Your text to link here... _(a+b) mod x = (a mod x+b mod x) mod _ 对于本题: (ai+aj) mod x = (ai mod x+aj mod x) mod x = 0 即ai mod x+aj mod x = 0 / x 变形为->ai mod x = (x-aj mod x) mod x (ai mod x+aj mod x=0时ai mod x = aj mod x = 0,依然满足上式)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int n,a[maxn],x,y;

void solve()
{
    cin >> n >> x >> y;
    for(int i = 1;i<=n;i++) cin >> a[i];
    long long ans = 0;
    map<pair<int,int>,int> mp;
    for(int i = 1;i<=n;i++)
    {
        int res1 = (x-a[i]%x)%x,res2 = a[i]%y;
        pair<int,int> p={res1,res2};
        ans += mp[p];
        p = {a[i]%x,a[i]%y};
        mp[p]++;
    }
    cout << ans << endl;
}

signed main()
{
    int t = 1;
    cin >> t;
    while(t--)
    {
        solve();
    }

    return 0;
}