This problem did not need so much trouble, you only need to look at the reasoning:
Suppose we have two numbers \ (the X-, the y-\) , first \ (x \) is set to a large value, \ (the y-\) is set to a smaller value. Now divided into three cases:
\ (1 \) . If the number is a multiple of two relations, one winning operation.
\ (2 \) . If the two quotient \ (> 1 \) , it is still operating the winning party.
\(why?\)
Take for example \ (25,7 \) is. In this case the operation side has three options: \ (18 \) \ (7 \) , \ (11 \) \ (7 \) , \ (4 \) \ (7 \)
If he chose \ (18 \) \ (7 \) , then the latter will face the \ (11 \) \ (7 \) or \ (4 \) \ (7 \) ; and if he does not choose \ (18 \) \ (7 \) , then his face or \ (11 \) \ (7 \) or \ (4 \) \ (7 \) .
At this point you will find that \ (11 \) \ (7 \) and \ (4 \) \ (7 \) is a must win, but both smart enough, so who has the right to choose who will be able win! That he can not choose \ (18 \) \ (7 \) !
Again give a chestnut: \ (31 \) \ (6 \)
Then you will find, just get party only option \ (7 \) \ (6 \) or \ (1 \) \ (6 \) to ensure control of his hands, but apparently \ (1 \) \ (6 \) is not enough, you can only choose \ (7 \) \ (6 \) . Thus \ (7 \) \ (6 \) \ (- \) \ (6 \) \ (1 \) \ (- \) \ (6 \) \ (0 \) , the result is the upper hand to win!
Then you should know: who has the right to choose (select two or more) who will win!
\ (3 \) . Quotient is \ (1 \) , continued
For chestnuts, such as \ (6 \) \ (4 \) , then the upper hand no choice, it will only continue to argue, as \ (2 \) \ (4 \) .
\(Code:\)
#include<bits/stdc++.h>
using namespace std;
bool check(int x,int y){
for(int i=1;;i++){
int ma=max(x,y);
int mi=min(x,y);
x=ma,y=mi;
//x为两数较大值，y为两数较小值
if(x%y==0){
return i%2;//若两数为倍数关系，操作的一方赢
}else if(x/y>1){
return i%2;//若两数商>1，那么还是操作一方赢
}else{
x-=y;//否则说明商为1，那就继续
}
}
}
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++){
int x,y;
cin>>x>>y;
if(check(x,y))cout<<"Stan wins"<<endl;
else cout<<"Ollie wins"<<endl;
}
return 0;
}