Problem:


Score: 600 points

Problem Statement

How many strings can be obtained by applying the following operation on a string S exactly K times: "choose one lowercase English letter and insert it somewhere"?

The answer can be enormous, so print it modulo (10^9+7).

Constraints

K is an integer between 1 and 10^6 (inclusive).

S is a string of length between 1 and 10^6 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

K
S

Output

Print the number of strings satisfying the condition, modulo (10^9+7).

Sample Input 1

5
oof

Sample Output 1

575111451

For example, we can obtain proofend, moonwolf, and onionpuf, while we cannot obtain oofsix, oofelevennn, voxafolt, or fooooooo.

Sample Input 2

37564
whydidyoudesertme

Sample Output 2

318008117



Sample input:
5
oof


Sample output:
575111451

Submission status: Wrong Answer

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, combinations_with_replacement # 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
    def gcd(a,b):
        # from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
        # => 最近の環境でfractions.gcdを使うとTLEする。
        if a<b:
            a, b = b, a
        while a%b !=0 :
            a, b = b, a%b
        return b

    # def gcd(p,q):
    #     while q != 0:
    #         r = p % q
    #         p = q
    #         q = r
    #     return p

    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)

    # Eratosthenesの篩により、整数end未満の素数を列挙する
    def sieve_of_eratosthenes(end, typecode="L"):
        # len(sieve_of_eratosthenes(10**7)) => 664579; 514ms in pypy
        # len(sieve_of_eratosthenes(10**6)) => 78498; 134ms in pypy

        # https://hamukichi.hatenablog.jp/entry/2016/02/17/215948#Eratosthenes%E3%81%AE%E7%AF%A9%E3%81%A8%E5%8C%BA%E9%96%93%E7%AF%A9
        import array
        import textwrap
        assert end > 1
        # 整数iが素数であるかをis_prime[i]が示す
        # 最初はすべてTrueで初期化しておく
        # 最終的にprimesではなくこれを返してもよい
        is_prime = array.array("B", (True for i in range(end)))
        # 0, 1はいずれも素数ではない
        is_prime[0] = False
        is_prime[1] = False
        # 素数を格納する配列
        primes = array.array(typecode)
        # 篩う
        for i in range(2, end):
            if is_prime[i]:
                primes.append(i)
                for j in range(2 * i, end, i):
                    is_prime[j] = False  # 素数ではないため除外する
        return primes

    class PrimeFactorization:
        '''初期化にO(NloglogN)/ 因数分解クエリをO(log(k))/ 素数判定クエリをO(1)
        >>> pf = PrimeFactorization(100)
        >>> pf.query(100)
        ((2, 2), (5, 2))
        
        '''
        # 自分で実装した。
        # https://atcoder.jp/contests/abc177/editorial/82
        # 高速素因数分解
        # 問題： A 以下の数が N 個与えられる。全て素因数分解せよ。
        # 前計算としてエラトステネスの篩を行い、「その数をふるい落とした素数」を配列 D に記録します。
        # 例えば D[4]=D[6]=2,D[35]=5 です。x が素数のときは D[x]=x としておきます。この配列はエラトステネスの篩と同様 O(AloglogA) で構築できます。
        # D[x] は x を割り切る最小の素数なので、この配列 D を利用すると素因数分解を行うときに「試し割り」をする必要がなくなり(D[x]で割ればよい)、1つの数の素因数分解が素因数の個数である O(logA) でできるようになります。
        def __init__(self, Num=10**6):
            from collections import defaultdict
            from math import floor
            self.Num = Num
            self.D = [-1] * (Num+1)
            for i in range(2, Num+1):
                if self.D[i] != -1:
                    continue
                self.D[i] = i
                for j in range(1, floor(Num/i)+1):
                    if self.D[j*i] != -1:
                        continue
                    else:
                        self.D[j*i] = i
        
        def query(self, k):
            ans = defaultdict(int)
            tmp = k
            while self.D[tmp] != -1:
                div = self.D[tmp]
                ans[div] += 1
                tmp = tmp//div
            return tuple(ans.items())
        
        def is_prime(self, k):
            return k == self.D[k]

    # first create factorial_list
    # fac_list = modfactorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrstuvwxyz') # 26
    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 like:
        >>      pass
        >>  else:
        >>      do something like:
        >>      L[find_root(a)] = L[find_root(a)] + L[find_root(b)]
        >>      L[find_root(b)] = L[find_root(a)]
        >>      
        >>      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]

