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.Scanner;

public class Main {

	public static void main(String[] args) {
	/*
		int pp=7;
		for (int i=1;i<=10000;i++) {
			String ss = String.valueOf(i);
			long count = 0;
			for (int j=0;j<ss.length();j++) {
				for (int k=j+1;k<=ss.length();k++) {
					if (Integer.parseInt(ss.substring(j,k))%pp==0) {
					count++;
					}
				}
			}
			if (count != ans(ss.length(), pp, ss)) {
				System.out.println(i);
			}
		}
		
		*/
		

		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int p = sc.nextInt();
		String s = sc.next();
	
		if (p==2 || p==5) {
			long ans = 0;
			for (int i=0;i<n;i++) {
				if ((s.charAt(i)-'0')%p==0) {
					ans+=(i+1);
				}
			}
			System.out.println(ans);
			
		} else {
			long[] count = new long[p];
			int rem = 0;
			count[0]=1;
			int digit = 1;
			for (int i=n-1; i>=0; i--) {
				rem=(rem+digit*(s.charAt(i)-'0'))%p;
				count[rem]++;
				digit=(digit*10)%p;
			}
			
			long ans = 0;
			for(int i=0;i<p;i++) {
				ans += (count[i]*(count[i]-1)/2);
			}
			System.out.println(ans);
		}

		
	}
	
	static long ans(int n, int p, String s) {
		if (p==2 || p==5) {
			long ans = 0;
			for (int i=0;i<n;i++) {
				if ((s.charAt(i)-'0')%p==0) {
					ans+=(i+1);
				}
			}
			return ans;
			
		} else {
			long[] count = new long[p];
			int rem = 0;
			count[0]=1;
			int digit = 1;
			for (int i=n-1; i>=0; i--) {
				rem=(rem+digit*(s.charAt(i)-'0'))%p;
				count[rem]++;
				digit=(digit*10)%p;
			}
			
			long ans = 0;
			for(int i=0;i<p;i++) {
				ans += (count[i]*(count[i]-1)/2);
			}
			return ans;
		}
	}
	
}