Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.Arrays;
import java.util.HashMap;
import java.util.Optional;
import java.util.Scanner;

public class Main {
    void run() {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int P = sc.nextInt();
        String s = sc.next();

        HashMap<Integer, Integer> cnt = new HashMap<>();
        if (P == 2 || P == 5) {
            long ans = 0;
            for (int i = 0; i < N; i++) {
                int x = s.charAt(i) - '0';
                if (x % P == 0) {
                    ans += i + 1;
                }
            }
            System.out.println(ans);
            return;
        }

        long ans = 0;
        int cur = 0;
        int pow = 1;
        for (int i = N - 1; i >= 0; i--) {
            int x = s.charAt(i) - '0';
            cur = (cur + x * pow) % P;
            int v = cnt.getOrDefault(cur, 0);
            cnt.put(cur, v + 1);
            ans += v;
            if (cur == 0) ans++;
            pow = (pow * 10) % P;
        }

        System.out.println(ans);
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }

}