Problem:


Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.



Sample input:
12 18


Sample output:
3

Submission status: Accepted

Solution code:
def makelist(n, m):
    return [[0 for _ in range(m)] for _ in range(n)]

A, B = map(int, input().split())

def gcd(a, b):  # a >= b
    if b == 0:
        return a
    else:
        return gcd(b, a % b)

hoge = gcd(A, B)

ans = 1
i = 2
while i * i <= min(A, B):
    if hoge % i == 0:
        ans += 1
        while hoge % i == 0:
            hoge /= i
    i += 1
if hoge != 1:
    ans += 1
    
print(ans)

Submission status: Accepted

Solution code:
import math
from functools import reduce
from collections import deque
import sys
sys.setrecursionlimit(10**7)

def s(generator, splitter, mapper):
    return [ mapper(s) for s in generator().split(splitter) ]

# スペース区切りの入力を読み込んで数値リストにして返します。
def get_nums_l():
    return [ int(s) for s in input().split(" ")]

# 改行区切りの入力をn行読み込んで数値リストにして返します。
def get_nums_n(n):
    return [ int(input()) for _ in range(n)]

def gcd(m, n):
    # 入力を m, n (m ≧ n) とする。
    tmp = max(m,n)
    n = min(m,n)
    m = tmp
    # n = 0 なら、 m を出力してアルゴリズムを終了する。
    if n == 0:
        return m
    # m を n で割った余りを新たに n とし、更に 元のnを新たにm とし 2. に戻る。
    return gcd(n, m % n)

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

a,b = get_nums_l()

_gcd = gcd(a,b)

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

print(len(factorization(_gcd))+1)

Submission status: Accepted

Solution code:
import fractions
import collections


def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a

a,b = map(int, input().split())

g = fractions.gcd(a,b)
c = collections.Counter(prime_factorize(g))
print(len(c)+1)

Submission status: Accepted

Solution code:
import fractions
import math
def pFactorize(num):
    factor_list = []
    for i in range(2, math.floor(math.sqrt(num))+ 1):
        if num % i == 0:
            factor_list.append([i,1])
            num //= i
        while num % i == 0:
            factor_list[-1][1] += 1
            num //= i
    if not num == 1:
        factor_list.append([num,1])
    return factor_list
a,b=[int(_) for _ in input().split()]
c=pFactorize(fractions.gcd(a,b))
print(len(c)+1)

Submission status: Accepted

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

def divisor(n):
    i = 1
    table = []
    while i * i <= n:
        if n%i == 0:
            table.append(i)
            table.append(n//i)
        i += 1
    table = list(set(table))
    return table

def is_prime(n):
    for i in range(2, n + 1):
        if i * i > n:
            break
        if n % i == 0:
            return False
    return n != 1

aa=divisor(a)
bb=divisor(b)
c=set(aa)&set(bb)
d=[i for i in c if is_prime(i)]
print(len(d)+1)

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # 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 = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    A, B = R()
    
    U = make_divisors(gcd(A, B))
    cnt = 0
    for u in U:
        if len(make_divisors(u)) == 2:
            cnt += 1
    print(cnt+1)

    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

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

def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)

g = gcd(a, b)

i = 2
count = 1
m = gcd(a, b)
while i * i <= g:
    if m % i == 0:
        count += 1
    while m % i == 0:
        m = m // i
    i += 1

if m > 1:
    count += 1

print(count)

Submission status: Accepted

Solution code:
import sys
input=sys.stdin.readline
a,b=map(int,input().split())
import fractions
c=fractions.gcd(a,b)
res=[]
i=2
while i**2<=c:
    if c%i==0:
        c=c//i
        res.append([i,1])
    while c%i==0:
        c=c//i
        res[-1][1]+=1
    i=i+1
if not c==1:
    res.append([c,1])
m=len(res)
ans=m+1
print(ans)

Submission status: Accepted

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

common_divisors = []
for n in range(1, int(min(A, B) ** 0.5) + 1):
    if (A % n == 0) & (B % n == 0):
        common_divisors.append(n)

    if A % n == 0:
        x = A // n
        if B % x == 0:
            common_divisors.append(x)


# ある自然数の素数判定
def is_prime(n):
    if n == 1:
        return False

    for i in range(2, int(n ** 0.5) + 1):
        if n % i == 0:
            return False
    else:
        return True


ans = 0
for cd in common_divisors:
    ans += is_prime(cd)

print(ans + 1)

