题目描述：
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
分析：
若存在x=k*y+r(k>1)则谁先面对x=k*y+r(k>1)，即x>2*y的局势谁赢,
因为此时可以拿成 (x%y,y) 或 (x%y+y,y)这两种局势，而这两种局势中要达到最终（num,0）步数必然
一奇一偶(相差一步)，则此时只要选择自己取胜的最优策略就可以了
当然若一开始x<2*y，则看到达(num,0)的步数的奇偶性来决定胜负
而过程类似于欧几里得算法（估计也是这道题为何叫欧几里得的游戏的原因所在），递归就行了。
伪代码：
f(a,b)
if a%b == 0 --->先手赢
else if a<2*b ---> 递归
else if a>2*b --->当前者赢
end if
代码：
#include<cstdio> #include<algorithm> using namespace std; int ice(int a,int b){ if(a < b) swap(a,b); if(a % b == 0) return 1; else if(a < 2*b)//a=b+r,k==1;记步数 return !(ice(a-b,b));//先手赢返回1，否则返回0； return 1;//a>2*y,直接先手赢。 } int main(){ int a,b,c; while(scanf("%d%d",&a,&b) != EOF){ if(a == 0 && b == 0) break; c = ice(a,b); if(c) printf("Stan wins "); else printf("Ollie wins "); } return 0; }