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.util.*;

public class Main {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int P = sc.nextInt();
        String S = sc.next();
        Map<Integer, Integer> mp = new HashMap<>();
        int[] mods = new int[S.length()+1];
        mods[S.length()] = 0;
        mp.put(0, 1);
        for(int i=S.length()-1;i>=0;i--){
            int num = S.charAt(i)-'0';
            mods[i] = mod10(num, S.length()-i-1, P) + mods[i+1];
            mods[i] %= P;
//            mods[i] = (mods[i+1]+num*10) % P;
            mp.put(mods[i],mp.computeIfAbsent(mods[i], (k)->0)+1);
        }
        long sum = 0;
        for(Integer t:mp.keySet()){
            sum += mp.get(t)*(mp.get(t)-1)/2;
        }
        System.out.println(sum);
    }
    private static int mod10(int a, int keta, int P){
        for(int i=0;i<keta;i++){
            a = a * 10;
            a = a % P;
        }
        return a;
    }
}