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 (Python):
def main(sample_file = ''):

    """ convenient functions
    
    
    
    
    
    """

    """convenient decorator
    
    
        
        
        
        
        
        
        
        
        
        
    """
    
    
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product 
    
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  
        b, e = 2, 0  
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict 
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest 
    from copy import deepcopy, copy 
    import operator
    from operator import itemgetter 
    
    
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd 
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    
    
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) 
    def modinv(a, p = MOD):
        
        
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        
        
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    
    
    
    
    

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    
        
        
            
            
    
        
        
        
        
        
        
        
    
        
        
        
        
    
        
        
    """

    
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    
    
    
    
    
    
    
    """

    
    if sample_file:
        sys.stdin = open(sample_file)
        
    
    
    
    N, P = R()
    S = r()
    if P == 2:
        ans = 0
        for i,s in enumerate(reversed(S)):
            if int(s)%2 == 0:
                ans += combinations_count(N-i, 1)
        print(ans)
        return
    elif P == 5:
        ans = 0
        for i, s in enumerate(reversed(S)):
            if int(s)%5 == 0:
                ans += combinations_count(N-i, 1)
        print(ans)
        return
    
    L = [[0]*(len(S)+2) for _ in range(10)]
    for i in range(10):
        L[i][0] = i % P
    for i in range(len(S)+1):
        for j in range(10):
            L[j][i+1] = (L[j][i] * 10) % P
    
    l = [0] * len(S)
    m = [0] * len(S)
    for i, s in enumerate(reversed(S)):
        ss = int(s)
        l[-1-i] = L[ss][i]
    
    tmp = 0
    for i, ll in enumerate(reversed(l)):
        m[i] = (tmp + ll) % P
        tmp = m[i]
    m.append(0)
    c = Counter(m)
    ans = 0
    for i in c.values():
        ans += combinations_count(i, 2)
    print(ans)
    
    
    
    
    
    
    
    
    
    
    
    

    """memo: how to use defaultdict of list
    
    Dic = defaultdict(list)
    
    Dic[x].append(y)
    
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    
    >>> a = 5
    >>> bin_str_a = format(a, '
    >>> print(bin_str_a)
    0b0101 
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Solution (Python):
n, p = map(int,input().split())
s = str(input())
ans = 0

if p == 2 or p == 5:
    for i in range(n):
        if int(s[i])%p == 0:
            ans += i+1
    print(ans)
    exit()

tmp = [0] * p
tmp[0] = 1
mod = 0
r = 1
for i in range(1,n+1):
    mod += (int(s[-i]) * r) % p
    mod %= p
    tmp[mod] += 1
    r *= 10
    r %= p

for i in tmp:
    ans += i*(i-1)//2

print(ans)

Solution (Python):
n, p = map(int, input().split())
s = input()

if 10%p==0:
    ans = 0
    for r in range(n):
        if int(s[r])%p == 0:
            ans += r+1
    print(ans)
    exit()

d = [0]*(n+1)
ten = 1

for i in range(n-1, -1, -1):
    a = int(s[i])*ten%p
    d[i] = (d[i+1]+a)%p
    ten *= 10
    ten %= p

cnt = [0]*p
ans = 0
for i in range(n, -1, -1):
    ans += cnt[d[i]]
    cnt[d[i]] += 1
print(ans)

Solution (Python):
import collections

n,p=map(int,input().split())
s=input()
arr=[int(s[i]) for i in range(len(s))]
if p==2 or p==5:
  ans=0
  for i in range(n):
    if arr[i]%p==0:
      ans+=i+1
  print(ans)
else:
  dic=collections.defaultdict(int)
  tmp=0
  for i in range(n-1,-1,-1):
    tmp+=arr[i]*pow(10,n-1-i,p)
    tmp%=p
    dic[tmp]+=1
  ans=0
  rem=0
  for i in range(n-1,-1,-1):
    ans+=dic[rem]
    rem+=arr[i]*pow(10,n-1-i,p)
    rem%=p
    dic[rem]-=1
  print(ans)

Solution (Python):
N, P = map(int, input().split())
S = input()
 
ans = 0
if P == 2 or P == 5:
	for i in range(N):
		if int(S[i]) % P == 0:
			ans += i+1
else:
	t = [0] * 20000
	t[0] = 1
	now = 0
	po = 1
	for i in range(N):
		n = N - i - 1
		now = (int(S[n]) * po + now) % P
		po = 10 * po % P
		ans += t[now]
		t[now] += 1
print(ans)

