Hello everyone this is my first time posting, so please bare with me :P

So I was solving this problem and came up with a solution.

Idea —

say for example x = 7. if ai is of the form 7d + 2 then aj needs to be of the form 7q + '5'. Basically the remainders of ai % x and aj % x needs to add up to x for the sum to be divisible by x. also for ai — aj to be divisible by y, ai % y must be same as aj % y.

Initialize a 2d vector(rems) with all values set to 0.

Loop through the input vector and calculate input%y(remy) and input%x(remx) increment rems[remy][remx].

calculate all possibilities by multiplying rems[remy][remx] * rems[remx][x-remx] at every iteration and add that to ans variable. also subtract the result of product of last time when we iterated through the same rems[remy][remx].

if say x = 8 and remx = 4; the we have to select combinations of 2 elements so we do Nc2 for such values.

    long long n,x,y;
    cin >> n >> x >> y;
    vector<vector<long long>> v(n);
    for(int i = 1;i < n; ++i) cin >> v[i];

    vector<vector<long long>> rems(y, vector<long long> (x, 0)); //init a 2d array
    ll ans = 0;
    ll remx,remy;
    ll curr,curr2;

    for(int i = 0; i < n;++i){
    	remy = v[i]%y;
    	remx = v[i]%x;
    	rems[remy][remx]++;
    	curr = rems[remy][remx];	
    	if(remx && remx!= (x- remx)){
    		curr2 = rems[remy][x-remx];
	    	ans += (curr * curr2) - ((curr - 1) * (curr2)); // on simplification = curr2
    	}
    	else{
    		ans+= (curr * (curr - 1))/2 - ((curr - 1) * (curr -2)/2); //Nc2 - (N-1)C2
    	}
    }

    cout << ans << '\n';	

I am unable to identify where am I going wrong? Exactly where is it taking so long? Specifically it gives TLE on test case 3. Also I don't wanna spoil the solution to me by reading tutorial.

If I'm unclear in any part of solution please do let me know. I'll try my best to explain again. Thanks!