Problem:


Score : 400 points

Problem Statement

Given is a string S. Each character in S is either a digit (0, ..., 9) or ?.

Among the integers obtained by replacing each occurrence of ? with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.

Since the answer can be enormous, print the count modulo 10^9+7.

Constraints

S is a string consisting of digits (0, ..., 9) and ?.

1 \leq |S| \leq 10^5

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of integers satisfying the condition, modulo 10^9+7.

Sample Input 1

??2??5

Sample Output 1

768

For example, 482305, 002865, and 972665 satisfy the condition.

Sample Input 2

?44

Sample Output 2

1

Only 044 satisfies the condition.

Sample Input 3

7?4

Sample Output 3

0

We may not be able to produce an integer satisfying the condition.

Sample Input 4

?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???

Sample Output 4

153716888



Sample input:
??2??5


Sample output:
768

Submission status: Time Limit Exceeded

Solution code:
def main():

    """ 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
    # 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)

    # 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]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # open(0).read() is a convenient method:
        # 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 used
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = 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)]
    """

    if __name__ != '__main__':
        # for test
        sys.stdin = open('sample.txt')
        
    # ----------------------------------
    # main

    S = r()
    lenS = len(S)
    target_digit_S = lenS - 1
    dp = [[0]*13 for _ in range(lenS)] 
    reslist = [10, 9, 12, 3, 4, 1]
    # initialize
    
    res = reslist[(target_digit_S-1) % 6] # (10**target_digit_S)%13
    if S[0] == '?':
        for i in range(10):
            dp[0][(i*res)%13] += 1
    else:
        intS = int(S[0])
        dp[0][(intS*res)%13] += 1
    
    for i in range(1, lenS):
        target_digit_S -= 1
        res = reslist[(target_digit_S-1) % 6] # res = (10**target_digit_S)%13
        if S[i] == '?':
            for k in range(10):
                for j in range(13):
                    dp[i][(k*res+j)%13] = (dp[i][(k*res+j)%13] + dp[i-1][j]) % MOD
        else:
            intS = int(S[i])
            for j in range(13):
                dp[i][(intS*res+j)%13] += (dp[i][(intS*res+j)%13] + dp[i-1][j]) % MOD
    print(dp[lenS-1][5])
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # for
    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
    """

if __name__ == '__main__':
    main()

Submission status: Time Limit Exceeded

Solution code:
def main():

    """ 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
    # 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)

    # 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]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # open(0).read() is a convenient method:
        # 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 used
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = 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)]
    """

    if __name__ != '__main__':
        # for test
        sys.stdin = open('sample.txt')
        
    # ----------------------------------
    # main

    S = r()
    lenS = len(S)
    target_digit_S = lenS - 1
    dp = [[0]*13 for _ in range(lenS)] 
    
    # initialize
    res = (10**target_digit_S)%13
    if S[0] == '?':
        for i in range(10):
            dp[0][(i*res)%13] += 1
    else:
        intS = int(S[0])
        dp[0][(intS*res)%13] += 1
    
    for i in range(1, lenS):
        target_digit_S -= 1
        res = (10**target_digit_S)%13
        if S[i] == '?':
            for k in range(10):
                for j in range(13):
                    dp[i][(k*res+j)%13] = (dp[i][(k*res+j)%13] + dp[i-1][j]) % MOD
        else:
            intS = int(S[i])
            for j in range(13):
                dp[i][(intS*res+j)%13] += (dp[i][(intS*res+j)%13] + dp[i-1][j]) % MOD
    print(dp[lenS-1][5])
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # for
    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
    """

if __name__ == '__main__':
    main()

Submission status: Time Limit Exceeded

Solution code:
# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 10 ** 9 + 7

s = input()[::-1]

dp = [0 for i in range(N)]
nextdp = [0 for i in range(N)]
zeros = [0 for i in range(N)]
dp[0] = 1

for i, si in enumerate(s):
    nextdp = zeros[:]
    r = 10 ** i % N

    for j in range(10):
        jj = j
        if si == "?":
            pass
            # c = (j * 10 ** i) % N
        elif j == 0:
            jj = int(si)
        else:
            break

        for k in range(N):
            l = (k + r * jj) % N
            nextdp[l] = (nextdp[l] + dp[k]) % MOD

    dp = nextdp[:]