Solution (Python):
import sys
sys.setrecursionlimit(10**7)
INTMAX = 9223372036854775807
INTMIN = -9223372036854775808
DVSR = 1000000007
def POW(x, y): return pow(x, y, DVSR)
def INV(x, d=DVSR): return pow(x, d - 2, d)
def DIV(x, y, d=DVSR): return (x * INV(y, d)) % d
def LI(): return [int(x) for x in input().split()]
def LF(): return [float(x) for x in input().split()]
def LS(): return input().split()
def II(): return int(input())

N,P=LI()
S=input()

if P == 2 or P == 5:
  res = 0
  for i in range(N):
    if int(S[i])%P == 0:
      res += i+1
  print(res)
else:
  RESTS=[0]*(P+1)
  RESTS[0] = 1
  POWS=[0]*(N+1)
  POWS[0]=1
  res = 0
  for i in range(N): POWS[i+1] = (POWS[i]*10)%P

  cur = 0
  for i in range(N):
    cur += (int(S[N-1-i])*POWS[i])
    cur %= P
    res += RESTS[cur]
    RESTS[cur] += 1
  print(res)

Solution (Python):
n, p = map(int, input().split())
s = list(input())
s.reverse()
ans = 0

if p == 2 or p == 5:
    for i in range(n):
        if int(s[i]) % p == 0:
            ans += n - i
else:
    modlist = [0] * (n + 1)
    ten = 1
    for i in range(n):
        mod_now = (int(s[i]) * ten) % p
        modlist[i + 1] = (modlist[i] + mod_now) % p
        ten *= 10
        ten %= p
    countlist = [0] * p
    for i in range(n + 1):
        ans += countlist[modlist[i]]
        countlist[modlist[i]] += 1
print(ans)

Solution (Python):
from collections import defaultdict

N, P = map(int, input().split())
S = input()

"""
S = "s0 s1 ... s(N-1)"に対して"sl s(l+1) ... s(r-1)"が素数Pで割り切れるかを考える。
これはa_i := int("si s(i+1) ... s(N-1)")と定めることで、
    (a_l - a_r) / 10^{N-r} ¥cong 0  (mod P)
と定式化できる。

（整数関連の区間の問題は後ろからの累積和が相性良さそう。頭からの累積を考えたのが敗因な気がする）

Pが2でも5でもない場合は、上式は
    a_l - a_r ¥cong 0 (mod P)
に同値であるから、各a_iのmod Pでの値を辞書にまとめておけば良さそう。
Pが2, 5のいずれかの場合は偶然チェックが簡単なので解ける。
"""

ans = 0
if P == 2:
    for i in range(N):
        a = int(S[i])
        if a % 2 == 0:
            ans += i + 1
elif P == 5:
    for i in range(N):
        a = int(S[i])
        if a == 0 or a == 5:
            ans += i + 1
else:
    S += "0" 
    d = defaultdict(int)
    r = 0
    d[0] += 1
    for i in range(N - 1, -1, -1):
        """
        ここでS[i:]を毎回Pで割るとTLEする（桁の大きい演算はやはり負担が大きいのか？）
        (a_i % P)をi = N, N-1,...の降順に求めていくことを考えれば、
        前の計算結果が使える、特に繰り返し二乗法が使えるのでだいぶ早くなるみたい。
        """
        r += int(S[i]) * pow(10, N - i - 1, P)
        r %= P
        d[r] += 1

    for num in d.values():
        ans += num * (num - 1) // 2

print(ans)

Solution (Python):
import collections

n,p=map(int,input().split())
s=input()
arr=[int(s[i]) for i in range(len(s))]
if p==2 or p==5:
  ans=0
  for i in range(n):
    if arr[i]%p==0:
      ans+=i+1
  print(ans)
else:
  dic=collections.defaultdict(int)
  match=[]
  pows=[1]
  for _ in range(n):
    pows.append((pows[-1]*10)%p)
  tmp=0
  for i in range(n-1,-1,-1):
    tmp+=arr[i]*pows[n-1-i]
    tmp%=p
    match.append(tmp)
    dic[tmp]+=1
  match=match[::-1]
  ans=0
  rem=0
  rev=pow(10,p-2,p)
  for i in range(n-1,-1,-1):
    ans+=dic[rem]
    rem+=arr[i]*pows[n-1-i]
    rem%=p
    dic[match[i]]-=1
  print(ans)

Solution (Python):
n,p = tuple(int(x) for x in input().split())
s = [int(s) for s in input()][::-1]
modlist = [0]*p
modlist[0] = 1
if p!= 2 and p != 5:
    mod = 0
    pwrmod = 1
    for i in s:
        mod = (mod + pwrmod * i) % p
        pwrmod = (pwrmod * 10) % p
        modlist[mod] += 1
    ans = 0
    for i in modlist:
        ans += (i * (i-1)) // 2

    print(ans)
else:
    ans = 0
    for i in range(n):
        if s[i] % p == 0:
            ans += n-i
    print(ans)