Submission status: Accepted

Solution code:
#!/usr/bin/env python
# -*- coding: utf-8 -*-


# 最大公約数
def gcd(a, b):
  while b != 0:
    a, b = b, a % b

  return a


# 素因数分解
def factorization(n):
  arr = []
  temp = n

  for i in range(2, int(-(-n ** 0.5 // 1)) + 1):
    if temp % i == 0:
      cnt = 0
      while temp % i == 0:
        cnt += 1
        temp //= i
      arr.append([i, cnt])

  if temp != 1:
    arr.append([temp, 1])

  if arr == []:
    arr.append([n, 1])

  return arr


def main():
  a, b = map(int, input().split())
  g = gcd(a, b)
  fs = [v for v, _ in factorization(g) if v != 1]
  print(len(fs) + 1)


if __name__ == '__main__':
  main()

Submission status: Accepted

Solution code:
from collections import deque
import sys
sys.setrecursionlimit(10**7)
input = sys.stdin.readline
def isprime(n):
  for i in range(2,int(n**0.5)+1):
    if n%i == 0:
      return False
  return True
 
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
from fractions import gcd
def coprime(a, b):
    return gcd(a, b) == 1


def solve():
  a,b = (int(i) for i in input().split())
  ad = make_divisors(a)
  bd = make_divisors(b)
  act = []
  bct = []
  for i in ad:
    if isprime(i):
      act.append(i)
  for j in bd:
      if isprime(j):
        bct.append(j)
  ct = 0
  for i in act:
    if i in bct:
      ct += 1
  print(ct)
solve()

Submission status: Accepted

Solution code:
import fractions
def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a
a,b=map(int,input().split())
ans=0
c=fractions.gcd(a,b)
d=prime_factorize(c)
d=set(d)
print(len(d)+1)

Submission status: Accepted

Solution code:
def prime_decomposition(n):
  i = 2
  d = {}
  while i * i <= n:
    while n % i == 0:
      n //= i
      if i not in d:
          d[i] = 0
      d[i] += 1
    i += 1
  if n > 1:
    if n not in d:
        d[n] = 1
  return d

def gcd(a, b):
	while b:
		a, b = b, a % b
	return a

A,B = map(int,input().split())
d = gcd(A,B)
p = prime_decomposition(d)
ans = 1
for i in p:
    ans += 1
print(ans)

Submission status: Accepted

Solution code:
from fractions import gcd

A, B = map(int, input().split())

d = gcd(A, B)
d_ori = d
ans = 1
k = 2
while d > 2:
    plus = 0
    while d % k == 0:
        d //= k
        plus = 1
    ans += plus
    k += 1
    if k > d_ori**(1/2):
        ans += 1
        break
print(ans)

Submission status: Accepted

Solution code:
inpl = lambda: list(map(int,input().split()))

def gcd(a, b):
    if a < 0: a = -a
    if b < 0: b = -b
    while b > 0:
        (a, b) = (b, a % b)
    return a

import math
def primes(N):
    P = []
    for i in range(2,N):
        sieve = 1
        for p in P:
            if p*p > i:
                break
            elif i % p == 0:
                sieve = 0
                break
        if sieve == 0:
            continue
        else:
            P.append(i)
    return P

def factorize(n, P=None):
    if P is None:
        P = primes(int(math.sqrt(n))+1)
    factor = []
    power = []
    for p in P:
        if p * p > n:
            break
        elif n % p == 0:
            k = 0
            while n % p == 0:
                n //= p
                k += 1
            factor.append(p)
            power.append(k)
    if n > 1:
        factor.append(n)
        power.append(1)
    return factor, power

A, B = inpl()
C = gcd(A,B)
factor, _ = factorize(C)
print(len(factor)+1)

Submission status: Accepted

Solution code:

(N,M) = (int(i) for i in input().split(" "))
(N,M)=(min(M,N),max(M,N))
while M%N!=0:
	M=M%N
	(N,M)=(min(M,N),max(M,N))
i = 2
ans = 0
DN = N
while i*i <= DN :
	if N%i==0:
		ans += 1
		while N%i==0:
			N/=i
	i+=1
if i <= N:
	ans +=1
print(ans+1)

Submission status: Accepted

Solution code:
def gcd(a,b):
  while b!=0:
    a,b=b,a%b
  return a

def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a

def solve():
  A, B = map(int, input().split())
  return len(set(prime_factorize(gcd(A, B)))) + 1

print(solve())

Submission status: Accepted

Solution code:
import math

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


def gcd(a, b):
    if a < b:
        a, b = b, a
    if b == 0:
        return a
    else:
        return gcd(b, a % b)


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 is_prime(n):
    if n == 1:
        return True

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

    return True


print(sum([is_prime(n) for n in make_divisors(gcd(A, B))]))

Submission status: Accepted

Solution code:
def divisor(n): #nの約数を全て求める
    i = 1
    table = []
    while i * i <= n:
        if n%i == 0:
            table.append(i)
            table.append(n//i)
        i += 1
    table = list(set(table))
    return table

