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 <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <string>
#include <cmath>
using namespace std;

int n, p;
string s;

int main()
{
  cin >> n >> p >> s;

  if (p == 2)
  {
    int ans = 0;
    for (int i = 0; i < n; i++)
    {
      if ((s[i] - '0') % 2 == 0)
      {
        ans += i + 1;
      }
    }
    cout << ans << endl;
  }
  else if (p == 5)
  {
    int ans = 0;
    for (int i = 0; i < n; i++)
    {
      if ((s[i] - '0') % 5 == 0)
      {
        ans += i + 1;
      }
    }
    cout << ans << endl;
  }
  else
  {
    // u[i]: s[0]~s[i]を繋いでできるintをpで割った余り
    vector<int> u(n);
    u[0] = (s[0] - '0') % p;
    for (int i = 1; i < n; i++)
    {
      u[i] = (u[i - 1] * 10 + (s[i] - '0')) % p;
    }

    // v[i]: u[0]~u[n-1]のうち、pで割った余りがiであるものの個数
    vector<int> v(p, 0);
    for (int i = 0; i < n; i++)
    {
      v[u[i]]++;
    }

    // for (int i = 0; i < p; i++)
    // {
    //   cout << v[i] << endl;
    // }

    int ans = 0;
    ans += v[0];
    for (int i = 0; i < p; i++)
    {
      ans += v[i] * (v[i] - 1) / 2;
    }

    cout << ans << endl;
  }
}