TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: 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.

Submission attempts (in order):
# ABC142D - Disjoint Set of Common Divisors
def divisors(n: int) -> set:
    div = set()
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            div.add(i), div.add(n // i)
    return div


def main():
    A, B = map(int, input().split())
    divs = sorted(divisors(A) & divisors(B))
    divs.pop(0)
    groups = set()
    for i in divs:
        if all(i % x for x in groups):
            groups.add(i)
    ans = len(groups) + 1
    print(ans, groups)


if __name__ == "__main__":
    main()

# ABC142D - Disjoint Set of Common Divisors
from fractions import gcd


def divisors(n: int) -> set:
    div = set()
    for i in range(1, int(n ** 0.5) + 1):
        if n % i == 0:
            div.add(i), div.add(n // i)
    return div


def get_primes(n: int) -> set:
    # return a list of primes <= n
    n += 1
    sieve = [1] * n
    for i in range(3, int(n ** 0.5) + 1, 2):
        if sieve[i]:
            sieve[i * i :: 2 * i] = [0] * ((n - i * i - 1) // (2 * i) + 1)
    return {2} | {i for i in range(3, n, 2) if sieve[i]}


def main():
    A, B = map(int, input().split())
    g = gcd(A, B)
    ans = len(divisors(g) & get_primes(g)) + 1
    print(ans)


if __name__ == "__main__":
    main()

# ABC142D - Disjoint Set of Common Divisors
def gcd(x: int, y: int) -> int:
    while y:
        x, y = y, x % y
    return x


def prime_factors(x: int) -> list:
    if x % 2:
        max_prime, factors = 0, []
    else:
        max_prime, factors = 2, [2]
        while x % 2 == 0:
            x //= 2
    for i in range(3, int(x ** 0.5) + 1, 2):
        if x % i == 0:
            factors.append(i)
            while x % i == 0:
                x //= i
    if x > 2:  # when x is prime greater than 2
        factors.append(x)
    return factors


def main():
    # compute the number of prime factors of gcd(A, B)
    A, B = map(int, input().split())
    g = gcd(A, B)
    ans = len(prime_factors(g)) + 1
    print(ans)


if __name__ == "__main__":
    main()


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
import numpy as np
a, b = map(int, input().split())
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a
gcd = gcd(a, b)

tmp = []
for i in range(2, int(gcd**0.5)):
    if gcd%i == 0:
        soinsu = True
        for j in tmp:
            if i%j == 0:
                soinsu = False
                break
        if soinsu:
            tmp.append(i)
    
print(len(tmp)+1)

import numpy as np
a, b = map(int, input().split())
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a
n = gcd(a, b)

table = []
i = 2
while i*i <= n:
    if n%i == 0:
        table.append(i)
        n //= i
    else:
        i += 1
if n > 1:
    table.append(n)

print(len(set(table))+1)

Verdicts: Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
a, b = map(int, input().split())

a_set = set()
b_set = set()

i = 2
while a!=1 and i<a**(1/2):
  if a%i==0:
    while a%i==0: 
      a//=i
    a_set.add(i)
  i+=1

i = 2
while b!=1 and i<b**(1/2):
  if b%i==0:
    while b%i==0: 
      b//=i
    b_set.add(i)
  i+=1  
  
print(len(a_set&b_set)+1)

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

a_set = set()
b_set = set()

i = 2
while a!=1 and i<a**(1/2)+1:
  if a%i==0:
    while a%i==0: 
      a//=i
    a_set.add(i)
  i+=1

i = 2
while b!=1 and i<b**(1/2)+1:
  if b%i==0:
    while b%i==0: 
      b//=i
    b_set.add(i)
  i+=1  
  
print(len(a_set&b_set)+1)

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

def factorize(n):
    b = 2
    fct = []
    while b * b <= n:
        while n % b == 0:
            n //= b
            fct.append(b)
        b = b + 1
    if n > 1:
        fct.append(n)
    return fct

print(len(set(factorize(a))&set(factorize(b)))+1)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
a,b = map(int,input().split())
from math import gcd
print(gcd(a,b))

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

from itertools import combinations

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

gcd_n = gcd(a,b)



