Problem:


Score : 400 points

Problem Statement

Given is a number sequence A of length N.

Find the number of integers i \left(1 \leq i \leq N\right) with the following property:

For every integer j \left(1 \leq j \leq N\right) such that i \neq j , A_j does not divide A_i.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^6

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the answer.

Sample Input 1

5
24 11 8 3 16

Sample Output 1

3

The integers with the property are 2, 3, and 4.

Sample Input 2

4
5 5 5 5

Sample Output 2

0

Note that there can be multiple equal numbers.

Sample Input 3

10
33 18 45 28 8 19 89 86 2 4

Sample Output 3

5



Sample input:
5
24 11 8 3 16


Sample output:
3

Submission status: Time Limit Exceeded

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 = modfactorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz') # 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
        >>  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]

    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)
        """
        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
    #%%
    N = r_int()
    A = R()
    A.sort()
    s = set()
    ans = 0
    for i, a in enumerate(A):
        if i == N-1 or (i != N-1 and A[i] != A[i+1]):
            fac_list = make_divisors(a)
            flag = 0
            for fa in fac_list:
                if fa in s:
                    flag = 1
            if flag == 0:
                ans += 1
        s.add(a)

    print(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: Time Limit Exceeded

Solution code:
N = int(input())
A = list(map(int, input().split()))
T = [1] * (N)
for i in range(N):
  for j in range(N):
    if T[j] == 1 and i != j and A[j] % A[i] == 0:
      T[j] = 0
print(sum(T))

Submission status: Time Limit Exceeded

Solution code:
n = int(input())
in_data = [int(x) for x in input().split(' ')]
result = 0

for i in range(n):
  ok_flg = True
  for j in range(n):
#    print('in_data[i]:{}'.format(in_data[i]))
#    print('in_data[j]:{}'.format(in_data[j]))
    if i == j:
#      print('i == j')
      continue
    if in_data[i]%in_data[j]==0:
      ok_flg = False
      break
  if ok_flg:
    result += 1

print(result)

Submission status: Time Limit Exceeded

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

A = sorted(A)
cnt = 0
devide_list = []
i = 0
while i <= len(A)-1:
  
  if i < N-1:
    if A[i] == A[i+1]:
      if A[i] not in devide_list:
        devide_list.append(A[i])
      i += 1
      continue
    
  flag = 1
  for d in devide_list:
    if A[i]%d==0:
      flag = 0
      break
  if flag == 1:
    cnt += 1
  if A[i] not in devide_list:
      devide_list.append(A[i])
  i += 1
  
  
print(cnt)

Submission status: Time Limit Exceeded

Solution code:
import sys
import math
from collections import deque
import heapq
import itertools
from decimal import Decimal
import bisect
from operator import itemgetter
MAX_INT = int(10e18)
MIN_INT = -MAX_INT
mod = 1000000000+7
sys.setrecursionlimit(1000000)
def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(sys.stdin.readline())
def S(): return input()

def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)

    return divisors

N = I()
a = IL()
a.sort()

ans = 0
for i in a:
  data = make_divisors(i)
  for j in data:
    l = bisect.bisect_left(a, j)
    r = bisect.bisect_right(a, j)
    if i == j:
      if r - l >= 2:
        ans += 1
        break
    else:
      if r - l >= 1:
        ans += 1
        break
print(N - ans)

Submission status: Time Limit Exceeded

Solution code:
from collections import Counter
import sys
input = sys.stdin.buffer.readline

N = int(input())
A = list(map(int, (input().split())))
cntA = Counter(A)
A = sorted(list(set(A)))
setA = set()
for k, v in cntA.items():
    if v==1:
        setA.add(k)
# print(setA)
for i in range(len(A)-1):
    for j in range(i+1, len(A)):
        if A[j] % A[i] == 0:
            if A[j] in setA:
                setA.remove(A[j])
print(len(setA))

Submission status: Time Limit Exceeded

Solution code:
import sys
import itertools
sys.setrecursionlimit(1000000000)
from heapq import heapify,heappop,heappush,heappushpop
import math
import collections
import copy

def make_divisors(n):
    divisors = []
    for i in range(1,int(n**0.5)+1):
        if n%i == 0:
            divisors.append(i)
            if i != n//i:
                divisors.append(n//i)
    divisors.sort()
    return divisors


if __name__ == "__main__":
    n = int(input())
    a = sorted(list(map(int,input().split())))
    cnt = 0
    pattarn = []
    for i in range(n):
        b = make_divisors(a[i])
        pattarn.append(set(b))
    queue = collections.deque(a)
    for i in range(n):
        flag = 0
        p = queue.popleft()
        for j in queue:
            if j in pattarn[i]:
                flag = 1
        if flag == 0:
            cnt+=1
        queue.append(p)
    print(cnt)

