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:
from math import sqrt
def Divisor_Set(n):
    s = set()
    for i in range(1, int(sqrt(n))+2):
        if n%i == 0:
            s.add(i)
            s.add(n//i)
    return s

def main():
    n = int(input())
    a = list(map(int, input().split()))
    dic = {}
    for x in a:
        if x in dic:
            dic[x] += 1
        else:
            dic[x] = 1
    ans = 0
    for v in a:
        if dic[v] > 1:
            continue
        tmp = Divisor_Set(v)
        tmp.remove(v)
        f = True
        for x in tmp:
            if x in dic:
                f = False
                break
        if f:
            ans += 1
    print(ans)

if __name__ == "__main__":
    main()

Submission status: Time Limit Exceeded

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

result = 0
for i in a:
  tmp = list(map(lambda x: i%x, a))
  if tmp.count(0) == 1:
    result += 1
print(result)

Submission status: Time Limit Exceeded

Solution code:
N = int(input())
A = list(map(int, input().split()))
count = 0
A.sort()
c= 0
if N>1:
    if A[0]%A[1] >0:
        count+=1

for k in range(1, len(A)):
    if k<N-1:
        if A[k] == A[k+1]:
            continue
    for j in range(c, k):
        if A[k]%A[j] ==0:
            del A[k]
            A.insert(0, 0)
            c +=1
            break
    else:
        count +=1
        #print(A[k])

    
if N ==1:
    count = 1
print(count)

Submission status: Time Limit Exceeded

Solution code:
from collections import Counter


def main():
    N = int(input())
    A = list(map(int, input().split(' ')))
    counter = Counter(A)
    A.sort()
    nums = list()
    for a in A:
        is_target = True
        for n in nums:
            if a % n == 0:
                is_target = False
                break
            if a // 2 < n:
                break
        if is_target:
            nums.append(a)
    print(len(list(filter(lambda n: counter[n] == 1, nums))))


if __name__ == '__main__':
    main()

Submission status: Time Limit Exceeded

Solution code:
from math import sqrt
def divisors(n):
    divisors = set()
    for i in range(1, int(sqrt(n))+1):
        q, r = divmod(n, i)
        if r == 0:
            divisors.add(i)
            divisors.add(q)
    return divisors

n, *A = map(int, open(0).read().split())
X = [0] * (max(A) + 1)
c = 0
for a in A:
    X[a] += 1
for a in A:
    X[a] -= 1
    c += all(X[b] == 0 for b in divisors(a))
    X[a] += 1
print(c)

Submission status: Time Limit Exceeded

Solution code:
def solve(A,N):
    if N == 1:
        return 1
    else:
        A.sort()
        cand = [A[0]]
        cand_del = [-1]
        for i in range(N):
            push = True
            for c in cand:
                if A[i] % c == 0:
                    push = False
                    break
            if push:
                cand.append(A[i])
            if i >= 1:
                if cand[-1] == A[i-1]:
                    cand_del.append(A[i])
        cand,cand_del = set(cand),set(cand_del)
        return len(cand - cand_del)

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

    print(solve(A,N))

if __name__ == "__main__":
    main()

Submission status: Time Limit Exceeded

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

A.sort()
ans = 0
flag = True
prime = []

while A:
  p = A[0]
  if len(A) >= 2 and A[0] == A[1]:
    A = [e for e in A if e % p != 0]
  else:
    prime.append(p)
    A = [e for e in A if e % p != 0]


print(len(prime))

Submission status: Time Limit Exceeded

Solution code:
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**6)
from numba import jit

@jit
def Ed(num):
    ret=set()
    for i in range(1,int(num**0.5)+1):
        if num%i==0:
            ret.add(i)
            if num//i!=i:
                ret.add(num//i)
    return ret

