A Multiplication Game

A Multiplication Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2264    Accepted Submission(s): 1300

Problem 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.

 

 分析：

大致题意：Stan和Ollie玩相乘游戏，一开始Stan在1的基础上乘2~9的整数，然后轮到Ollie·····谁先将得数超过或等于所给的数，就算赢

思路分析：先来看看基本的规律

给的数为2~9 Stan wins

           10~18 Ollie wins

那下一个呢？

由上面可得：

9=9；

18=2×9；

就是Stan尽量小，而Ollie为了超过所给的数，尽量大

那数再大一点，超过了18，Ollie就没辙了

因为Stan取2，而Ollie最大取9，才只能到达18；那如果Ollie也取2，Stan取9也能到36

如果数再大一点，超过36，Stan也可以根据第一次的取值来增大后来的乘数

那也总有个值，Stan取不到吧。其实就是9(S)×2(O)×9(S)=162;

所以后面也就这样轮流寻找了

19~162 Stan wins   162=9×2×9；

163~324 Ollie wins   324=2×9×2×9;

 `````````````````

而且很有规律的，Stan和AOllie就这样交换着，我们可就这一点将所给的数重复除以18，将结果控制到18里来讨论

例如：19~162  除以18后，就是1.055556~9，正是在Stan赢的范围内

代码如下：

# include<stdio.h>
int main()
{
    double num;
    while(scanf("%lf",&num)!=EOF)
    {
        while(num>18)
        num/=18;
        if(num>9)
            printf("Ollie wins.\n");
        else
            printf("Stan wins.\n");
    }
    return 0;
}