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



Sample input:
4 3
3543


Sample output:
6

Submission status: Runtime Error

Solution code:
n,p = map(int, input().split())
s = input()
x = 0
if(p == 5):
    for i in range(n):
        if(s[i]%5 == 0):
            x += i+1
else:
    for j in range(1,n+1):
        if(int(s[-j:])%p == 0):
            x += 1
        #print(s[-j:])
    for i in range(1,n):
        for j in range(1,n-i+1):
            if(int(s[-j:])%p == 0):
                x += 1
            #print(s[-i-j:-i])
print(x)

Submission status: Runtime Error

Solution code:
n,p = map(int, input().split())
s = input()
x = 0
if(p == 5):
    for i in range(n):
        if(s[i]%5 == 0):
            x += i+1
else:
    if()
    for j in range(1,n+1):
        if(int(s[-j:])%p == 0):
            x += 1
        #print(s[-j:])
    for i in range(1,n):
        for j in range(1,n-i+1):
            if(int(s[-j:])%p == 0):
                x += 1
            #print(s[-i-j:-i])
print(x)

Submission status: Runtime Error

Solution code:
def i1():
 return int(input())
def i2():
 return [int(i) for i in input().split()]
[n,p]=i2()
s=input()

  
if p==2 or p==5:
   x1=0
   for i in range(n):
     if int(s[i])%p==0:
        x1+=i+1  
   print(x1)
else:
  t=0
  x2=0
  d=[0]
  for i in range(n)[::-1]:
     t+=int(s[i])*a
     t%=p
     a*=10
     d.append(t)
  from collections import Counter
  c= Counter(d)
  x=0
  for i in c.values():
    if i>1:
      x+=i*(i-1)//2
  print(x)

Submission status: Runtime Error

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    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 = []  # prime factor
        b, e = 2, 0  # base, exponent
        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
        """
        # TODO: How should I do when n - r is negative?
        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 # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    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 # Deprecated since version 3.5: Use math.gcd() instead.
    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)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        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):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        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]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    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
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    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
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    
    N, P = R()
    S = r()
    if P == 2:
        assert 1 == 2
    if P == 5:
        assert 1 == 2
    # len(S) * 10 の行列に各桁各数値をPで割った時の余りを入力
    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
    # Sの右端（1桁目）から順に、その桁のその数値をPで割ったあまりをcumで足していく
    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)
    
    # 0から9までの数をそれぞれカウント
    
    # nC2の計算を0~9までそれぞれ計算し、和をtmpとおく
    
    # ans = tmp
    
    # 
    
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Runtime Error

Solution code:
def substr(s):
  sub=[]
  for i in range(len(s)):
    x=[]
    for j in range(i, len(s)):
      x.append(s[i])
    sub.append(x)
  return sub

n, p = map(int, input().split())
s = list(input().split())
x=substr(s)
c=0
for i in x:
  if int(i)%p == 0:
    c+=1
print(c, x)

Submission status: Runtime Error

Solution code:
N,P=map(int,input().split())
S=input()
L=[]
M=0
for i in range(len(S)):
    L.append((M*10+int(S[i]))%P)
    M=L[-1]
#print(L)
if P==5:
    n=0
    ans=0
    for i in range(len(S)):
        if L[i]==0 or L[i]==5:
            ans+=n
            n+=1
    if n>0:
        ans+=n
elif P==2:
    n=0
    ans=0
    for i in range(len(S)):
        if L[i]%2==0:
            ans+=n
            n+=1
    if n>0:
        ans+=n
else:    
    D={}
    ans=0
    for i in range(len(L)):
        if L[i] not in D:
            D[L[i]]=1
        else:
            ans+=D[L[i]]
            D[L[i]]+=1
    if D[0]>0:
        ans+=D[0]
print(ans)

Submission status: Runtime Error

Solution code:
n,p=map(int,input().split())
s=input()
arr=[int(s[i]) for i in range(len(s))]
dic={}
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)
  if tmp not in dic:
    dic[tmp]=1
  else:
    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*=rev
  rem+=arr[i]
  rem%=p
  dic[match[i]]-=1
print(ans)