Solution (Python):
from collections import defaultdict

N, P = map(int, input().split())
S = input()

if P in (2, 5):
    count = 0
    for i in range(N):
        if int(S[i]) % P == 0:
            count += i + 1
    print(count)
else:
    d = defaultdict(int)
    r = 0
    for i in range(N):
        r += int(S[N - 1 - i]) * pow(10, i % (P - 1), P)
        r %= P
        d[r] += 1

    count = d[0]
    for num in d.values():
        count += num * (num - 1) // 2
    print(count)

Solution (Python):
n,p = tuple(map(int,input().split()))
s = input()
modlist = [0]*p
modlist[0] = 1
if p!= 2 and p != 5:
    mod = 0
    pwrmod = 1
    for i in s[::-1]:
        mod = (mod + pwrmod * int(i)) % p
        pwrmod = (pwrmod * 10) % p
        modlist[mod] += 1
    ans = 0
    for i in modlist:
        ans += (i * (i-1)) // 2

    print(ans)
else:
    ans = 0
    for i in range(n):
        if int(s[i]) % p == 0:
            ans += i+1
    print(ans)

Solution (Python):
from collections import defaultdict


def main():
    N, P = map(int, input().split())
    S = input()
    ans = 0
    if P in [2, 5]:
        for i, d in enumerate(S[::-1]):
            if int(d) % P == 0:
                ans += N - i
    else:
        total = defaultdict(int)
        total[0] = 1
        cur = 0
        for i, d in enumerate(S[::-1]):
            d = int(d) % P
            prev = cur
            cur = (prev + pow(10, i, P) * d) % P
            ans += total[cur]
            total[cur] += 1
    print(ans)


if __name__ == "__main__":
    main()

Solution (Python):
import sys
from collections import Counter
readline = sys.stdin.buffer.readline

n,m = map(int,readline().split())
s = readline().rstrip().decode('utf-8')

lst1 = [0]*len(s)
res = ""
for i in range(n):
    res = int(str(res)+s[i])
    res %= m
    lst1[i] = res

ans = 0

if m == 2 or m == 5:
    for i in range(n):
        if lst1[i] == 0:
            ans += i+1


else:
    s = s[::-1]
    dig = 1
    lst1 = [0]*n
    for i in range(n):
        if i == 0:
          lst1[i] = int(s[i])%m
          continue
          
        dig *= 10
        dig %= m
        lst1[i] = lst1[i-1] + dig*int(s[i])
        lst1[i] %= m
    c = Counter(lst1)
    
    for i in c.values():
        
        ans += i*(i-1)//2
    ans += c[0]
print(ans)

Solution (Python):
import sys


N,P = map(int,input().split())
S = list(input())

A = [0]*(N)
A[-1] = int(S[-1])
A[-1] %= P
base = 10 % P
for i in reversed(range(N-1)):
    A[i] = A[i+1] + base*int(S[i])
    A[i] %= P
    base *= 10
    base %= P


ans = 0
if P==2 or P==5:
    for i in reversed(range(N)):
        if int(S[i])%P==0:
            ans += i+1
else:
    nums = {}
    for a in A:
        if not (a in nums):
            nums[a] = 1
        else:
            nums[a] += 1
    
    for n in nums.keys():
        k = nums[n]
        if n==0:
            ans += k*(k+1)//2
        else:
            ans += k*(k-1)//2
    
print(ans)

Solution (Python):
n,p=map(int,input().split())
s=list(map(int,input()))[::-1]
sum=0
cnt=[0]*p
cnt[0]=1
for i in range(n):
    d=s[i]*pow(10,i,p)
    sum=(sum+d)%p
    cnt[sum]+=1
ans=0
if p in [2,5]:
    if p==2:
        ok_list=[0,2,4,6,8]
    if p==5:
        ok_list=[0,5]
    for i in range(n):
        d=s[i]
        if d in ok_list:
            ans+=n-i 
else:
    for i in range(p):
        ans+=cnt[i]*(cnt[i]-1)//2
print(ans)

