http://codeforces.com/contest/1029/problem/D

You are given an array $\mathrm{aa,\; consisting\; of\; nn\; positive\; integers.}$

Let‘s call a concatenation of numbers $\mathrm{xx\; and\; yy\; the\; number\; that\; is\; obtained\; by\; writing\; down\; numbers\; xx\; and\; yy\; one\; right\; after\; another\; without\; changing\; the\; order.\; For\; example,\; a\; concatenation\; of\; numbers}$

Count the number of ordered pairs of positions $(i,j)(i,j)\; (i\ne ji\ne j)\; in\; array\; aa\; such\; that\; the\; concatenation\; of\; aiai\; and\; ajaj\; is\; divisible\; by\; kk.$

The first line contains two integers $\mathrm{nn\; and\; kk\; (1\le n\le 2?1051\le n\le 2?105,\; 2\le k\le 1092\le k\le 109).}$

The second line contains

Print a single integer — the number of ordered pairs of positions $(i,j)(i,j)\; (i\ne ji\ne j)\; in\; array\; aa\; such\; that\; the\; concatenation\; of\; aiai\; and\; ajaj\; is\; divisible\; by\; kk.$

6 11
45 1 10 12 11 7

7

4 2
2 78 4 10

12

5 2
3 7 19 3 3

0

代碼：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 10;
int N, K;
int num[maxn];
map<long long, long long> mp[15];

int main() {
    scanf("%d%d", &N, &K);
    for(int i = 1; i <= N; i ++) {
        scanf("%d", &num[i]);
        long long a = num[i];
        for(int j = 1; j <= 10; j ++) {
            a *= 10;
            a %= K;
            mp[j][a] ++;
        }
    }
    long long cnt = 0;
    for(int i = 1; i <= N; i ++) {
        int t = num[i] % K;
        int len = log10(num[i]) + 1;
        cnt += mp[len][(K - t) % K];
        long long x = 1;
        for(int j = 1; j <= len; j ++)
            x = (x * 10) % K;
        if(((num[i] * x) % K + num[i] % K) % K == 0)
            cnt --;
    }
    printf("%I64d\n", cnt);
    return 0;
}

CodeForces D. Concatenated Multiples