Submission status: Runtime Error

Solution code:
N,P=map(int,input().split())
S=input()
if P==2 or P==5:
    ans=0
    for i in range(N):
        if int(S[i]) % P == 0:
            ans += i+1
    print(ans)            
else:
    mod_sum = 0
    mod_dic = {}
    for i in range(N):
        mod = (10**i * int(S[-1-i])) % P
        mod_sum = (mod_sum + mod) % P
        mod_dic[mod_sum]= mod_dic[mod_sum]+1 if mod_sum in mod_dic else 1
    ans = mod_dic[0]
    for v in mod_dic.values():
        ans += v*(v-1)//2
    print(ans)

Submission status: Runtime Error

Solution code:
import sys


N,P = map(int,input().split())
S = list(sys.stdin.readline().rstrip())

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


ans = 0
if P==2 or P==5:
    for i in reversed(range(N)):
        if int(S[i])%P==0:
            ans += i+1
else:
    a = A[10**9]
    nums = {}
    for a in A:
        nums[a] = 0
    for a in A:
        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)

Submission status: Runtime Error

Solution code:
import sys
input = sys.stdin.readline

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

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


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:
        nums[a] = 0
    for a in A:
        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)

Submission status: Runtime Error

Solution code:

import sys
input = sys.stdin.readline

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

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


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:
        nums[a] = 0
    for a in A:
        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)

Submission status: Runtime Error

Solution code:
n, p = map(int, input().split())
 
s = list(input())
ans = 0
sum=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
        sum+=a
        ten *= 10
        ten %= p
        d[i]= sum % 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(d)
print(int(ans))

Submission status: Runtime Error

Solution code:
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
        sum+=a
        ten *= 10
        ten %= p
        d[i]= sum % 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(d)
print(int(ans))

Submission status: Runtime Error

Solution code:
n, p = map(int,input().split())
D = input()
count = [0] * p
out = 0

if 10 % p == 0:
    for i in range(n):
        if int(D[i]) % p == 0:
            out += i + 1
else:
    ten = 1
    count[0] = 1
    for i in range(n):
        mod = (mod + int(D[n-i-1]) * ten) % p
        count[mod] += 1
        ten *= 10
    for c in count:
        out += (c * (c - 1)) // 2
print(out)

Submission status: Runtime Error

Solution code:
n, p = map(int,input().split())
D = input()
count = [0] * p
out = 0

if 10 % p == 0:
    for i, d in enumerate(D):
        if int(d) % p == 0:
            out += i + 1
else:
    ten = 1
    count[0] = 1
    for i in range(n):
        mod = (mod + int(D[n-i-1]) * ten) % p
        count[mod] += 1
        ten *= 10
    for c in count:
        out += (c * (c - 1)) // 2
print(out)

Submission status: Runtime Error

Solution code:
n, p = map(int,input().split())
D = input()
count = [0] * p
out = 0

if 10 % p == 0:
    for i, d in enumerate(D):
        if int(d) % p == 0:
            out += i + 1
else:
    ten = 1
    count[0] = 1
    for i in range(n):
        mod = (mod + int(D[n-i-1]) * ten) % p
        count[mod] += 1
        ten *= 10
    for c in count:
        out += (c * (c - 1)) // 2
print(int(out))

Submission status: Runtime Error

Solution code:
n, p = map(int,input().split())
D = input()
count = [0] * p
out = 0

if 10 % p == 0:
    for i, d in enumerate(D):
        if int(d) % p == 0:
            out += i + 1
else:
    ten = 1
    for i in range(n):
        mod = (mod + int(D[n-i-1]) * ten) % p
        count[mod] += 1
        ten *= 10
    for c in count:
        out += (c * (c - 1)) // 2
print(int(out) + count[0])