@jit
def main():
    n=int(input().rstrip())
    ans=0
    a=sorted([int(i) for i in input().split()])
    for i in range(n):
        if i!=n-1:
            if a[i+1]==a[i]:continue
        yaku=Ed(a[i])
        for j in range(i):
            if a[j] in yaku:
                break
        else:
            #print(a[i])
            ans+=1
    print(ans)

if __name__=="__main__":main()

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq

ri = lambda: int(sys.stdin.readline())
rl = lambda: list(map(int, sys.stdin.readline().split()))

n = ri()
li = rl()
def heapsort(iterable):
  h = []
  for value in iterable:
    heapq.heappush(h, value)
  l = []
  for _ in range(len(h)):
    l.append(heapq.heappop(h))
  return l
li = heapsort(li)
cnt = 0
for i in range(n):
  lii = li[i]
  if i < n-1:
    if lii == li[i+1]:
      continue
  for j in range(i):
    lij = li[j]
    if lii % lij == 0:
      break
    if lij > lii**0.5:
      continue
  else:
    cnt += 1
print(cnt)

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)
            for fa in fac_list:
                if fa in s:
                    break
            else:
                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()))
ans = 0
for i in range(n):
    count = 0
    for j in range(n):
        if i == j:
            pass
        elif count == 0:
            if a[i] % a[j] == 0:
                count = 1
    else:
        if count == 0:
            ans += 1
print(ans)

Submission status: Time Limit Exceeded

Solution code:
from itertools import combinations

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

for ai,aj in combinations(A,2):
    if ai%aj==0:
        if ai in ans:
            ans.remove(ai)
    if aj%ai==0:
        if aj in ans:
            ans.remove(aj)
print(len(ans))

Submission status: Time Limit Exceeded

Solution code:
from collections import Counter as ct


def solve(n):
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            if i in a:
                if i != n:
                    return 0
            if i != n // i:
                if n // i in a:
                    if n // i != n:
                        return 0
    return 1


n = int(input())
a = ct(list(map(int, input().split())))
cnt = 0
for i, j in a.items():
    if j == 1:
        cnt += solve(i)
print(cnt)

Submission status: Time Limit Exceeded

Solution code:
def divisor(n,A):
    ass = []
    for i in range(1,int(n**0.5)+1):
        if n%i == 0:
            if i in A:
                return False
            ass.append(i)
            if i**2 == n:
                continue
            if n//i in A:
                return False
    return True

def main():
    N = int(input())
    A = list(map(int, input().split()))
    A.sort(reverse=True)
    cnt = 0
    for i in range(N):
        if i > 0 and A[i] == A[i-1]:
            continue
        if divisor(A[i],A[i+1:]):
            cnt += 1

    print(cnt)


if __name__ == "__main__":
    main()

Submission status: Time Limit Exceeded

Solution code:
import sys


def Sieve(N, AList):
    AList.sort()
    N_NotDivisible = 0
    while( len(AList) > 0 ):
        if(len(AList) > 1):
            if(AList[0]!= AList[1]):
                N_NotDivisible+= 1
                value = AList[0]
                AList.pop(0)
                AList = [ i for i in AList if i % value != 0 ]
            else:
                indices_last = [i for i, x in enumerate(AList) if x == AList[0] ]
                Times = indices_last[-1]+1
                for i in range(Times):
                    AList.pop(0)

        else:
            N_NotDivisible+= 1
            break
    #print(Mins)
    return N_NotDivisible

if(__name__ == "__main__"):

    N = int( input().strip() )
    AList_str = input().strip().split()

    AList = [0] * N
    for i in range(N):
        AList[i] = int(AList_str[i])
    N_NotDivisible= Sieve(N, AList)
    print(N_NotDivisible)

Submission status: Time Limit Exceeded

Solution code:
from bisect import bisect_left, bisect_right


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

def main():
    N = int(input())
    A = list(map(int, input().split()))
    A.sort()
    ans = 0
    B = set(A)
    for a in B:
        flag = False
        left = bisect_left(A, a)
        right = bisect_right(A, a)

        if right-left > 1:
            continue

        divs = make_divisors(a)
        for d in divs:
            if d in B and d != a:
                flag = True
                break

        if flag == False:
            ans += 1
    
    print(ans)
    
if __name__ == "__main__":
    main()

