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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
//error_reporting(0);
//$s = trim(fgets(STDIN));
//list($A,$B) = explode(" ",trim(fgets(STDIN)));
//$a[] = array_map('intval',explode(" ",trim(fgets(STDIN))));
list($N,$P) = array_map('intval',explode(" ",trim(fgets(STDIN))));
$S = trim(fgets(STDIN));

https://atcoder.jp/contests/abc158/submissions/10605418
$S = strrev($S);
$slen = strlen($S);
$ans=0;
$pow10=1;
$sum = 0;
$cnt=array();

if ($P != 2 && $P != 5) {
  $cnt[0]=1;
  for($i=0;$i<$slen;$i++) {
    $sum = ($sum + $S[$i]*$pow10) % $P;
    $ans += $cnt[$sum];
    if (isset($cnt[$sum])) {
      $cnt[$sum]++;
    }
    else {
      $cnt[$sum]=1;
    }
    $pow10 = $pow10 * 10 % $P;
  }
}
else {
  $z = 0;
  for($i=0;$i<$slen;$i++) {
    if ($S[$i] % $P == 0) $z++;
    $ans += $z;
  }
}
printf("%d\n",$ans);