和一个数位dp里的取模方式很像

// A[i] * 10^len(j) + A[j] == 0(mod k)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int N = 100005;
LL n, k, a[N], s[N][11]; // x*10^j % k == i 个数
LL mod(LL x, LL p){
    return (x % p + p) % p;
}
int main(){
    scanf("%lld%lld", &n, &k);
    for(int i = 0; i < n; i ++ )
        scanf("%lld", &a[i]);

    for(int i = 0; i < n; i ++ ){
        LL t = a[i] % k;
        for(int j = 0; j < 11; j ++ ){
            s[t][j] ++;
            t = t * 10 % k;
        }
    }

    LL res = 0;
    for(int i = 0; i < n; i ++ ){
        int len = log10(a[i]) + 1;
        LL t = a[i] % k;
        LL x = mod(-t, k);
        res += s[x][len];

        while(len -- ) t = t * 10 % k;
        if(t == x) res --;
    }
    cout << res;
    return 0;
}