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  • poj 2505 A multiplication game(博弈)

    A multiplication game
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 5622   Accepted: 2811

    Description

    Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n.

    Input

    Each line of input contains one integer number n.

    Output

    For each line of input output one line either
    Stan wins.
    or
    Ollie wins.
    assuming that both of them play perfectly.

    Sample Input

    162
    17
    34012226

    Sample Output

    Stan wins.
    Ollie wins.
    Stan wins.
    

    Source

    【思路】

           博弈

           2<=n<=9,先手必胜

           10<=n<=18,先手必败

           19<=n<=162,先手必胜

           163<=n<=324,先手必败

           …

           =-=

    【代码】

     1 #include<cstdio>
     2 
     3 int main() {
     4     double n;
     5     while(scanf("%lf",&n)==1) {
     6         while(n>18) n/=18;
     7         if(n<=9) puts("Stan wins.");
     8         else puts("Ollie wins.");
     9     }
    10     return 0;
    11 }
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  • 原文地址:https://www.cnblogs.com/lidaxin/p/5171496.html
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