Solution (Python):
N, P = map(int, input().split())
S = list(map(int, input()))
ans = 0
if P == 2:
    for i in range(N):
        if S[i]%2 == 0:
            ans +=i+1
    print(ans)

elif P ==5:
    for i in range(N):
        if S[i]%5 == 0:
            ans +=i+1
    print(ans)
else:
    data =[0]*P
    tens = 1
    k = 0
    for i in range(N-1,-1,-1):
        k += S[i] *tens %P
        k = k%P
        data[k] += 1
        tens *=10
        tens %=P

    for i in data:
        ans += i*(i-1)/2
    ans +=data[0]
    print(int(ans))

Solution (Python):
N, P = map(int, input().split())
S = list(map(int, input()))
ans = 0
if P == 2:
    for i in range(N):
        if S[i]%2 == 0:
            ans +=i+1
    print(ans)

elif P ==5:
    for i in range(N):
        if S[i]%5 == 0:
            ans +=i+1
    print(ans)
else:
    data =[0]*P
    tens = 1
    k = 0
    for i in range(N-1,-1,-1):
        k += S[i] *tens %P
        k = k%P
        data[k] += 1
        tens *=10
        tens %=P

    for i in data:
        ans += i*(i-1)/2
    ans +=data[0]
    print(int(ans))

Solution (Python):
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by enraged(rating - 5 star at CodeChef and Specialist at Codeforces).
"""
import sys
import bisect
import heapq

from collections import defaultdict as dd  
from collections import deque  
from collections import Counter as c  
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect



mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()


def out(var): sys.stdout.write(var)


def l(): return list(map(int, data().split()))


def sl(): return list(map(str, data().split()))


def sp(): return map(int, data().split())


def ssp(): return map(str, data().split())


def l1d(n, val=0): return [val for i in range(n)]


def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


n, p = sp()
s = data()
answer = 0
if p == 2:
    for i in range(n):
        if (ord(s[i])-ord('0')) % 2 == 0:
            answer += (i+1)
elif p == 5:
    for i in range(n):
        if int(s[i]) % 5 == 0:
            answer += (i+1)
else:
    dp = dd(int)
    dp[0] += 1
    a, tens = 0, 1
    for i in range(n-1, -1, -1):
        a = (a + int(s[i]) * tens) % p
        answer += dp[a]
        dp[a] += 1
        tens = (tens * 10) % p
out(str(answer))
exit(0)

Solution (Python):
from operator import mul
from functools import reduce
import sys

def cmb(n,r):
    r=min(n-r,r)
    if r==0:
        return 1
    over=reduce(mul,range(n,n-r,-1))
    under=reduce(mul,range(1,r+1))
    return over//under

N,P=map(int,input().split())
S=list(map(int,list(input())))

if P==2:
    ans=0
    for i in range(N):
        if S[i]%2==0:
            ans+=i+1
    print(ans)
    sys.exit()

if P==5:
    ans=0
    for i in range(N):
        if S[i]==0 or S[i]==5:
            ans+=i+1
    print(ans)
    sys.exit()


data=[1]*N
for i in range(N-2,-1,-1):
    data[i]=data[i+1]*10%P

data[-1]=data[-1]*S[-1]%P

for i in range(N-2,-1,-1):
    data[i]=(data[i]*S[i]+data[i+1])%P

data.sort()
cnt=1
ans=data.count(0)
for i in range(1,N):
    if data[i]==data[i-1]:
        cnt+=1
    else:
        if cnt>=2:
            ans+=cmb(cnt,2)
        cnt=1
if cnt>=2:
    ans+=cmb(cnt,2)
    
print(ans)

Solution (Python):
N, P = map(int, input().split())
S = list(input())

if P != 2 and P != 5:
    count = [0] * P
    ans = 0
    tmp = 0
    for i in range(N - 1, -1, -1):
        tmp += int(S[i]) * pow(10, (N - 1 - i), P)
        tmp %= P
        ans += count[tmp]
        count[tmp] += 1
    ans += count[0]

    
    print (ans)

if P == 2:
    count = 0
    for index, s in enumerate(S):
        if int(s) % 2 == 0:
            count += (index + 1)
    print (count)

if P == 5: 
    count = 0
    for index, s in enumerate(S):
        if int(s) % 5 == 0:
            count += (index + 1)
    print (count)

Solution (Python):
N, P = map(int, input().split())
S = list(input())

if P != 2 and P != 5:
    count = [0] * P
    ans = 0
    tmp = 0
    for i in range(N - 1, -1, -1):
        tmp += int(S[i]) * pow(10, (N - 1 - i), P)
        tmp %= P
        ans += count[tmp]
        count[tmp] += 1
    ans += count[0]

    
    print (ans)

if P == 2:
    count = 0
    for index, s in enumerate(S):
        if int(s) % 2 == 0:
            count += (index + 1)
    print (count)

if P == 5: 
    count = 0
    for index, s in enumerate(S):
        if int(s) % 5 == 0:
            count += (index + 1)
    print (count)

