POJ 2505

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.

因为每个人都是聪明的，所以如果当前的p乘以9之后还不能>=n,那么他将选择乘以2，选择其他的数都将是对方赢的概率增加。

故我们可以从给定的数n倒推回来，首先[(n+8)/9,n-1] 的区间是必胜的区间，然后[(n+8)/9/2,(n+8)/9-1]是必败区间，以此类推下去，记录经过多少次就可以达到起始状态。

根据奇偶性来判断输赢。

#include <iostream>
using namespace std;
int main()
{
	__int64 p;
	while(scanf("%I64d",&p))
	{
		int t=0;
		while(p!=1)
		{
			if(t%2==0)
				p=(p+8)/9;
			else
				p=(p+1)/2;
			t++;
		}
		puts(t%2?"Stan wins.":"Ollie wins.");
	}
	return 0;
}