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 10 ** 9 + 7

s = input()
rs = reversed(s)

dp = [0 for i in range(N)]
dp[0] = 1

for i, si in enumerate(rs):
    nextdp = [0 for i in range(N)]
    r = 10 ** i % N

    for j in range(10):
        jj = j
        if si == "?":
            pass
            # c = (j * 10 ** i) % N
        elif j == 0:
            jj = int(si)
        else:
            break

        for k in range(N):
            l = (k + r * jj) % N
            nextdp[l] = (nextdp[l] + dp[k]) % MOD

    dp = nextdp

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 10 ** 9 + 7

s = input()
rs = reversed(s)

dp = [0 for i in range(N)]
dp[0] = 1

for i, si in enumerate(rs):
    nextdp = [0 for i in range(N)]
    r = 10 ** i % N

    for j in range(10):
        jj = j
        if si == "?":
            pass
            # c = (j * 10 ** i) % N
        elif j == 0:
            jj = int(si)
        else:
            break

        for k in range(N):
            l = (k + r * jj) % N
            nextdp[l] = (nextdp[l] + dp[k]) % MOD

    dp = copy.copy(nextdp)

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 10 ** 9 + 7

s = input()
rs = reversed(s)

dp = [0 for i in range(N)]
dp[0] = 1
zeros = [0 for i in range(N)]

for i, si in enumerate(rs):
    nextdp = zeros[:]

    for j in range(10):
        c = 0
        if si == "?":
            c = (j * 10 ** i) % N
        elif j == 0:
            c = (int(si) * 10 ** i) % N
        else:
            break

        for k in range(N):
            nextdp[(k + c) % N] = (nextdp[(k + c) % N] + dp[k]) % MOD

    dp = nextdp

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 10 ** 9 + 7

s = input()
rs = reversed(s)

dp = [0 for i in range(N)]
dp[0] = 1
nextdp = [0 for i in range(N)]

for i, si in enumerate(rs):
    nextdp = [0 for j in range(N)]

    for j in range(10):
        c = 0
        if si == "?":
            c = (j * 10 ** i) % N
        elif j == 0:
            c = (int(si) * 10 ** i) % N
        else:
            break

        for k in range(N):
            nextdp[(k + c) % N] = (nextdp[(k + c) % N] + dp[k]) % MOD

    dp = copy.copy(nextdp)

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

S = input()

dp = [0] * 13

dp[0] = 1

mod = 10 ** 9 + 7
mul = 1
for i in range(len(S)-1, -1, -1):
    # 二次元dpの代わり
    ndp = [0] * 13
    
    if S[i] == '?':
        for k in range(10):
            for j in range(13):
                # 次の桁の余りが加わった時に、そこへ前のdp[j]の値を入れる
                # ただし、0-9までの値をその桁が取る可能性があるので、
                #　それぞれの場合全てを加算する
                ndp[(k * mul + j) % 13] += dp[j]
                ndp[(k * mul + j) % 13] %= mod
    else:
        k = int(S[i])
        for j in range(13):
            # 次の桁の余りが加わった時に、そこへ前のdp[j]の値を入れる
            ndp[(k * mul + j) % 13] += dp[j]
            ndp[(k * mul + j) % 13] %= mod

    mul *= 10
    mul %= 13
    dp = copy.deepcopy(ndp)

# 13で割って5で余るものを出力
print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
# import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 10 ** 9 + 7
s = input()
rs = reversed(s)

dp = [0 for i in range(N)]
dp[0] = 1

for i, si in enumerate(rs):
    nextdp = [0 for i in range(N)]

    for j in range(10):
        c = 0
        if si == "?":
            c = (j * 10 ** i) % N
        elif j == 0:
            c = (int(si) * 10 ** i) % N
        else:
            break

        for k in range(N):
            nextdp[(k + c) % N] += dp[k]
            nextdp[(k + c) % N] %= MOD

    for j in range(N):
        dp[j] = nextdp[j]
    # dp = copy.copy(nextdp)

