题干：

Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7): 

25 7 
11 7 
4 7 
4 3 
1 3 
1 0 

an Stan wins. 
 

Input

The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.

Output

For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed. 
 

Sample Input

34 12
15 24
0 0

Sample Output

Stan wins
Ollie wins

题目大意：

   给出两个数，a和b，将大的数中，减去若干b的倍数，最终有一个数为0的话就胜了。

解题报告：

贴bin巨博客：%bin巨

假设两个数为a,b（a>=b)

如果a==b.那么肯定是先手获胜。一步就可以减为0,b

如果a%b==0.就是a是b的倍数，那么也是先手获胜。

如果a>=2*b.  那么 那个人肯定知道a%b,b是必胜态还是必败态。如果是必败态，先手将a,b变成a%b,b,那么先手肯定赢。如果是必胜态，先手将a,b变成a%b+b,b.那么对手只有将这两个数变成a%b,b,先手获胜。

如果是b<a<2*b  那么只有一条路：变成a-b,b  (这个时候0<a-b<b).这样一直下去看谁先面对上面的必胜状态。

所以假如面对b < a <2*b的状态，就先一步一步走下去。直到面对一个a%b==0 || a >=2*b的状态。

AC代码：

#include<bits/stdc++.h>using namespace std;int main()
{int a,b;while(scanf("%d%d",&a,&b)) {if(a == 0 && b == 0)break;if(a < b) swap(a,b);int win = 0;while(b) {if(a%b == 0 || a/b >= 2) break;a = a-b;swap(a,b);win ^= 1;}if(win == 0) puts("Stan wins");else puts("Ollie wins");}return 0;
}

总结： 

   找必败点就好了。看题解总是感觉有道理，，，但是就是自己想不出来。。。难受难受