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: Accepted

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:
        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
    # 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: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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" # 上の定式化の都合上、r = Nのケースが必要なので
    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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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()

Submission status: Accepted

Solution code:
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)
    #zero sum range
    for i in c.values():
        #余りがその値だった時の個数iについて、割った値が0になる状態数はi*(i-1)//2
        ans += i*(i-1)//2
    ans += c[0]#割った余りが0の時はもちろん条件を満たす
print(ans)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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):#左から●桁めまでの整数を(mod p)でもっておく
    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 #dで終わる整数はすべてpで割り切れる
else:
    for i in range(p):
        ans+=cnt[i]*(cnt[i]-1)//2
print(ans)

Submission status: Accepted

Solution code:
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))

Submission status: Accepted

Solution code:
"""
    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 math import *
from collections import defaultdict as dd  # defaultdict(<datatype>) Free of KeyError.
from collections import deque  # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb
from bisect import bisect_left as bl, bisect_right as br, bisect
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
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)

Submission status: Accepted

Solution code:
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)

Submission status: Accepted

Solution code:
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 (count)
    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)

Submission status: Accepted

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
        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))