Solution (Python):
n, p = map(int, input().split())
 
s = list(input())
ans = 0
if p == 2 or p == 5:
    for i in range(n):
        if int(s[i]) % p == 0:
            ans += i + 1
else:
    d = [0] * (n + 1)
    ten = 1
    for i in range(n - 1, -1, -1):
        a = int(s[i]) * ten % p
        ten *= 10
        ten %= p
        d[i - 1] = (d[i] + a) % p
    cnt = [0] * p
    for i in range(n, -1, -1):
        cnt[d[i]] += 1
    for  i in cnt:
        ans+=i*(i-1)/2

print(int(ans))

Solution (Python):
import sys

def solve():
    input = sys.stdin.readline
    N, P = map(int, input().split())
    S = list(input().strip("\n"))

    if P == 2 or P == 5:
        count = 0
        for i in range(N):
            s = int(S[N - 1 - i])
            if s % P == 0: count += N - i
        print(count)
    else:
        modDict = dict()
        modDict[0] = 1
        sSum = 0
        for i in range(N):
            s = int(S[N - 1 - i])
            sSum += s * pow(10, i, P)
            sSum %= P
            if sSum in modDict: modDict[sSum] += 1
            else: modDict[sSum] = 1
        count = 0
        for key in modDict:
            count += modDict[key] * (modDict[key] - 1) // 2
        print(count)
            

    return 0

if __name__ == "__main__":
    solve()

Solution (Python):
n, p = map(int, input().split())
s = input()
ans = 0

if 10 % p == 0:
    for r in range(n):
        if int(s[r]) % p == 0:
            ans += r + 1
    print(ans)
    exit()

d = [0 for _ in range(n + 1)]
ten = 1

for i in range(n - 1, -1, -1):
    a = int(s[i]) * ten % p
    d[i] = (d[i + 1] + a) % p
    ten *= 10
    ten %= p

cnt = [0 for _ in range(p)]
for i in range(n, -1, -1):
    ans += cnt[d[i]]
    cnt[d[i]] += 1



print(ans)

Solution (Python):
N, P = map(int, input().split())
S = input()

T = [0] * (N + 1)
T[0] = 0
e = 1
for i in range(N):
  T[i + 1] = T[i] + int(S[N - i - 1]) * e
  T[i + 1] %= P
  e *= 10
  e %= P

a = {}
for i in range(N + 1):
  if T[i] in a:
    a[T[i]] += 1
  else:
    a[T[i]] = 1

num = 0
if P % 2 == 0 or P % 5 == 0:
  for i in range(N):
    if int(S[i]) % P == 0:
      num += 1 + i
else:
  for k in a:
    num += a[k] * (a[k] - 1) // 2

print(num)

Solution (Python):
from collections import Counter


N, P = map(int, input().split())
S = list(input())

cnt = 0
if P in [2, 5]:
    for i, s in enumerate(S):
        if int(s) % P == 0:
            cnt += (i + 1)
else:
    num = 0
    mods = Counter()
    for i, s in enumerate(reversed(S)):
        num += int(s) * pow(10, i, P)
        num %= P
        mods[num] += 1
    mods[0] += 1
    for i, s in enumerate(S):
        mods[num] -= 1
        cnt += mods[num]
        num -= int(s) * pow(10, N - 1 - i, P)
        num %= P

print(cnt)

Solution (Python):
from collections import Counter


N, P = map(int, input().split())
S = input()

cnt = 0
if P in [2, 5]:
    for i, s in enumerate(S):
        if int(s) % P == 0:
            cnt += (i + 1)
else:
    T = [0] * (N + 1)
    mods = Counter({0: 1})
    for i, s in enumerate(reversed(S)):
        T[N - 1 - i] = (T[N - i] + int(s) * pow(10, i, P)) % P
        mods[T[N - 1 - i]] += 1
    for t in T:
        mods[t] -= 1
        cnt += mods[t]

print(cnt)

Solution (Python):
n, p = map(int, input().split())

s = list(input())

ans = 0
if p == 2 or p == 5:
    for i in range(n):
        if int(s[i]) % p == 0:
            ans += i + 1
else:
    d = [0] * (n + 1)
    ten = 1
    for i in range(n - 1, -1, -1):
        a = int(s[i]) * ten % p
        ten *= 10
        ten %= p
        d[i - 1] = (d[i] + a) % p
    cnt = [0] * p
    for i in range(n, -1, -1):
        ans += cnt[d[i]]
        cnt[d[i]] += 1

print(ans)