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

l = div(gcd_n)
cnt = 0
pre = 0
for i in range(len(l)):
    if pre==l[i]:
        continue
    pre = l[i]
    cnt += 1
print(cnt)


Verdicts: Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
def prime_factorize(n):
    a = set([1])
    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 not in a:
        a.add(n)
    return a
def gcd(a, b):
	while b:
		a, b = b, a % b
	return a
a, b = list(map(int, input().split()))
x = gcd(a, b)
print(len(prime_factorize(x)))


def prime_factorize(n):
    a = set([1])
    while n % 2 == 0:
        a.add(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.add(f)
            n //= f
        else:
            f += 2
    a.add(n)
    return a
def gcd(a, b):
	while b:
		a, b = b, a % b
	return a
  
a, b = list(map(int, input().split()))
x = gcd(a, b)
print(len(prime_factorize(x)))

Verdicts: Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
a, b = map(int, input().split())
a_prime_factor = []
b_prime_factor = []
for i in range(2, int(a ** 0.5) + 1):
    if a % i == 0:
        a_prime_factor.append(i)
    while a % i == 0:
        a //= i
if a != 1:
    a_prime_factor.append(a)
for i in range(2, int(b ** 0.5) + 1):
    if b % i == 0:
        b_prime_factor.append(i)
    while b % i == 0:
        b //= i
if b != 1:
    b_prime_factor.append(b)
ans = 1
for i in a_prime_factor:
    if i in b_prime_factor:
        ans += 1
print(a_prime_factor)
print(b_prime_factor)
print(ans)

a, b = map(int, input().split())
a_prime_factor = []
b_prime_factor = []
for i in range(2, int(a ** 0.5) + 1):
    if a % i == 0:
        a_prime_factor.append(i)
    while a % i == 0:
        a //= i
a_prime_factor.append(a)
for i in range(2, int(b ** 0.5) + 1):
    if b % i == 0:
        b_prime_factor.append(i)
    while b % i == 0:
        b //= i
b_prime_factor.append(b)
ans = 1
for i in a_prime_factor:
    if i in b_prime_factor:
        ans += 1
print(ans)

a, b = map(int, input().split())
a_prime_factor = []
b_prime_factor = []
for i in range(2, int(a ** 0.5) + 1):
    if a % i == 0:
        a_prime_factor.append(i)
    while a % i == 0:
        a //= i
if a != 1:
    a_prime_factor.append(a)
for i in range(2, int(b ** 0.5) + 1):
    if b % i == 0:
        b_prime_factor.append(i)
    while b % i == 0:
        b //= i
if b != 1:
    b_prime_factor.append(b)
ans = 1
for i in a_prime_factor:
    if i in b_prime_factor:
        ans += 1
print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
a,b = map(int,input().split())
d = []
j = 1
if a>=b:
  x = a
else:
  x = b
for i in range (1,x):
  if a%i==0 and b%i==0:
    d+=[i]
while j<len(d):
  k=j+1
  while d[k]<math.sqrt(d[len(d)-1]):
    if d[k]%d[j]==0:
      d.pop(k)
      k-=1
    else:
      k+=1
  j+=1
print(len(d))

import math
a,b = map(int,input().split())
d = []
j = 1
if a>=b:
  x = a
else:
  x = b
for i in range (1,x):
  if a%i==0 and b%i==0:
    d+=[i]
while j<len(d):
  k=j+1
  while k<len(d) and d[k]<math.sqrt(d[len(d)-1]):
    if d[k]%d[j]==0:
      d.pop(k)
      k-=1
    else:
      k+=1
  j+=1
print(len(d))

a,b = map(int,input().split())
d = []
j = 1
if a>=b:
  x = a
else:
  x = b
for i in range (1,x):
  if a%i==0 and b%i==0:
    d+=[i]
while j<len(d):
  k=j+1
  while k<len(d) and d[k]<math.sqrt(d[len(d)-1]):
    if d[k]%d[j]==0:
      d.pop(k)
      k-=1
    else:
      k+=1
  j+=1
print(len(d))