Submission status: Time Limit Exceeded

Solution code:
n = int(input())
a=list(map(int,input().split()))
a.sort(reverse = True)

from collections import deque
from collections import Counter as cnt

aaa = dict(cnt(a))

ans = []
a = deque(a)
while True:
    if len(a) <1:
        break
    i = a.pop()
    ans.append(i)
    b = deque()
    while len(a)>0:
        j = a.pop()
        if j%i !=0:
            b.appendleft(j)
    a = b.copy()




mans = 0
for i,v in aaa.items():
    if (i in ans) & (v == 1) :
        mans+=1

print(mans)

Submission status: Time Limit Exceeded

Solution code:
num = int(input())
a_i = list(map(int, input().split()))
a_i.sort()
ans = 0
for x in range(num):
    val = a_i[x]
    if (x != 0 and val == a_i[x - 1]) or (x != num - 1 and val == a_i[x + 1]):
            ans += 1
            continue
    for y in range(num):
        val2 = a_i[y]
        if x == y: break
        elif val // 2 < val2: break
        elif val % val2 == 0:
            ans += 1
            break
print(num - ans)

Submission status: Time Limit Exceeded

Solution code:
n,*a = map(int,open(0).read().split())
s = sorted(list(set(a)),reverse=True)
m = len(s)
ans = 0
for i in range(m):
  if a.count(s[i]) != 1:
    continue
  for j in range(i+1,m):
    if s[i]%s[j]==0:
      break
  else:
    ans+=1
print(ans)

Submission status: Time Limit Exceeded

Solution code:
N = int(input())
A = list(map(int, input().split()))
A.sort()
c = 0

for i in range(N):
	d = 0

	if not A[i] == -1:
		for j in range(i+1, N):
			if A[j] == A[i]:
				d = 1
			if A[j] % A[i] == 0 and A[j] != -1:
				A[j] = -1
				c += 1
		c += d
		
if c == 0:
	print(N)
	

else:
	print(N-c)

Submission status: Time Limit Exceeded

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

dic = {}
for ai in a:
    if ai not in dic:
        dic[ai] = 0
    dic[ai] += 1

ans = 0

for ai in a:
    cnt = 0
    f = 1
    for i in range(1, int(ai**0.5)+1):
        if ai % i == 0:
            d = i
            if d in dic:
                cnt += dic[d]
            if d == ai:
                cnt -= 1
            if cnt > 0:
                f = 0
                break
            if i != ai // i:
                d = ai // i
                if d in dic:
                    cnt += dic[d]
                if d == ai:
                    cnt -= 1
                if cnt > 0:
                    f = 0
                    break

    ans += f

print(ans)

Submission status: Time Limit Exceeded

Solution code:
import numpy as np
N = int(input())
An = list(map(int, input().split()))
ans = 0
ng = set()
a_An = np.array(An)
for i, a in enumerate(An):
    temp = a_An / a
    temp[i] = 0.1
    check = np.where(temp % 1 == 0)
    check = np.delete(check, np.where(check == i))
    # check = check[check != i]
    if len(check) > 0:
        ng = ng | set(check)
print(N-len(ng))

Submission status: Time Limit Exceeded

Solution code:
N = int(input())
AA = input().split()
A = [int(i) for i in AA]
count = 0
flag =  False
for i in range(N):
    for j in range(N):
        # print(A[i],i,j)
        if i != j and A[i] % A[j] == 0:
            # print(i, j, A[i], A[j])
            flag = True
            break
    if flag:
        flag = False
    else:
        count+= 1
        # print("count")
        # print(i,j,A[i], count)
print(count)

Submission status: Time Limit Exceeded

Solution code:

n = int(input())

num_list = list(map(int, input().split()))

length = int(len(num_list))

check_list = [0]*length

count = 0

for i in range(0, length):
    if check_list[i] == 0:
        for j in range(0, length):
            if j != i:
                if num_list[i] % num_list[j] == 0:
                    check_list[i] = 1

                if num_list[j] % num_list[i] == 0:
                    check_list[j] = 1


result = check_list.count(0)

print(result)

