1931D - Divisible Pairs - CodeForces Solution

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <string>

using namespace std;

#define ll long long int
#define pb push_back
#define mp make_pair
#define vi vector<ll>
#define vs vector<string>
#define v(t) vector<t>
#define mii map<ll, ll>
#define umii unordered_map<ll, ll>
#define si set<ll>
#define usi unordered_set<ll>
#define pq priority_queue<ll>
#define pqs priority_queue<ll, vi, greater<ll>>
#define all(v) v.begin(), v.end()
#define ff first
#define ss second
#define loop(i, a, b) for (ll i = a; i < b; i++)
#define loopr(i, a, b) for (ll i = a; i > b; i--)
#define print(v) for (auto x : v) cout << x << " ";
#define endl "\n"
#define fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);

#define MOD 1000000007
#define PI 3.1415926535897932384626
ll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

void solve() {
    ll n, x, y, a, a_x, a_y, b_x, b_y, ans = 0;
    cin >> n >> x >> y;
    map<pair<int, int>, int> count;
    loop(i, 0, n) {
        cin >> a;
        a_x = a % x, a_y = a % y;
        if (a_x == 0) b_x = 0; else b_x = x - a_x;
        if (a_y == 0) b_y = 0; else b_y = a_y;
        ans += count[make_pair(b_x, b_y)];
        count[make_pair(a_x, a_y)]++;
    }

    cout << ans << endl;
}

int main() {
    fast

    #ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	#endif

    ll t;
    cin >> t;

    while (t--) {
        solve();
    }

    // solve();

    return 0;
}