Count length-N digit strings whose value is divisible by B, using only the allowed digits. The first digit may be zero. A state is a remainder modulo B; appending digit d to remainder r produces (10r + d) % B.
Define T[next][current] as the number of allowed digits that move current to next. Thus T[next][current] counts digits d satisfying next = (10 * current + d) % B. The vector starts with one empty prefix at remainder zero. Multiplying by T once appends one digit, so after N transitions the answer is the count at remainder zero. Binary exponentiation computes T^N applied to the start vector. Repeated values in the input digit list are treated as one allowed digit.
Matrix multiplication and exponentiation take O(B^3 log N) time; reading the allowed digits takes O(K). The matrices use O(B^2) space, which is practical for B <= 100.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
import java.io.*;
import java.util.*;
public class Main {
static final long MOD = 1_000_000_007L;
static long[][] multiply(long[][] a, long[][] b) {
int size = a.length;
long[][] product = new long[size][size];
for (int row = 0; row < size; row++) {
for (int middle = 0; middle < size; middle++) {
if (a[row][middle] == 0) continue;
for (int column = 0; column < size; column++) {
product[row][column] =
(product[row][column] + a[row][middle] * b[middle][column]) % MOD;
}
}
}
return product;
}
static long[] multiply(long[][] matrix, long[] vector) {
int size = vector.length;
long[] product = new long[size];
for (int row = 0; row < size; row++) {
for (int column = 0; column < size; column++) {
product[row] = (product[row] + matrix[row][column] * vector[column]) % MOD;
}
}
return product;
}
public static void main(String[] args) throws IOException {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer firstLine = new StringTokenizer(input.readLine());
long n = Long.parseLong(firstLine.nextToken());
int b = Integer.parseInt(firstLine.nextToken());
int k = Integer.parseInt(firstLine.nextToken());
boolean[] allowed = new boolean[10];
StringTokenizer digits = new StringTokenizer(input.readLine());
for (int i = 0; i < k; i++) {
allowed[Integer.parseInt(digits.nextToken())] = true;
}
long[][] transition = new long[b][b];
for (int current = 0; current < b; current++) {
for (int digit = 0; digit <= 9; digit++) {
if (allowed[digit]) {
int next = (10 * current + digit) % b;
transition[next][current]++;
}
}
}
long[] ways = new long[b];
ways[0] = 1;
while (n > 0) {
if ((n & 1) != 0) {
ways = multiply(transition, ways);
}
n >>= 1;
if (n > 0) {
transition = multiply(transition, transition);
}
}
System.out.println(ways[0]);
}
}