import math
a,b = map(int,input().split())
p = []
while(b!=0):
  a,b=b,a%b
x = a
for i in range(a+1):
  p.append(True)
p[0],p[1] = False,True
for i in range(2,a+1):
  j=0
  if p[i]:
    while x%i==0:
      p[i] = True
      x//=i
    j=i*2
    if a%i!=0:
      p[i] = False
    while j<=a:
      p[j] = False
      j+=i
  if x==1:
    break
print(p.count(True))

a,b = map(int,input().split())
f = [1]
while(b!=0):
  a,b=b,a%b
while a%2==0:
  f.append(2)
  a//=2
p = 3
while p*p<=a:
  if a%p==0:
    f.append(p)
    a//=p
  else:
    p+=2
if a!=1:
  f.append(a)
print(len(set(f)))

Verdicts: Runtime Error

Wrong Answer

Runtime Error

Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):

def kouyakusu(x1,x2):
    cf=[1]
    for i in range(2,min(x1,x2)+1):
        if x1 % i == 0 and x2 % i == 0:
            cf.append(i)
    return cf


def yakusu_tapulu(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            syou,amari =divmod(n,i)
            divisors.append((syou,i))
    return divisors


x_yakusu =[]
x_kouyaku = kouyakusu(A,B) 
x_yakusu =[yakusu_tapulu(i) for i in x_kouyaku  if len(yakusu_tapulu(i)) ==1]
print(len(x_yakusu))


A,B = (int(x) for x in input().split())


def soinsu_bunkai(m):
    pf={}

    for i in range(2,int(m**0.5)+1):
        while m%i==0:
            pf[i]=pf.get(i,0)+1
            m//=i
    if m>1:
        pf[m]=1
    return pf

A_soinsu =soinsu_bunkai(A)
B_soinsu =soinsu_bunkai(B)


kyotu_soinsu ={}
a_sosu ={i for i in A_soinsu.keys()}
b_sosu ={i for i in B_soinsu.keys()}
kotae = a_sosu & b_sosu

print(len(kotae)+1) 

Verdicts: Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
a,b=map(int,input().split())
import numpy as np
def gcd(x,y):
  if x%y==0:
    return y
  else:
    return gcd(x,x%y)
  
  
  
z=np.ceil(np.sqrt(gcd(a,b)))
j=0
ans=[]
for i in range(1,z+1):
  while z%i==0:
    z/=i
    ans.append(i)
  

  
print(len(ans))
    


a,b=map(int,input().split())
import numpy as np
def gcd(x,y):
  if x%y==0:
    return y
  else:
    return gcd(x,x%y)
  
  
  
z=int(np.ceil(np.sqrt(gcd(a,b))))
j=0
ans=[]
for i in range(1,z+1):
  while z%i==0:
    z/=i
    ans.append(i)
  

  
print(lenset((ans)))
    


a,b=map(int,input().split())
import numpy as np
def gcd(x,y):
  if x%y==0:
    return y
  else:
    return gcd(x,x%y)
  
  
n=gcd(a,b)  
i=2
ans=[1]

while i * i <= n:
  if n % i == 0:
    n /= i
    ans.append(i)
  else:
    i += 1

  
print(len(set(ans)))
    


a,b=map(int,input().split())
def gcd(x,y):
  if x%y==0:
    return y
  else:
    return gcd(x,x%y)
  
z=gcd(a,b)
j=0
ans=[]
for i in range(0,z):
  if z%i==0:
    z/=i
    ans.append(i)
  else:
    j+=1
print(len(set(ans)))    
    

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

def gcd(x,y):
  if x%y==0:
    return y
  else:
    return gcd(y,x%y)
  
  
n=gcd(a,b)  
i=2
ans=[1]

while i * i <= n:
  if n % i == 0:
    n /= i
    ans.append(i)
  else:
    i += 1

ans.append(n)    
  
print(len(set(ans)))