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 1000000000 + 7
s = input()
dp = [0 for i in range(N)]
nextdp = [0 for i in range(N)]
dp[0] = 1
rs = reversed(s)

for i, si in enumerate(rs):
    nextdp = [0 for j in range(N)]

    for j in range(10):
        c = 0
        if si == "?":
            c = (j * 10 ** i) % N
        elif j == 0:
            c = (int(si) * 10 ** i) % N
        else:
            break

        for k in range(N):
            nextdp[(k + c) % N] += dp[k]
            nextdp[(k + c) % N] %= MOD

    dp = copy.copy(nextdp)

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 1000000000 + 7
s = input()
dp = [0 for i in range(N)]
nextdp = [0 for i in range(N)]
dp[0] = 1

for i, si in enumerate(reversed(s)):
    nextdp = [0 for j in range(N)]

    for j in range(10):
        c = 0
        if si == "?":
            c = (j * 10 ** i) % N
        elif j == 0:
            c = (int(si) * 10 ** i) % N
        else:
            break

        for k in range(N):
            nextdp[(k + c) % N] += dp[k]
            nextdp[(k + c) % N] %= MOD

    dp = copy.copy(nextdp)

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
import copy

# standart input
# n = int(input())
# a = list(map(int, input().split()))

N = 13
MOD = 1000000000 + 7
s = input()
dp = [0 for i in range(N)]
dp[0] = 1

for i, si in enumerate(reversed(s)):

    nextdp = [0 for i in range(N)]
    for j in range(10):
        c = 0
        if si == "?":
            c = (j * 10 ** i) % N
        elif j == 0:
            c = (int(si) * 10 ** i) % N
        else:
            break

        for k in range(N):
            nextdp[(k + c) % N] += dp[k]
            nextdp[(k + c) % N] %= MOD

    dp = copy.copy(nextdp)

print(dp[5])

Submission status: Time Limit Exceeded

Solution code:
d=[0]*13
d[0]=1
mul=1
mulmod=1
t=input()[::-1]
mod=10**9+7
zero=[0]*13

for c in t:
    e=zero[:]
    if c=="?":
        for i in range(10):
            for j in range(13):
                k=(i*mulmod+j)%13
                e[k]+=d[j]
                e[k]%=mod
    else:
        for j in range(13):
            k=(int(c)*mulmod+j)%13
            e[k]+=d[j]
            e[k]%=mod
    d=e
    mul*=10
    mulmod=mul%13
print(d[5])

Submission status: Time Limit Exceeded

Solution code:
d=[0]*13
d[0]=1
mul=1
mulmod=1
t=input()[::-1]
mod=10**9+7
zero=[0]*13

for c in t:
    e=zero[:]
    if c=="?":
        for i in range(10):
            for j in range(13):
                k=(i*mulmod+j)%13
                e[k]+=d[j]
                e[k]%=mod
    else:
        for j in range(13):
            k=(int(c)*mulmod+j)%13
            e[k]+=d[j]
            e[k]%=mod
    d=e[:]
    mul*=10
    mulmod=mul%13
print(d[5])

Submission status: Time Limit Exceeded

Solution code:
d=[0]*13
d[0]=1
mul=1
mulmod=1
t=input()[::-1]

zero=[0]*13
l=len(t)
modlist=[10**i%13 for i in range(l)]
for n,c in enumerate(t):
    e=zero[:]

    for i in (range(10) if c=="?" else[int(c)]):
        for j in range(13):
            #e[(i*mul+j)%13]+=d[j]
            k=(i*mulmod+j)%13
            e[k]+=d[j]
            e[k]%=1000000007
    d=e
    mul*=10
    mulmod=mul%13
print(d[5])