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;
typedef long long LL;
typedef pair<LL, LL> PII;

// chmax, chmin
template <class T>
inline bool chmax(T &a, T b)
{
  if (a < b)
  {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
  if (a > b)
  {
    a = b;
    return 1;
  }
  return 0;
}

template <typename T>
function<T(T, T)> op_max = [](T a, T b) -> T { return max(a, b); };
template <typename T>
function<T(T, T)> op_min = [](T a, T b) -> T { return min(a, b); };

template <typename T>
function<T(T, T)> op_sum = [](T a, T b) -> T { return a + b; };

namespace detail
{
  template <typename T, std::size_t NDIMS>
  struct vector_builder
  {
    using type = std::vector<typename vector_builder<T, NDIMS - 1>::type>;

    static type make(std::vector<std::size_t> dims, const T &v = {})
    {
      const auto vec = std::vector<T>(dims.empty() ? 0 : dims.back(), v);
      if (!dims.empty())
        dims.pop_back();
      return vector_builder<std::vector<T>, NDIMS - 1>::make(dims, vec);
    }
  };

  template <typename T>
  struct vector_builder<T, 1>
  {
    using type = std::vector<T>;

    static type make(std::vector<std::size_t> dims, const T &v = {})
    {
      return type(dims.empty() ? 0 : dims.back(), v);
    }
  };
} // namespace detail

template <typename T, typename... SIZE_T>
auto exvector(const T &v, SIZE_T... dims)
{
  static_assert(sizeof...(dims) != 0, "invalid dimension");
  std::vector<std::size_t> vec_dims{dims...};
  return detail::vector_builder<T, sizeof...(dims)>::make(vec_dims, v);
}

using namespace std;
//********************  dumps ************************//

template <typename T>
void dump(const T &data, vector<int> &iter)
{
  cout << data << " ";
}
template <typename T>
void dump(const vector<T> &data, vector<int> &iter)
{
  for (auto elem : data)
    dump(elem, iter);
  cout << endl;
}

template <typename T>
void dump(const vector<vector<T>> &data, vector<int> &iter)
{
  for (auto elem : iter)
  {
    cout << "[" << elem << "]";
  }
  cout << endl;
  for (int i = 0; i < data.size(); i++)
  {
    iter.push_back(i);
    dump(data[i], iter);
    iter.pop_back();
  }
  cout << endl;
}

template <typename T>
void dump(const vector<T> &data, int dummy)
{
  for (int i = 0; i < data.size(); i++)
  {
    cout << "[" << i << "] " << data[i] << endl;
  }
}
template <typename T>
void dump(const T &data)
{ // T : data, U = base type val.
  vector<int> iter;
  dump(data, iter);
}
///////////////////////////////////////////////

void calc(LL P, string s)
{
  LL ans = 0;
  for (LL i = 0; i < s.size(); i++)
  {
    bool flag;
    if (P == 2 && (s[i] - '0') % 2 == 0)
      flag = true;
    if (P == 5 && (s[i] - '0') % 5 == 0)
      flag = true;
    if (flag)
      ans += (i + 1);
  }
  cout << ans << endl;
}

void solve(long long N, long long P, std::string S)
{

  if (P == 2 || P == 5)
  {
    calc(P, S);
    return;
  }
  string s = S;
  LL ans = 0;
  map<int, int> vals;
  LL cur = 0;
  vals[0] = 1;
  for (LL i = s.size() - 1; i >= 0; i--)
  {
    cur *= 10;
    cur += s[i] - '0';
    cur %= P;
    ans += vals[cur];
    vals[cur]++;
  }
  cout << ans << endl;
}

int main()
{
  cout << setprecision(10);
  long long N;
  scanf("%lld", &N);
  long long P;
  scanf("%lld", &P);
  std::string S;
  std::cin >> S;
  solve(N, P, S);
  return 0;
}