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> P;
typedef pair<int, int> i_i;

#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define REP(i,n) reps(i,0,n)
#define ALL(v) v.begin(), v.end()
#define SZ(x) ((int)(x).size())

const double PI  = acos(-1.0);
const double EPS = 1e-9;
const ll INF = 1e9+10;
const ll mod = 1e9+7;
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; }
#define print(x) cerr << #x << " is " << x << endl;
//#define print(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl; 

int B[5001][26];
int lower[200001];
int upper[200001];
int ketamod[200001];
int cnt[10001];
int main() {
    int N, P;
    string S;
    cin >> N >> P;
    cin >> S;
    reverse(ALL(S));
    ll ans = 0;
    if (P == 2 || P == 5){
        rep(i, N){
            if (S[i] % P == 0){
                ans += N - i;
            }
        }
        //cout << ans;
        return 0;
    }
    
    int num = 0;
    lower[0] = 0;
    cnt[0]++;
    rep(i, N){
        if (i == 0){
            ketamod[i] = 1;
        } else {
            ketamod[i] = (ketamod[i - 1] * 10)%P;
        }
        lower[i + 1] = (lower[i] + (S[i] - '0') *  ketamod[i])%P;
        if (i < N - 2){
            cnt[lower[i + 1]]++;
            //cout << i << " lower: " << lower[i] << endl;
        }
            
    }
    //cout << N << " lower: " << lower[N] << endl;
    upper[N - 1] = lower[N];
    
    //rep(i, P)
    //    cout << "cnt[" << i << "]" << cnt[i] << endl;
    
    //cout << N - 1<< " upper: " << upper[N - 1] << endl;
    //cout << "ans = " << cnt[upper[N - 1]] << endl;
    ans += cnt[upper[N - 1]];
    for (int i = N - 2; i >= 0; i--){
        int tmp = ((S[i + 1] - '0')* ketamod[i + 1])%P;
        upper[i] = (upper[i + 1] + P - tmp)%P;
        // upper[i] = (upper[i + 1] + P - S[i + 1] * ketamod[i + 1])%P;
        //cout << i << " upper: " << upper[i] << endl;
        ans += cnt[upper[i]];
        //cout << "ans = " << cnt[upper[i]] << endl;
        cnt[lower[i]]--;
    }
    cout << ans;
    return 0;
}