Submission status: Runtime Error

Solution code:
n, p = map(int,input().split())
D = input()
count = [0] * p
out = 0

if 10 % p == 0:
    for i, d in enumerate(D):
        if int(d) % p == 0:
            out += i + 1
else:
    ten = 1
    for i in range(n):
        mod = (mod + int(D[n-i-1]) * ten) % p
        count[mod] += 1
        ten *= 10
    print(count)
    for c in count:
        out += (c * (c - 1)) // 2
print(int(out) + count[0])

Submission status: Runtime Error

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

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

    # print (count)
    ans = 0
    for i in count:
        ans += (i * (i - 1)) // 2

    print (ans)

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

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

Submission status: Runtime Error

Solution code:
import collections

n, p = map(int,input().split())
D = input()
s = [0] * n
mod = 0
out = 0
if 10 % p == 0:
    for i, d in enumerate(D):
        if int(d) % p == 0:
            out += i + 1
else:
    M = [(int(d) * 10 ** i) % p for i, d in enumerate(D)]
    s[0] = M[0]
    for i in range(1, n):
        s[i] = (s[i - 1] + M[i]) % p
    C = collections.Counter(s)
    out += C[0]
    for c in C.values():
        out += (int(c) * (int(c) - 1)) / 2
print(int(out)

Submission status: Runtime Error

Solution code:
n,p=map(int,input().split())
s=input()
if p==2 or p==5:
    a=0
    for i in range(n):
        if int(s[i])%p==0:
            ans+=n
    print(a)
else:
    a=0
    g=[0]*(n+1)
    t=[0]*p
    t[0]=1
    d=1
    for i in range(n-1,-1,-1):
        g[i]=(g[i+1]+int(s[i])*d)%p
        a+=t[g[i]]
        t[g[i]]+=1
        d*=10
        d%=p
    print(a)

Submission status: Runtime Error

Solution code:
#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, P;
  string S;
  cin >> N >> P >> S;
  int64_t res = 0;
  if (P == 2 || P == 5) {
    for (int i = 0; i < N; i++) {
      if ((S[i] - '0') % P == 0) res += i + 1;
    }
  } else {
    map<int, int> mp;
    mp[0] = 1;
    int cur = 0;
    for (char s : S) {
      cur = cur * 10 + (s - '0');
      cur %= P;
      res += mp[cur];
      mp[cur]++;
    }
  }
  cout << res << '\n';
  return 0;
}

Submission status: Runtime Error

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


  def solve1(s, p):
      v = 0
      pow10 = 1
      ans = 0
      cnt = [0] * p
      cnt[0] = 1

      for i in range(n - 1, -1, -1):
          v = (pow10 * int(s[i]) + v) % p
          pow10 = (pow10 * 10) % p
          ans += cnt[v]
          cnt[v] += 1
      return ans


  def solve25(s, p):
      cnt = 0
      ans = 0
      for i in range(len(s) - 1, -1, -1):
          if int(s[i]) % p == 0:
              cnt += 1
          ans += cnt
      return ans



  if p in (2,5):
      print(solve25(s, p))
  else:
      print(solve1(s, p))

Submission status: Runtime Error

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

if 10 % p ==0:
  print(sum(i for i, x in enumerate(s,1) if int(x)%p == 0))
else:
  d = [1]+[0]*(p-1)
  t,y = 0, 1
  for i in reversed(s):
    t = (i*y + t)%p
    y *= 10%p
    d[t] += 1
  print(sum(i*(i-1)/2 for i in d))

Submission status: Runtime Error

Solution code:
from collections import Counter
n,p=map(int,input().split())
s=input()[::-1]
a=0
t=0
d=1
m=[0]*p
m=[0]
for i in range(n):
  x=int(s[i])*d%p
  m.append((m[-1]+x)%p)
  d=(d*10)%p
c=Counter(m)
for i,j in c.items():
  a+=j*(j-1)//2

if p==2 or p==5:
  a=0
  for i in range(n):
    if s[n-1-i]%p==0:
      a+=n-1-i
  print(a)
else:
  print(a)