will be in the range of [1,6]. The string contains only digit characters.

    Proceed to the end of file.
     

    Output
    Output Yes in a line if the first player can win the game, otherwise output No.
     

    Sample Input
    00 
    20
    Sample Output
    Yes 
    Yes 
    No 
    No
    组合博弈,暴力打表
    #include
    #include
    int len;
    int win[1000000];
    char s[10];
    int solve(int num)
    {
        int i,j,k,k2=num,p=1;
        while(k2)
        {

            k=num/p;
            if(k==0)
            {
                if(!win[num/p/10])
                  return 1;
            }
            else
            {
                for(i=1;i<=k;i++)
                {
                    if(!win[num-i*p] &&  !(i==k && k2/10==0))
                     return 1;
                }

            }
            k2/=10;
            p*=10;
        }
        return 0;
    }
    int main()
    {
        int i,j,k;
        win[0]=1;
        for(i=1;i<1000000;i++)
           if(solve(i))
             win[i]=1;
        while(~scanf("%s",s+1))
        {
            if(s[1]=='0')
            {
                printf("Yes ");
                continue;
            }
            len=strlen(s+1);
            k=0;
            for(i=1;i<=len;i++)
               k=k*10+s[i]-'0';
            if(win[k])
              printf("Yes ");
            else
              printf("No ");
        }
    }

  • 相关阅读:
    jQuery 间歇式无缝滚动特效分享(三张图片平行滚动)
    百度网页分享js代码
    如何在linux中搭建JEECMS系统
    Python菜鸟之路:Python基础-类(2)——成员、成员修饰符、异常及其他
    Python菜鸟之路:Python基础-类(1)——概念
    Python菜鸟之路:Python基础-生成器和迭代器、递归
    Python菜鸟之路:Python基础-逼格提升利器:装饰器Decorator
    Python菜鸟之路:Python基础-内置函数补充
    Python菜鸟之路:Python基础——函数
    Python菜鸟之路:Python基础(三)
  • 原文地址:https://www.cnblogs.com/kuroko-ghh/p/9363388.html
Copyright © 2020-2023  润新知