    # c++ set
    # https://nagiss.hateblo.jp/entry/2020/09/08/203701
    
    class SegTree:
        """
        ref: https://qiita.com/takayg1/items/c811bd07c21923d7ec69
        init(init_val, ide_ele): 配列init_valで初期化 O(N)
        update(k, x): k番目の値をxに更新 O(logN)
        query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
        
        ex: seg = SegTree([14, 5, 9, 13], min, float('inf'))
        """
        def __init__(self, init_val, segfunc, ide_ele):
            """
            init_val: 配列の初期値
            segfunc: 区間にしたい操作
            ide_ele: 単位元
            n: 要素数
            num: n以上の最小の2のべき乗
            tree: セグメント木(1-index)
            """
            n = len(init_val)
            self.segfunc = segfunc
            self.ide_ele = ide_ele
            self.num = 1 << (n - 1).bit_length()
            self.tree = [ide_ele] * 2 * self.num
            # 配列の値を葉にセット
            for i in range(n):
                self.tree[self.num + i] = init_val[i]
            # 構築していく
            for i in range(self.num - 1, 0, -1):
                self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

        def update(self, k, x):
            """
            k番目の値をxに更新
            k: index(0-index)
            x: update value
            """
            k += self.num
            self.tree[k] = x
            while k > 1:
                self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
                k >>= 1

        def query(self, l, r):
            """
            [l, r)のsegfuncしたものを得る
            l: index(0-index)
            r: index(0-index)
            """
            res = self.ide_ele
            l += self.num
            r += self.num
            while l < r:
                if l & 1:
                    res = self.segfunc(res, self.tree[l])
                    l += 1
                if r & 1:
                    res = self.segfunc(res, self.tree[r - 1])
                l >>= 1
                r >>= 1
            return res

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # 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]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ 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
    #%%
    K = r_int()
    S = len(r())
    
    ans = 0
    MOD = 10**9+7
    l = [1] * (K+1)
    r = [1] * (K+1)
    m = [1] * (K+1)
    for i in range(1, K+1):
        l[i] = (l[i-1] * 25)%MOD
        r[i] = (r[i-1] * 26)%MOD
        m[i] = (m[i-1] * (S-1+i) / (i))%MOD
    
    for i in range(K+1):
        ans = (ans + m[i]*l[i]*r[K-i])%MOD
    
    print(int(ans))
    
    #%%
    # 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: Wrong Answer

Solution code:
MOD = 10**9 + 7
list_size = 2000001
 
f_list = [1] * list_size
f_r_list = [1] * list_size
 
for i in range(list_size-1):
	f_list[i+1] = (f_list[i] * (i+1)) % MOD
 
f_r_list[-1] = pow(f_list[-1], MOD - 2, MOD)
 
for i in range(list_size-2, -1, -1):
	f_r_list[i] = (f_r_list[i+1] * (i+1)) % MOD
 
def comb(n, r):
	if n < r or r < 0:
		return 0
	elif n == 0 or r == 0 or n == r:
		return 1
	else:
		return (f_list[n] * f_r_list[n-r] * f_r_list[r]) % MOD

Submission status: Wrong Answer

Solution code:
from sys import stdin
input = stdin.readline
def nCr(n,r):
    return fact[n]*(inv[n-r]%M)*(inv[r]%M)
M = 1000000007
k = int(input())
s = len(input())
t = k+s
fact = [1]
p25 = [1]
for i in range(1,t+1):
    fact += [(fact[-1] * i)%M]
    if i<=k:
        p25  += [(p25[-1] * 25) % M]

inv = [1]*(t+1)
inv[t] = pow(fact[t], M-2, M)
for i in range(t-1, -1, -1):
    inv[i] = inv[i+1] * (i+1) % M


res = 0
for i in range(k+1):
    res+=nCr(t,t-i)*p25[i]
    res%=M
print(res)

Submission status: Wrong Answer

Solution code:
def main():
  k = int(input())
  s = input()
  n = 1
  for i in range( len(s)-1, k+1):
    n = n * i
  
if __name__ == '__main__':
  main()

