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: Wrong Answer)


import java.io.*;
import java.text.*;
import java.util.*;

public class Main {

    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        solver s = new solver();
        int t = 1;
        while (t > 0) {
            s.solve();
            t--;
        }
        out.close();
    }

    static class InputReader {

        public BufferedReader br;
        public StringTokenizer token;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String sn() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int ni() {
            return Integer.parseInt(sn());
        }

        public String snl() throws IOException {
            return br.readLine();
        }

        public long nlo() {
            return Long.parseLong(sn());
        }

        public double nd() {
            return Double.parseDouble(sn());
        }

        public int[] na(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.ni();
            return a;
        }

        public long[] nal(int n) {
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nlo();
            return a;
        }
    }

    static int inf = 5000013;

    static class solver {
        DecimalFormat df = new DecimalFormat("0.00000000");
        long max = 1000000000000000l, min = -10000000000000000l;
        long mod = (long) 1e9 + 7;

        long pro(long a, long b) {
            return (a % mod * b % mod) % mod;
        }

        long add(long a, long b) {
            return (a + b) % mod;
        }

        long sub(long a, long b) {
            return ((a % mod - b % mod) + mod) % mod;
        }

        long ncr3(long n) {
            return (n * (n - 1) * (n - 2)) / 6l;
        }

        long ncr2(long n) {
            return n * (n - 1) / 2l;
        }

        int modPow(int x, int y) {
            int res = 1;
            while (y > 0) {
                if (y % 2 != 0)
                    res *= x;
                x = x * x;
                y = y >> 1;
            }
            return res;
        }

        int gcd(int a, int b) {
            if (b == 0)
                return a;
            return gcd(b, a % b);
        }

        final int maxn = 1000005;

        void solve() {
            int n = sc.ni();
            int p = sc.ni();
            char s[] = sc.sn().toCharArray();
            HashMap<Integer, Integer> map = new HashMap<>();
            long ans = 0;
            if (p == 2) {
                for (int i = 0; i < n; i++) {
                    if ((s[i] - '0') % 2 == 0)
                        ans += i + 1;
                }
            } else if (p == 5) {
                for (int i = 0; i < n; i++) {
                    if ((s[i] - '0') % 5 == 0)
                        ans += i + 1;
                }
            } else {
                int t = 0;
                map.put(0, 1);
                for (int i = 0; i < n; i++) {
                    t = (t * 10 + (s[i] - '0')) % p;
                    if (map.containsKey(t))
                        ans += map.get(t);
                    map.put(t, map.getOrDefault(t, 0) + 1);
                }
            }
            out.println(ans);
        }
    }
}