2505 A multiplication game

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本文介绍了一个名为“乘法游戏”的简单博弈问题，通过玩家轮流将整数与2到9之间的数字相乘来决定胜负。文章提供了一段C++代码，用于判断在完美策略下哪位玩家能够首先达到或超过预先设定的阈值。

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A multiplication game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3560		Accepted: 1669

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.

#include<iostream>
using namespace std;
int check(long long n)
{
    if(n<=18&&n>9) return 1;
    else if(n<=9&&n>1) return 0;
    if(n%9==0) n=n/9;
    else n=n/9+1;
    if(n%2==0) n=n/2;
    else n=n/2+1;
    return check(n);
}

int main()
{
    long long n;
    while(cin>>n)
    {
        if(check(n)==1)  cout<<"Ollie wins."<<endl;
        else cout<<"Stan wins."<<endl;
    }
    cout<<endl;
    return 0;
}