T*log n复杂度都会TLE最后一点

#include <bits/stdc++.h>
using namespace std;
#define int long long
int n;
signed main()
{
	int T;
	cin>>T;
	while(T--)
	{
		scanf("%d",&n);
		while(n%2==0)
		{
			n/=2;
		}
		if(n==1||n-2==1) cout<<"0\n";
		else cout<<"1\n";
	}
	return(0);
}