Submission status: Wrong Answer

Solution code:
from sys import stdin,stdout
input = stdin.readline
print = stdout.write
from math import factorial as f
def nCr(n,r):
    return f(n)//f(r)//f(n-r)
M = 1000000007
k = int(input())
ls = len(input())
total = pow(26,ls+k,M)
noSubsequence = 0
for i in range(ls):
    noSubsequence += (nCr(k+ls,i)%M)*pow(25,k+ls-i,M)
    noSubsequence %= M
subsequence = str(total - noSubsequence)
print(subsequence+'\n')

Submission status: Wrong Answer

Solution code:
from numba import njit
MOD = 10**9 + 7


def prepare(n, MOD):
 
    # 1! - n! の計算
    f = 1
    factorials = [1]  # 0!の分
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    # n!^-1 の計算
    inv = pow(f, MOD - 2, MOD)
    # n!^-1 - 1!^-1 の計算
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
     
    return factorials, invs

@njit
def main(k,n,fact,inv):
    ans = 0
    for i in range(k+1):
        c25 = 25**i
        c25 % MOD
        f = fact[i+n-1] * inv[i] % MOD * inv[n-1] % MOD
        c25 *= f
        c25 %= MOD
    
        c26 = 26**(k-i)
        c26 % MOD
        ans += (c25 * c26 % MOD)
        ans %= MOD
    return ans


k = int(input())
s = input()
n = len(s)
fact, inv = prepare(n+k, MOD)
print(main(k,n,fact,inv))

Submission status: Wrong Answer

Solution code:
MOD = 10**9 + 7

def prepare(n, MOD):
 
    # 1! - n! の計算
    f = 1
    factorials = [1]  # 0!の分
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    # n!^-1 の計算
    inv = pow(f, MOD - 2, MOD)
    # n!^-1 - 1!^-1 の計算
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
     
    return factorials, invs

'''
k = int(input())
s = input()
'''
k = 5
s = 'oop'
n = len(s)
ans = 0

fact, inv = prepare(10**6, MOD)
for i in range(k+1):
    c25 = 1
    for j in range(i):
        c25 *= 25
        c25 % MOD
    f = fact[i+n-1] * inv[i] % MOD * inv[n-1] % MOD
    c25 *= f
    c25 %= MOD
    c26 = 1
    for h in range(k-i):
        c26 *= 26
        c26 % MOD
    ans += (c25 * c26 % MOD)
    ans %= MOD
print(ans)

Submission status: Wrong Answer

Solution code:
k = int(input())
s = input()

mod = 10**9+7
n = len(s) + k

frac = [1]*(n+1)
finv = [1]*(n+1)
for i in range(n):
    frac[i+1] = (i+1)*frac[i]%mod
finv[-1] = pow(frac[-1],mod-2,mod)
for i in range(1,n+1):
    finv[n-i] = finv[n-i+1]*(n-i+1)%mod
print(finv,frac)
def nCr(n,r,m):
    if n<0 or r<0 or n<r: return 0
    r = min(r,n-r)
    return frac[n]*finv[n-r]*finv[r]%mod
a,b = [1],[1]
for _ in range(k+1):
    a.append(a[-1]*25%mod)
    b.append(b[-1]*26%mod)
n = len(s)
ans = 0
for i in range(k+1):
    ans += a[i]*b[k-i]*nCr(i+n-1,n-1,mod)
    ans %= mod
print(ans)

Submission status: Wrong Answer

Solution code:
import sys
import time


K = int(sys.stdin.readline())
S = sys.stdin.readline().rstrip('\n')

# ## COMBINATION (MOD) ###
N_MAX = 10**6  # 問題サイズに合わせて変えておく
MOD = 10**9 + 7

st = time.time()
inv = [0] * (N_MAX + 2)
inv[0] = 0  # 逆元テーブル計算用テーブル
inv[1] = 1

for i in range(2, N_MAX + 2):
    q, r = divmod(MOD, i)
    inv[i] = -inv[r] * q % MOD

en = time.time()
print(en-st)

# K 文字追加
ans = 0

ln = len(S)

p = pow(26, K, MOD)

st = time.time()
for i in range(1, K + 2):

    ans += p % MOD
    ans %= MOD

    # pre
    p = p * (ln + i - 1) * inv[i] * 25 * inv[26] % MOD

    # suf
    # s2 = (s2 * inv[26]) % MOD
en = time.time()

print(ans)
print(en-st)

