暴力O(n * n)

#include <iostream>
#include <cmath>
using namespace std;
const int N = 1E5 + 10;
typedef long long ll;
int arr[N];
int main(){
    int n, k;   cin >> n >> k;
    for(int i = 0; i < n; i ++ )    cin >> arr[i];
    ll cnt = 0;
    for(int i = 0; i < n - 1; i ++ ){
        for(int j = i + 1; j < n; j ++ ){
            if(arr[i] % k == 0 && arr[j] % k == 0){
              cnt += 2;
              continue;
            }  
            ll f = log10(arr[i]), s = log10(arr[j]);
            ll num1 = arr[i] * ll(pow(10, s + 1)) + arr[j];
            ll num2 = arr[j] * ll(pow(10, f + 1)) + arr[i];
            if(num1 % k == 0)   cnt ++ ;
            if(num2 % k == 0)   cnt ++ ;

        }
    }
    cout << cnt << endl;
    return 0;
}

优化

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
typedef long long ll;
ll arr[N];
ll cnt[11][N];
ll res = 0;
int n, k;  
void work(){
    for(int i = 0; i < n; i ++ ){
        int cnt_a = log10(arr[i]) + 1; // a的位数
        res += cnt[cnt_a][(k - (arr[i] % k)) % k];  // 寻找余数为 k - (arr[i] % k) 的数的个数
        for(ll j = 1, power = 10; j <= 10; j ++ ){
            ll num = (arr[i] % k) * power % k;
            cnt[j][num] ++ ;
            power = power * 10; 
        }
    }
}
int main(){
    cin >> n >> k;
    for(int i = 0; i < n; i ++ )    cin >> arr[i];
    work();
    reverse(arr, arr + n);
    memset(cnt, 0, sizeof cnt);
    work();
    cout << res << endl;
    return 0;
}