Verdicts: Runtime Error

Time Limit Exceeded

Wrong Answer

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
import sys

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

a, b = na()

def gcd(x, y):
    while(x > 0):
        x, y = y % x, x
    return y

g = gcd(a, b)
ans = 0
for d in range(2,1000001):
    e = 0
    while g % d == 0:
        g //= d
        e += 1
    if e > 0:
        ans += 1
if g > 1:
    ans += 1
print(ans)


import sys

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

a, b = na()

def gcd(x, y):
    while(x > 0):
        x, y = y % x, x
    return y

g = gcd(a, b)
ans = 1
for d in range(2,1000001):
    e = 0
    while g % d == 0:
        g //= d
        e += 1
    if e > 0:
        ans += 1
if g > 1:
    ans += 1
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
import math

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

list_a = []
list_b = []
sqrt_a = math.sqrt(a)
sqrt_b = math.sqrt(b)


for i in range(2, float(sqrt_a) ):  
  if (a % (i) == 0):
    list_a.append( i )
    while( a % (i) == 0 ):
      a = a//(i)
      
for i in range(2, float(sqrt_b) ):  
  if (b % (i) == 0):
    list_b.append( i )
    while( b % (i ) == 0 ):
      b = b//(i)

#print(list_a)
#print(list_b)

set_a = set(list_a)
set_b = set(list_b)

set_ab = set_a & set_b
  
print(len(set_ab) + 1)
      

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


max_val = max(a, b)
min_val = min(a, b)

while(1):
  if max_val % min_val ==0:
    break
  else:
    tmp = max_val % min_val
    max_val = min_val
    min_val = tmp
    
#print(min_val)

list = []

for i in range(1, min_val//2 + 1):
  if min_val % i == 0:
    list.append(i)
    
list.append(min_val)
    
#print(list)
sum = 0
for i in range(len(list)):
  for j in range(len(list)):
    if (list[j] % list[i]) == 1:
      sum += 1
 
print(sum)

import math

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

list_a = []
list_b = []
sqrt_a = math.sqrt(a)
sqrt_b = math.sqrt(b)


for i in range(2, int(sqrt_a) ):  
  if (a % (i) == 0):
    list_a.append( i )
    while( a % (i) == 0 ):
      a = a//(i)
if a > 1:
  list_a.append(a)
      
for i in range(2, int(sqrt_b) ):  
  if (b % (i) == 0):
    list_b.append( i )
    while( b % (i ) == 0 ):
      b = b//(i)
if b > 1:
  list_b.append(b)

#print(list_a)
#print(list_b)

set_a = set(list_a)
set_b = set(list_b)

set_ab = set_a & set_b
  
print(len(set_ab) + 1)
      


Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
import math
A, B = map(int, input().strip().split())

minAB = min(A,B);

hurui = [True] * (minAB+1)
"""
upper = int(math.sqrt(minAB+1))
ans = 1

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

    if not(hurui[i]):
        continue
    
    for j in range(2*i, minAB+1, i):
        hurui[j] = False

    if A % i == 0  and B % i == 0:
        ans += 1
"""
ans = 1
print(ans)

import math
A, B = map(int, input().strip().split())

minAB = min(A,B);


"""
upper = int(math.sqrt(minAB+1))
ans = 1

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

    if not(hurui[i]):
        continue
    
    for j in range(2*i, minAB+1, i):
        hurui[j] = False

    if A % i == 0  and B % i == 0:
        ans += 1
"""
ans = 1
print(ans)

import math
A, B = map(int, input().strip().split())

minAB = min(A,B);

hurui = [True] * (minAB+1)
"""
upper = int(math.sqrt(minAB+1))
ans = 1

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

    if not(hurui[i]):
        continue
    
    for j in range(2*i, minAB+1, i):
        hurui[j] = False

    if A % i == 0  and B % i == 0:
        ans += 1
"""
print(ans)

import math
A, B = map(int, input().strip().split())

minAB = min(A,B);

hurui = [True] * (minAB+1)

upper = int(math.sqrt(minAB+1))
ans = 1
"""
for i in range(2, upper+1):

    if not(hurui[i]):
        continue
    
    for j in range(2*i, minAB+1, i):
        hurui[j] = False

    if A % i == 0  and B % i == 0:
        ans += 1
"""
print(ans)

import math
A, B = map(int, input().strip().split())

divisors = []

minAB = min(A,B)
maxAB = max(A,B)
upper = int(math.sqrt(minAB+1) + 1)
ans = 1

for i in range(1, upper):
    if minAB % i == 0:
        divisors.append(i)
        if i != minAB//i:
            divisors.append(minAB//i)

divisors.sort()


flag = [True] * len(divisors)
for i, num in enumerate(divisors):
    if maxAB % num != 0:
        flag[i] = False
        continue
    for x in divisors:
        if x == 1 :
            continue
        if x > math.sqrt(num):
            break
        if num % x ==0:
            flag[i] = False
            break

print(flag.count(True))

