CodeForces 1417B Two Arrays

Topic linking
The question ：
Give you a group number and an unfortunate number T, It is required to divide the array into two parts , Marked as 1or0, A fellow 1 perhaps 0 The sum of the two numbers of is not equal to T, Find any kind of division .
Their thinking ：
T/2+T/2=T, In this case , Put all less than T/2 Put aside the number of , Greater than T/2 Put the number of on the other side , In this way, no matter how you add it, it will not be added T. Exactly equal to T/2 The data of is 0 One for 1 In this way, it is opened alternately .

#include<bits/stdc++.h>
#define maxn 1000001
using namespace std;
long long int a[maxn];//int Meeting TE
int main()
{
    
	int num;
	cin>>num;
	while(num--)
	{
    
		long long int n,T;
		cin>>n>>T;
		for(int i=1;i<=n;i++)
		{
    
			cin>>a[i];
		}
		
		int x=1;
		for(int i=1;i<=n;i++)
		{
    
			if(T%2!=0)
			{
    
				int t=T/2;
				if(a[i]>t)
				{
    
					a[i]=1;
				}
				else
				{
    
					a[i]=0;
				}
			}
			else 
			{
    
				
				int t=T/2;
				if(a[i]>t)
				{
    
					a[i]=1;
				}
				else if(a[i]<t)
				{
    
					a[i]=0;
				}
				else
				{
    
					a[i]=x;
					x=1-x;
				}			
			}
		}
		for(int i=1;i<=n;i++)	
		{
    
			cout<<a[i]<<" ";
		}
		cout<<endl;
	}
	
	
	return 0;
}