#a,bの最大公約数
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

def is_prime(n):
    for i in range(2, n + 1):
        if i * i > n:
            break
        if n % i == 0:
            return False
    return n != 1

a, b = map(int, input().split())

a_div = divisor(a)
b_div = divisor(b)
ans = set(a_div) & set(b_div)

count = 0
for i in ans:
    if is_prime(i):
        count += 1

print(count + 1)

Submission status: Accepted

Solution code:

import sys
from collections import deque, defaultdict
import copy
import bisect
sys.setrecursionlimit(10 ** 9)
import math
import heapq
from itertools import product, permutations,combinations
import fractions

import sys
def input():
	return sys.stdin.readline().strip()

def soinsu_bunkai(n):
	ans = []
	num = n
	i = 2
	while i <= n**0.5:
		if num % i == 0:
			if len(ans) == 0 or ans[-1][0] != i:
				ans.append([i, 1])
			else:
				ans[-1][1] += 1
			num = num//i
		else:
			i+=1
	if num != 1:
		ans.append([num, 1])

	return ans

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

C = fractions.gcd(A, B)

C_list = soinsu_bunkai(C)

print(len(C_list) + 1)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
from fractions import gcd
a, b = map(int, input().split())
g = gcd(a, b)
ans = {}
for i in range(2, int(g**0.5 + 1)):
    while g % i == 0:
        ans[i] = ans[i] + 1 if i in ans else 1
        g //= i
if g != 1:
    ans[g] = 1
print(len(ans) + 1)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
from collections import defaultdict, deque
from heapq import heappush, heapreplace
from itertools import count
from math import ceil, sqrt
from fractions import gcd


class Primes_C(object):

    def b_search(L,t):
        #増大列LについてL[:+1]の要素すべてがt以下であるような最大のi<len(L)を返す。
        search_range = [-1, len(L)]
        update_range = lambda mid: search_range.__setitem__(t<L[mid], mid)
        for _ in range(search_range[1].bit_length()):
            update_range(sum(search_range)//2)
        return search_range[0]
    
    def __init__(self):
        self.P = [2]
        self.Pappend = self.P.append
        self.Q = deque([])
        self.Qappend = self.Q.append
        self.Qpopleft = self.Q.popleft
        self.hq = [] 
        self.integer = count(3,2)
        self.current_i = 2
        self.valid_range = 1
        self.iter_index = 0

    def __iter__(self):
        return self

    def __next__(self):
        if self.iter_index >= len(self.P):
            self.step_calc(1)
        self.iter_index += 1
        return self.P[self.iter_index-1]

    def iter_reset(self, index=0):
        self.iter_index = index

    def list(self, max_int=None):
        if max_int is None:
            return self.P[:]

        elif  max_int <= self.current_i:
            return self.P[:Primes_C.b_search(self.P,max_int)+1]

        else:
            self.calc(max_int)
            return self.P[:]

    def extend(self, i):
        if i >= self.valid_range:
            if self.Q:
                q = self.Qpopleft()
                heappush(self.hq, (q**2,q))
            self.valid_range = self.Q[0]**2 if self.Q else i**2

    def check_update(self,i):
        m,r = self.hq[0] if self.hq else (0,0)
        flag =  (i != m)
        if flag:
            self.Pappend(i)
            self.Qappend(i)
        while m == i:
            heapreplace(self.hq, (m+2*r, r))
            m,r = self.hq[0]
        return flag

    def calc(self, max_int):
        for i in self.integer:
            self.extend(i)
            self.check_update(i) 
            self.current_i = i+1
            if max_int <= self.current_i:
                break

    def step_calc(self, step):
        for i in self.integer:
            self.extend(i)
            step -= self.check_update(i) 
            self.current_i = i+1
            if not step:
                break

    def factorization(self, m, F=defaultdict(int), break_flag=False):
        if m <= 0:
            raise  ValueError('m >= 1')
        iter_index = self.iter_index
        self.iter_reset()
        flag = True
        for p in self:
            if flag:
                try:
                    max_int = ceil(sqrt(m))
                except OverflowError:
                    max_int = m//p

                flag = False
            if m >= p > max_int:
                F[m] += 1
                break
            while not m%p:
                m //= p
                F[p] += 1
                flag = True
            if (flag and break_flag) or m == 1:
                break
        F_ = dict(F)
        F.clear()
        self.iter_reset(iter_index)
        return F_

    def is_prime(self,m):
        F = self.factorization(m)
        return m in F



def solve():
    A, B = map(int, input().split())
    C = gcd(A,B)    
    P = Primes_C()
    F = P.factorization(C)
    res = len(F)+1
    return str(res)

if __name__ == '__main__':
    print(solve())