Verdicts: Runtime Error

Wrong Answer

Runtime Error

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
def gcd(a,b): 
    if a%b==0: return b
    return gcd(b,a%b)

def ela(g):
    s={}
    i=2
    while i*i<=g:
        while g%i==0:
            g//=i
            s.add(i)
        i+=1
    return s

a,b=map(int,input().split())
G=gcd(max(a,b),min(a,b))
print(len(ela(G))+1)

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

def ela(g):
    s={1}
    i=2
    while i*i<=g:
        while g%i==0:
            g//=i
            s.add(i)
        i+=1
    if g!=1: s.add(g)
    return s

a,b=map(int,input().split())
G=gcd(max(a,b),min(a,b))
print(len(ela(G)))

Verdicts: Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
from fractions import gcd as bltin_gcd
#return arr
def cf(x1,x2):
    cf=[1]
    for i in range(2,min(x1,x2)+1):
        if x1 % i == 0 and x2 % i == 0:
            cf.append(i)
    return cf

def coprime(a, b):
    return bltin_gcd(a, b) == 1
def rmdb(n, li):
  lst = [x for x in li if x%n != 0]
  return lst

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

ng_set = set()
for i in range(1, len(cf_list)):
  for j in range(1, i):
    if cf_list[i] % cf_list[j] == 0:
      ng_set.add(cf_list[i])
      break
  
print(len(cf_list)-len(ng_set))


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())
a_list = prime_factorize(a)
b_list = prime_factorize(b)
a_set = set(a_list)
b_set = set(b_list)

ans_set = a_set & b_set
ans_set.add(1)
print(len(ans_set))


Verdicts: Time Limit Exceeded

Accepted
Problem: 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.

Submission attempts (in order):
import sys
import math

stdin = sys.stdin

ns = lambda: stdin.readline().rstrip()
ni = lambda: int(stdin.readline().rstrip())
nm = lambda: map(int, stdin.readline().split())

a, b = nm()

n = math.gcd(a,b)

m = int(n**(1/2))

print(n)

count = 1 # 1
for i in range(2, n+1):
    if n%i ==0:
        count +=1
        while n%i == 0:
            n = n/i
    if n ==1:
        break

print(count)

import sys
import math

stdin = sys.stdin

ns = lambda: stdin.readline().rstrip()
ni = lambda: int(stdin.readline().rstrip())
nm = lambda: map(int, stdin.readline().split())

a, b = nm()

n = math.gcd(a,b)

m = int(n**(1/2))

print(n)

count = 1 # 1
for i in range(2, n+1):
    if n%i ==0:
        count +=1
        while n%i == 0:
            n = n/i
    if n ==1:
        break

print(count)

import sys
import math

stdin = sys.stdin

nm = lambda: map(int, stdin.readline().split())

a, b = nm()

n = math.gcd(a,b)

count = 1 # 1
for i in range(2, int(n**(1/2))+1):
    if n%i ==0:
        count +=1
        while n%i == 0:
            n = n/i
    if n ==1:
        break
if n != 1:
    count += 1

print(count)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
import math

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


def dv(x1, x2):
    ret = []
    v = int(math.sqrt(min(x1, x2))) + 1

    check = [0] * (v + 1)

    for n in range(2, v + 1):
        if check[n] == 1:
            continue

        if x1 % n == 0 and x2 % n == 0:
            ret.append(n)

            cnt = 1
            while n * cnt <= v:
                check[n * cnt] = 1
                cnt += 1

    return ret

can = 0
if A >= B:
    if A % B == 0:
        can = B
else:
    if B % A == 0:
        can = A

cf = [1] + dv(A, B)

