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 Common Lisp (SBCL 1.1.14) (judge verdict: Accepted)

(eval-when (:compile-toplevel :load-toplevel :execute)
  (sb-int:defconstant-eqx OPT
    #+swank '(optimize (speed 3) (safety 2))
    #-swank '(optimize (speed 3) (safety 0) (debug 0))
    #'equal)
  #+swank (ql:quickload '(:cl-debug-print :fiveam) :silent t)
  #-swank (set-dispatch-macro-character
           ;; enclose the form with VALUES to avoid being captured by LOOP macro
           #\# #\> (lambda (s c p) (declare (ignore c p)) `(values ,(read s nil nil t)))))
#+swank (cl-syntax:use-syntax cl-debug-print:debug-print-syntax)
#-swank (disable-debugger) ; for CS Academy

;; BEGIN_INSERTED_CONTENTS
;; TODO: more efficient handling when modulus is (unsigned-byte 31) or
;; (unsigned-byte 32)
(declaim (inline mod-power))
(defun mod-power (base power modulus)
  "BASE := integer
POWER, MODULUS := non-negative fixnum"
  (declare ((integer 0 #.most-positive-fixnum) modulus power)
           (integer base))
  (labels ((recur (x p)
             (declare ((integer 0 #.most-positive-fixnum) x p)
                      #+sbcl (values (integer 0 #.most-positive-fixnum)))
             (cond ((zerop p) 1)
                   ((evenp p) (recur (mod (* x x) modulus) (ash p -1)))
                   (t (mod (* x (recur x (- p 1))) modulus)))))
    (recur (mod base modulus) power)))

;;;
;;; Modular arithmetic
;;;

;; Blankinship algorithm
;; Reference: https://topcoder-g-hatena-ne-jp.jag-icpc.org/spaghetti_source/20130126/ (Japanese)
(declaim (ftype (function * (values fixnum fixnum &optional)) %ext-gcd))
(defun %ext-gcd (a b)
  (declare (optimize (speed 3) (safety 0))
           (fixnum a b))
  (let ((y 1)
        (x 0)
        (u 1)
        (v 0))
    (declare (fixnum y x u v))
    (loop (when (zerop a)
            (return (values x y)))
          (let ((q (floor b a)))
            (decf x (the fixnum (* q u)))
            (rotatef x u)
            (decf y (the fixnum (* q v)))
            (rotatef y v)
            (decf b (the fixnum (* q a)))
            (rotatef b a)))))

;; Simple recursive version. A bit slower but more comprehensible.
;; https://cp-algorithms.com/algebra/extended-euclid-algorithm.html (English)
;; https://drken1215.hatenablog.com/entry/2018/06/08/210000 (Japanese)
;; (defun %ext-gcd (a b)
;;   (declare (optimize (speed 3) (safety 0))
;;            (fixnum a b))
;;   (if (zerop b)
;;       (values 1 0)
;;       (multiple-value-bind (p q) (floor a b) ; a = pb + q
;;         (multiple-value-bind (v u) (%ext-gcd b q)
;;           (declare (fixnum u v))
;;           (values u (the fixnum (- v (the fixnum (* p u)))))))))

;; TODO: deal with bignums
(declaim (inline ext-gcd))
(defun ext-gcd (a b)
  "Returns two integers X and Y which satisfy AX + BY = gcd(A, B)."
  (declare ((integer #.(- most-positive-fixnum) #.most-positive-fixnum) a b))
  (if (>= a 0)
      (if (>= b 0)
          (%ext-gcd a b)
          (multiple-value-bind (x y) (%ext-gcd a (- b))
            (declare (fixnum x y))
            (values x (- y))))
      (if (>= b 0)
          (multiple-value-bind (x y) (%ext-gcd (- a) b)
            (declare (fixnum x y))
            (values (- x) y))
          (multiple-value-bind (x y) (%ext-gcd (- a) (- b))
            (declare (fixnum x y))
            (values (- x) (- y))))))

(declaim (inline mod-inverse)
         (ftype (function * (values (mod #.most-positive-fixnum) &optional)) mod-inverse))

;; (defun mod-inverse (a modulus)
;;   "Solves ax ≡ 1 mod m. A and M must be coprime."
;;   (declare (integer a)
;;            ((integer 1 #.most-positive-fixnum) modulus))
;;   (mod (%ext-gcd (mod a modulus) modulus) modulus))

;; FIXME: Perhaps no advantage in efficiency? Then I should use the above simple
;; code.
(defun mod-inverse (a modulus)
  "Solves ax ≡ 1 mod m. A and M must be coprime."
  (declare ((integer 1 #.most-positive-fixnum) modulus))
  (let ((a (mod a modulus))
        (b modulus)
        (u 1)
        (v 0))
    (declare (fixnum a b u v))
    (loop until (zerop b)
          for quot = (floor a b)
          do (decf a (the fixnum (* quot b)))
             (rotatef a b)
             (decf u (the fixnum (* quot v)))
             (rotatef u v))
    (setq u (mod u modulus))
    (if (< u 0)
        (+ u modulus)
        u)))

