time limit per test
2 seconds
memory limit per test
256 megabytes
Polycarp has two favorite integers x and y (they can be equal), and he has found an array a of length n.
Polycarp considers a pair of indices ⟨i,j⟩ (1≤i<j≤n) beautiful if:
- ai+aj is divisible by x;
- ai−aj is divisible by y.
For example, if x=5, y=2, n=6, a=1,2,7,4,9,6, then the only beautiful pairs are:
- ⟨1,5⟩: a1+a5=1+9=10 (10 is divisible by 5) and a1−a5=1−9=−8 (−8 is divisible by 2);
- ⟨4,6⟩: a4+a6=4+6=10 (10 is divisible by 5) and a4−a6=4−6=−2 (−2 is divisible by 2).
Find the number of beautiful pairs in the array a.
Input
The first line of the input contains a single integer t (1≤t≤104) --- the number of test cases. Then the descriptions of the test cases follow.
The first line of each test case contains three integers n, x, and y (2≤n≤2⋅105, 1≤x,y≤109) --- the size of the array and Polycarp's favorite integers.
The second line of each test case contains n integers a1,a2,...,an (1≤ai≤109) --- the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.
Output
For each test case, output a single integer --- the number of beautiful pairs in the array a.
Example
Input
Copy
7
6 5 2
1 2 7 4 9 6
7 9 5
1 10 15 3 8 12 15
9 4 10
14 10 2 2 11 11 13 5 6
9 5 6
10 7 6 7 9 7 7 10 10
9 6 2
4 9 7 1 2 2 13 3 15
9 2 3
14 6 1 15 12 15 8 2 15
10 5 7
13 3 3 2 12 11 3 7 13 14
Output
Copy
2
0
1
3
5
7
0解题说明：此题是一道数学题，可以进行预处理出每个数分别摸上xy的值，用map存一下，然后遍历每个数，如果a + b是x的倍数的话，那么他们模x的值相加为x，如果a - b是y的倍数的话，那么他们的模y的值相等。
cpp
#include<bits/stdc++.h>
#include<algorithm>
#include<map>
#include<iostream>
using namespace std;
long long n, x, y, c, i;
int a[200001];
int main()
{
int t = 1;
cin >> t;
while (t--)
{
cin >> n >> x >> y;
map<int, map<int, int>>m;
for (i = c = 0; i < n;)
{
cin >> a[i];
c += m[(x - a[i] % x) % x][a[i] % y];
m[a[i] % x][a[i++] % y]++;
}
cout << c << '\n';
}
return 0;
}