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: Accepted)


#include <iostream>
#include <iomanip> //! for setprecision(10)
#include <math.h>
#include <algorithm>
#include <functional> 
#include <string>
#include <vector>

#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <utility>

using namespace std;

typedef long long LL;

#define dump(c) { for (auto it = c.begin(); it != c.end(); ++it) if (it == c.begin()) cout << *it; else cout << ' ' << *it; cout << endl; } 
#define rep(i,n) for (int i = 0; i < (n); ++i)

const int MOD = 1000000007;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(0);

    string S;
    int N, P;
    cin >> N >> P >> S;

    LL ans = 0;

    if (P == 2 || P == 5) {
        rep(i, N) {
            int d = S[N-1-i] - '0';
            if (d % P == 0) {
                ans += (N-i);
            }
        }
    } else {
        vector<int> m(N);
        int b = 1;
        int prev = 0;
        rep(i, N) {
            int d = S[N-1-i] - '0';
            m[i] = (prev + b * d) % P;
            prev = m[i];
            b = (b * 10) % P;
        }
        vector<int> hist(P);
        hist[0] = 1;
        rep(i, N) {
            ans += hist[m[i]];
            ++hist[m[i]];
        }
    }

    cout << ans << endl;

    return 0;
}