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 24 20:53:22 2020

@author: NEC-PCuser
"""

if __name__ == '__main__':
    K = int(input())
    n = len(input())
    mod = 10**9+7 #出力の制限
    #25 25 a 25 25 25 b 25 25  c  26 26
    cmb = 1
    ans = 0
    for i in range(0, K + 1):
        print(cmb)
        ans1 = (pow(25, i, mod) * cmb)
        ans2 = pow(26, K - i, mod)
        cmb = (cmb * (n + i) // (i + 1)) % mod  
        ans = (ans + (ans1 * ans2)) % mod
    
    
    print(ans)

Submission status: Wrong Answer

Solution code:
import sys

K = int(sys.stdin.readline())
S = sys.stdin.readline().rstrip('\n')

# ## COMBINATION (MOD) ###
N_MAX = 10**6  # 問題サイズに合わせて変えておく
MOD = 10**9 + 7

inv = [0, 1]  # 逆元テーブル計算用テーブル
K = max(K, 26)
for i in range(2, K + 2):
    inv.append((-inv[MOD % i] * (MOD // i)) % MOD)

# K 文字追加
ans = 0

ln = len(S)

p1 = 1
p2 = 1
s2 = pow(26, K, MOD)

for i in range(1, K + 2):

    ans += (p1 * p2 * s2) % MOD
    ans %= MOD

    # print(p1, p2, s2)

    # pre
    p1 = (p1 * (ln + i - 1) * inv[i]) % MOD
    p2 = (p2 * 25) % MOD

    # suf
    s2 = (s2 * inv[26]) % MOD

print(ans)

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 24 20:53:22 2020

@author: NEC-PCuser
"""

if __name__ == '__main__':
    K = int(input())
    n = len(input())
    mod = 10**9+7 #出力の制限
    #25 25 a 25 25 25 b 25 25  c  26 26
    cmb = 1
    ans = 0
    for i in range(0, K + 1):
        ans1 = (pow(25, i, mod) * cmb)
        ans2 = pow(26, K - i, mod)
        cmb = (cmb * (n + i) // (i + 1)) % mod  
        ans += (ans1 * ans2) % mod
    
    print(ans)

Submission status: Wrong Answer

Solution code:
from operator import mul
from functools import reduce


def combinations_count(n, r):
    r = min(r, n - r)
    numer = reduce(mul, range(n, n - r, -1), 1)
    denom = reduce(mul, range(1, r + 1), 1)

    return numer // denom


def main():
    k = int(input())
    s = input()

    len_s = len(s)

    ans = 0
    for i in range(k + 1):
        print(i)
        n = 26 ** (k - i)
        n *= 25 ** i
        n *= combinations_count(i + len_s - 1, len_s - 1)
        ans += n

    print(ans % (10**9 + 7))


if __name__ == '__main__':
    main()

Submission status: Wrong Answer

Solution code:
MOD = int(10e9+7)

def mod(x, m = MOD):
    tmp = x%m 
    if(tmp < 0): tmp += m 
    return tmp

def fpow(x, y, m = MOD):
    res = 1
    while y>0:
        if y%2:
            res =  mod(res*x)
        x = mod(x*x)
        y>>=1
    return res

fat = [1]
ifat = [1]
power25 = [1]

def precalc(lim):
    global fat 
    global ifat 
    global power25

    for i in range(1,lim):
        fat.append(mod(fat[i-1]*i))
        ifat.append(fpow(fat[i], MOD - 2))
        power25.append(mod(25*power25[i-1]))

def comb(n, k):
    return (mod(fat[n] * ifat[n-k]) * ifat[k])

k = int(input())
s = input()
n = len(s) + k 

ans = fpow(26,n)
tmp = 0

precalc(n+1)

for m in range(0,len(s)):
    tmp = mod(tmp + mod(fpow(25,n-m)*comb(n,m)))

ans = mod(ans-tmp)

print(ans)

Submission status: Wrong Answer

Solution code:
def prepare(n, MOD):

    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv

    return factorials, invs


k = 5
s = "oof"
length = len(s)
MOD = 10**9 + 7

fact, inv = prepare(k + length, MOD)

ans = 0

for i in range(length, length + k + 1):
    coef = fact[i - 1] * inv[i - length] * inv[length - 1]
    ans += coef * pow(25, i - length, MOD) * pow(26, length + k - i, MOD) % MOD

print(ans % MOD)