Submission status: Time Limit Exceeded

Solution code:
import numpy as np

n = int(input())
a = list(map(int, input().split()))
a_np = np.array(a)
a_np = np.roll(a_np, 1)
counter = [0 for _ in range(n)]
for i in range(n - 1):
    index = list(np.where(a % a_np == 0))
    for n in index[0]:
        counter[int(n)] += 1
    a_np = np.roll(a_np, 1)
print(counter.count(0))

Submission status: Time Limit Exceeded

Solution code:
def _main(n, a_list):
    a_list.sort()
    count = 0
    one_cnt = 0
    for i in range(n):
        i_value = a_list[i]
        if i_value == 1:
            one_cnt += 1
            if 2 <= one_cnt:
                print(0)
                return
        elif one_cnt == 1:
            print(1)
            return

        fail = False
        for j in range(0, min(i + 2, n)):
            if i == j:
                continue
            j_value = a_list[j]
            q, mod = divmod(i_value, j_value)
            if mod == 0:
                fail = True
                break
        if not fail:
            count += 1

    print(count)


# if __name__ == "__main__":
n_a = int(input())
a_list_a = list(map(int, input().split()))
_main(n_a, a_list_a)

Submission status: Time Limit Exceeded

Solution code:
import sys 

N = int(sys.stdin.readline().strip())
A = list(map(int, sys.stdin.readline().strip().split()))
A.sort()
# print(A)

ans = 0 
exists = set()
for i in range(N):
    for e in exists:
        if A[i] % e == 0:
            break
    else:
        if i + 1 == N or A[i] != A[i+1]:
            ans += 1
    exists.add(A[i])

print(ans)

Submission status: Time Limit Exceeded

Solution code:

import collections

def main():
  n = int(input())
  A = tuple(sorted(map(int, input().split())))
  D = collections.defaultdict(int)
  for a in A:
    D[a] += 1

  ans = n
  for i in range(n):
    a = A[i]
    if D[a] > 1:
      ans -= 1
      continue

    for b in A[:i]:
      if a % b == 0:
        ans -= 1
        break

  print(ans)


main()

Submission status: Time Limit Exceeded

Solution code:
n = int(input())
a = list(map(int, input().split()))
c = 0
for i in range(n):
    b = 0
    j = 0
    for j in range(n):
        if i != j and a[i] % a[j] != 0:
                b = b + 1
    if b != n - 1:
        c = c + 1
print(n - c)

Submission status: Time Limit Exceeded

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

for i in range(N):
    flag = False
    temp = A.pop(i)
    for j in range(N-1):
        if temp % A[j] == 0 :
            flag = True
            break
    if flag == False:
        result += 1
    A.insert(i,temp)

print(result)

Submission status: Time Limit Exceeded

Solution code:
n = int(input())
a = list(map(int,input().split()))
ans = [0 for _ in range(n)]

for i in range(n):
  for j in range(n):
    if i == j: continue
    elif a[i] % a[j] == 0:
      ans[i] = 1
      break
print(ans.count(0))

Submission status: Time Limit Exceeded

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

if N==1:
    print(1)
    exit()

A.sort()
ans=0

if A[0]!=A[1]:
    ans=1

if A[0]==1 and A[1]==1:
    print(0)
    exit()

elif A[0]==1 and A[1]!=1:
    print(1)
    exit()

for i in range(1,N):
    if A[i]==A[i-1] or ((i!=N-1) and (A[i]==A[i+1])):            
        continue
    for j in range(i):
        if A[i]%A[j]==0:
            break
        elif j==i-1 or A[i]<A[j]*2:
            ans+=1
            break

print(ans)