Submission status: Time Limit Exceeded

Solution code:
from collections import deque
import collections

def main():
    N = int(input())
    A = list(map(int, input().split()))

    c = collections.Counter(A)
    # 重複排除
    A.sort()
    dq = deque()
    before = 0
    for a in A:
        dq.append(a)

    ans = 0
    while dq:
        a = dq.popleft()
        if c[a] == 1:
            ans += 1
        dq2 = deque()
        while dq:
            a2 = dq.popleft()
            if a2 % a:
                dq2.append(a2)
        dq = dq2

    print(ans)

if __name__ == '__main__':
    main()

Submission status: Time Limit Exceeded

Solution code:
import numpy as np

N = int(input())
x_list = np.array(sorted(list(map(int, input().split())), reverse=True))

count = 0
for i, xi in enumerate(x_list):
  if np.sum(xi%x_list==0) == 1:
      count += 1
print(count)

Submission status: Time Limit Exceeded

Solution code:
N = int(input())
A = [int(n) for n in input().split()]

A.sort()

flags = [1] * N

for ia, a in enumerate(A):
    if flags[ia] == 0:
        continue
    if ia == N-1:
        break
    for i in range(ia+1, len(A)):
        if A[i] % a == 0:
            flags[i] = 0
            if A[i] == a:
                flags[ia] = 0
print(sum(flags))

Submission status: Time Limit Exceeded

Solution code:
n = int(input())
a = list(map(int, input().split()))
count = 0
for i in range(n):
    hantei = True
    for j in range(n):
        if i == j: continue
        elif a[i] % a[j] == 0: hantei = False
    if hantei == True:
        count += 1
print(count)

Submission status: Time Limit Exceeded

Solution code:
N = int(input())
A = sorted([int(i) for i in input().split()])
flags = [True] * N

for i in range(N):
    if flags[i]:
        k = A[i]
        for j in range(N):
            if i != j:
                if A[j] % k == 0:
                    flags[j] = False
                    if A[j]==k:
                        flags[i] = False
print(sum(flags))

Submission status: Time Limit Exceeded

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

A.sort()

count = 0
for i in range(len(A)):
    if A[i] in A[:i]:
        continue
    if A[i] in A[i+1:]:
        continue
    flag = False
    for j in range(0, i):
        if (A[i] % A[j] == 0):
            flag = True
            break
    if not flag:
        count += 1

print(count)

Submission status: Time Limit Exceeded

Solution code:
import bisect
N = int(input())
A = list(map(int, input().split()))

A.sort()
#print(A)
count = 0
for i in range(N):
    Ai = A[i]
    hantei = True
    # j
    k = bisect.bisect_right(A, Ai)
#    print(Ai, k)
    for j in range(k-1):
        if Ai % A[j] == 0:
            hantei = False
            break
    if hantei:
        count += 1
print(count)

Submission status: Time Limit Exceeded

Solution code:
n=int(input())
a=sorted(map(int,input().split()))
ans=0
for i,num in enumerate(a):
	count=0
	for j,nuum in enumerate(a):
		if i!=j:
			if num%nuum==0:
				count+=1
				break
	if count==0:
		ans+=1
print(ans)

Submission status: Time Limit Exceeded

Solution code:
from collections import Counter
import math


def is_prime(n):
    if n == 1:
        return False

    for k in range(2, int(math.sqrt(n)) + 1):
        if n % k == 0:
            return False

    return True


def make_divisors(n):

    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            if i != n and i in c:
                return False

            tmp = n // i
            if i != tmp:
                if tmp != n and tmp in c:
                    return False
    return True


N = int(input())
A = list(map(int, input().split()))

c = Counter(A)

# 1がある時
if 1 in c:
    if c[1] >= 2:
        print(0)
    else:
        print(1)
    exit(0)

ans = 0
for a in A:
    # 同じ数が複数ある
    if c[a] >= 2:
        continue

    # 1と同数がなくて素数なら割り切れる数はない
    if is_prime(a):
        ans += 1
        continue

    # 約数があるか
    if make_divisors(a):
        ans += 1

print(ans)

Submission status: Time Limit Exceeded

Solution code:
import bisect


def make_divisors(n):
    lower_divisors, upper_divisors = [], []
    tmp = 1
    while tmp * tmp <= n:
        if n % tmp == 0:
            lower_divisors.append(tmp)
            if tmp != n // tmp:
                upper_divisors.append(n // tmp)
        tmp += 1
    return lower_divisors + upper_divisors[::-1]


n = int(input())
*A, = map(int, input().split())
A.sort()
setA = set(A)
cnt = 0
for i, a in enumerate(A):
    if bisect.bisect_right(A, a) - bisect.bisect_left(A, a) != 1:
        cnt += 1
        continue
    X = make_divisors(a)
    if any([x != a and x in setA for x in X]):
        cnt += 1
print(n - cnt)