2022 Henan Mengxin League game (3): Henan University J - magic number

J - Magic number

solution ： If the three numbers are equal , The result of the remainder of any number is the same , Output $-1$. If you don't want to wait , Think of any number can be written $x\times y+w$, do $a-b=(x1-x2)\times y$, this when $a,b$ Yes $x1-x2$, y The remainder result is the same , If it is $a,b,c$ Then take $gcd(abs(a-b),abs(b-c),abs(a-c))$ Then output all the quality factors .

#include<bits/stdc++.h>
using namespace std;

int main()
{
    
	int T;cin>>T;
	while(T--)
	{
    
		int a,b,c;cin>>a>>b>>c;
		if(a==b&&a==c) cout<<"-1"<<endl;
		else
		{
    
			int d1=abs(a-b),d2=abs(b-c);
			int t=__gcd(d1,d2);
			set<int> s;
			for(int i=1;i*i<=t;i++)
				if(t%i==0)
				{
    
					s.insert(i);
					if(i*i!=t) s.insert(t/i);
				}	
			for(auto &p:s) 
				cout<<p<<" ";
			cout<<endl;
		}
	}
	return 0;
}