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
题意：
给出两个数a,b，每次操作，大的数减掉小的数的整数倍。一个数变为0 的时候结束。先把其中一个数减为0的获胜。Stan固定先手，输出赢的人。
题解：
假设 a>=b
如果a==b 或者说 a%b==0 也就是a是b的倍数，那么肯定先手获胜。
如果 a>=2*b 那么当前那个人肯定知道 （b,a%b） 是必胜态还是必败态。如果是必败态，先手将a,b变成（b,a%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（b,a%b）,先手获胜。也就是说 a >=2*b 时先手必胜。
如果是b<a<2*b 那么只有一条路：变成 （b,a%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，就先走下去。直到面对一个 a%b==0 || a >=2*b 的状态。
代码：
#include <cstdio>
#include <algorithm>
using namespace std;
int main(){
int A,B;
while(scanf("%d %d",&A,&B) && (A||B)){
if(A < B)swap(A,B);
bool flag = true;
while(1){
if(A%B==0 || A/B>=2)break;
else {
int t = A;
A = A % B;
swap(A,B);
flag = !flag;
}
}
if(flag)printf("Stan wins
");
else printf("Ollie wins
");
}
return 0;
}