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;

  //pが２または５のとき、s*10^n % p はかならずゼロになる
  if (10 % p == 0) { //ｐが２または５のとき
    long long ans = 0;
    for (int r = 0; r < n; r++) {
      if ((s[r] - '0')%p == 0) { ans += r + 1; } //右端がpの倍数になっていれば左側はなんでもよい。
    }
    cout << ans << endl;
    return 0;
  }


  vector<int> d(n+1, 0); //累積和を格納するベクトル
  int ten = 1;
  for (int i = n - 1; i >= 0; i--) {
    int a = (s[i] - '0') * ten % p; //各桁のmod p
    d[i] = (d[i+1] + 1) % p;
    ten *= 10; ten %= p;
  }

  vector<int> cnt(p);
  long long ans = 0;
  for (int i = n; i >= 0; i--) {
    ans += cnt[d[i]];
    cnt[d[i]]++;
  }
  cout << ans << endl;
  return 0;
}