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 <algorithm>
#include <vector>
#include <cmath>
using namespace std;

void solve(){
  int n, p; cin >> n >> p;
  int ans = 0;
  if(p == 2 || p == 5){
    vector<int> v(n);
    for(int i = 0; i < n; i++){
      char c; cin >> c;
      int x = c - 48;
      v.push_back(x);
    }
    for(int i = v.size() - 1; i >= 0; i--){
      if(v[i] % 2 == 0) ans += i + 1;
    }
  }
  else{
    vector<int> v(n);
    vector<int> v2(n + 1, 0);
    int ten = 1;
    string s; cin >> s;
    for(int i = n - 1; i >= 0; --i){
      int x = s[i] - '0';
      v[i] = x;
      v2[i] = x * ten % p;
      ten *= 10;
      ten %= p;
    }
    //for(int i = 0; i < v2.size(); i++) cout << v2[i] % p; cout << endl;
    for(int i = v2.size() - 2; i >= 0; --i){
      v2[i] += v2[i + 1] % p;
    }
    //for(int i = 0; i < v2.size(); i++) cout << v2[i] % p; cout << endl;
    vector<int> a(p, 0);
    for(int i = 0; i < v2.size(); i++){
      a[v2[i] % p]++;
    }
    for(int i = 0; i < a.size(); i++){
      ans += a[i] * (a[i] - 1) / 2;
    }
  }
  cout << ans << endl;

  return;
}

void solve2(){
  int n, p; cin >> n >> p;
  string s; cin >> s;
  vector<int> d(n+1);
  int ten = 1;
  for (int i = n-1; i >= 0; --i) {
    int a = (s[i]-'0') * ten % p;
    d[i] = (d[i+1]+a) % p;
    ten *= 10; ten %= p;
  }
  for(int i = 0; i < d.size(); i++) cout << d[i]; cout << endl;
  vector<int> cnt(p);
  long long int ans = 0;
  for (int i = n; i >= 0; --i) {
    ans += cnt[d[i]];
    cnt[d[i]]++;
  }
  cout << ans << endl;
  return;
}

int main(){
  solve();
  //solve2();
  return 0;
}