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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int N, P;
    cin >> N >> P;

    string S;
    cin >> S;

    if (P == 2 || P == 5) {
        int count = 0;
        for (int i = N-1; i >= 0; i--) {
            int digit = S[i] - '0';
            if (digit % P == 0) {
                count += i + 1;
            }
        }
        cout << count << "\n";
        return 0;
    }

    map<int, int> mods;

    int prev = 0;
    int ten = 1;

    // off to the side remainder is 0
    mods[0] += 1;
    for (int i = N-1; i >= 0; i--) {
        int digit = S[i] - '0';
        int my_mod = (prev + digit*ten) % P;
        ten = (ten * 10) % P;
        mods[my_mod] += 1;
        prev = my_mod;
    }

    int count = 0;

    for (auto k : mods) {
        int i = k.second;
        count += (i*(i-1))/2;
    }

    cout << count << "\n";
}