(defmacro dbg (&rest forms)
  #+swank
  (if (= (length forms) 1)
      `(format *error-output* "~A => ~A~%" ',(car forms) ,(car forms))
      `(format *error-output* "~A => ~A~%" ',forms `(,,@forms)))
  #-swank (declare (ignore forms)))

(defmacro define-int-types (&rest bits)
  `(progn
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "UINT~A" b)) () '(unsigned-byte ,b))) bits)
     ,@(mapcar (lambda (b) `(deftype ,(intern (format nil "INT~A" b)) () '(signed-byte ,b))) bits)))
(define-int-types 2 4 7 8 15 16 31 32 62 63 64)

(declaim (inline println))
(defun println (obj &optional (stream *standard-output*))
  (let ((*read-default-float-format* 'double-float))
    (prog1 (princ obj stream) (terpri stream))))

(defconstant +mod+ 1000000007)

;;;
;;; Body
;;;

(defun solve2 (n s)
  (let ((res 0))
    (dotimes (i n)
      (let ((d (aref s i)))
        (when (evenp d)
          (incf res (+ i 1)))))
    res))

(defun solve5 (n s)
  (let ((res 0))
    (dotimes (i n)
      (let ((d (aref s i)))
        (when (or (= d 0) (= d 5))
          (incf res (+ i 1)))))
    res))

(defun main ()
  (let* ((n (read))
         (p (read))
         (original-s (read-line))
         (s (make-array n :element-type 'uint8))
         (table (make-array p :element-type 'uint62 :initial-element 0))
         (powers (make-array (+ n 1) :element-type 'uint32 :initial-element 1)))
    (dotimes (i n)
      (setf (aref powers (+ i 1))
            (mod (* 10 (aref powers i)) p)))
    (dotimes (i n)
      (setf (aref s i) (- (char-code (aref original-s i)) 48)))
    (when (= p 2)
      (println (solve2 n s))
      (return-from main))
    (when (= p 5)
      (println (solve5 n s))
      (return-from main))
    (setf (aref table 0) 1)
    (let ((sum 0)
          (res 0))
      (loop for x from 1 to n
            for d = (aref s (- x 1))
            do (setq sum (mod (+ (* 10 sum) d) p))
               (let ((sum/10 (mod (* sum (mod-inverse (aref powers x) p)) p)))
                 (incf res (aref table sum/10))
                 (incf (aref table sum/10))))
      (println res))))

#-swank (main)

;;;
;;; Test and benchmark
;;;

#+swank
(defun io-equal (in-string out-string &key (function #'main) (test #'equal))
  "Passes IN-STRING to *STANDARD-INPUT*, executes FUNCTION, and returns true if
the string output to *STANDARD-OUTPUT* is equal to OUT-STRING."
  (labels ((ensure-last-lf (s)
             (if (eql (uiop:last-char s) #\Linefeed)
                 s
                 (uiop:strcat s uiop:+lf+))))
    (funcall test
             (ensure-last-lf out-string)
             (with-output-to-string (out)
               (let ((*standard-output* out))
                 (with-input-from-string (*standard-input* (ensure-last-lf in-string))
                   (funcall function)))))))

#+swank
(defun get-clipbrd ()
  (with-output-to-string (out)
    (run-program "powershell.exe" '("-Command" "Get-Clipboard") :output out :search t)))

#+swank (defparameter *this-pathname* (uiop:current-lisp-file-pathname))
#+swank (defparameter *dat-pathname* (uiop:merge-pathnames* "test.dat" *this-pathname*))

#+swank
(defun run (&optional thing (out *standard-output*))
  "THING := null | string | symbol | pathname

null: run #'MAIN using the text on clipboard as input.
string: run #'MAIN using the string as input.
symbol: alias of FIVEAM:RUN!.
pathname: run #'MAIN using the text file as input."
  (let ((*standard-output* out))
    (etypecase thing
      (null
       (with-input-from-string (*standard-input* (delete #\Return (get-clipbrd)))
         (main)))
      (string
       (with-input-from-string (*standard-input* (delete #\Return thing))
         (main)))
      (symbol (5am:run! thing))
      (pathname
       (with-open-file (*standard-input* thing)
         (main))))))

#+swank
(defun gen-dat ()
  (uiop:with-output-file (out *dat-pathname* :if-exists :supersede)
    (format out "")))

#+swank
(defun bench (&optional (out (make-broadcast-stream)))
  (time (run *dat-pathname* out)))

;; To run: (5am:run! :sample)
#+swank
(it.bese.fiveam:test :sample
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "4 3
3543
"
    "6
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "4 2
2020
"
    "10
"))
  (it.bese.fiveam:is
   (common-lisp-user::io-equal "20 11
33883322005544116655
"
    "68
")))