flag = False

for c in cf:
    if c % can == 0:
        flag = True
        break
        
if not flag:
    cf = cf + [can]

print(len(cf))

import math

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


def dv(x1, x2):
    ret = []
    v = int(math.sqrt(min(x1, x2))) + 1

    check = [0] * (v + 1)

    for n in range(2, v + 1):
        if check[n] == 1:
            continue

        if x1 % n == 0 and x2 % n == 0:
            ret.append(n)

            cnt = 1
            while n * cnt <= v:
                check[n * cnt] = 1
                cnt += 1

    return ret

can = 0
if A >= B:
    if A % B == 0:
        can = B
else:
    if B % A == 0:
        can = A

cf = [1] + dv(A, B)

flag = False

for c in cf:
    if c % can == 0:
        flag = True
        break
        
if not flag:
    cf = cf + [can]

print(len(cf))

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


def prime_factorization(n):
    from collections import Counter
    counter = Counter()

    if n < 2:
        return counter

    for i in range(2, int(n ** 0.5) + 1):
        if i * i > n:
            break

        while n % i == 0:
            counter[i] += 1
            n //= i

    if n > 1:
        counter[n] += 1

    return counter


x = prime_factorization(A)
y = prime_factorization(B)

z = set(x.keys()) & set(y.keys())
print(len(z) + 1)


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
a, b = map(int, input().split())
ans = []

for i in range(1,min(a,b)+1):
    if a%i == 0 and b%i ==0:
        while True:
            for j in range(2,int(math.sqrt(i))+1):
                if i%j==0:
                    break
            else:
                ans.append(i)
            break
            
print(len(ans))
        

import collections

def gcd(x, y):
    if y == 0:
        return x
    return gcd(y, x%y)

def trial_division(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 main():
    a, b = map(int, input().split())
    c = collections.Counter(trial_division(gcd(a, b)))
    print(len(c.keys())+1)

if __name__ == '__main__':
    main()

Verdicts: Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
import math

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

def get_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if ((i == 2 or i % 2 != 0) and n % i == 0):
            divisors.append(i)
            if (((n // i) == 2 or (n // i) % 2 != 0) and i != n // i):
                divisors.append(n//i)

    return divisors

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

gcd = get_gcd(a, b)
list = get_divisors(gcd)

print(list)

count = 0
length = len(list)
for i in range(length):
    is_ok = True
  
    for j in range(i+1, length):
        print((list[i], list[j]))
      
        if (get_gcd(list[i], list[j]) != 1):
            is_ok = False
            break
    
    if is_ok:
        count += 1

print(count)


import math

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

def get_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        #if ((i == 2 or i % 2 != 0) and n % i == 0):
        if (n % i == 0):
            divisors.append(i)
            #if (((n // i) == 2 or (n // i) % 2 != 0) and i != n // i):
            if (i != n // i):
                divisors.append(n//i)

    return sorted(divisors)[::-1]

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

gcd = get_gcd(a, b)
list = get_divisors(gcd)

#print(list)

count = 0
length = len(list)
for i in range(length):
    is_ok = True
  
    for j in range(i+1, length):
        if (get_gcd(list[i], list[j]) != 1):
            is_ok = False
            break
    
    if is_ok:
        count += 1

print(count)


Verdicts: Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
# -*- coding: utf-8 -*-
"""
Created on Wed Jun  3 14:11:35 2020

@author: NEC-PCuser
"""
def gcd(a, b):
    
    if (a % b == 0):
        return b
    return gcd(b, a % b)

def factorize(num):
    i = 2
    factorize_list = []
    num_tmp = num
    while i * i <= num_tmp:
        count = 0
        if num % i == 0:
            while num % i == 0:
                count += 1
                num /= i
            factorize_list.append([i, count])
        i += 1
    return factorize_list 
                


if __name__ == "__main__":
    A, B =  map(int, input().split())
    big_num = max(A, B)
    short_num = min(A, B)
    gcd_num = gcd(big_num, short_num)
    print(len(factorize(gcd_num) + 1))
            

# -*- coding: utf-8 -*-
"""
Created on Wed Jun  3 14:11:35 2020

@author: NEC-PCuser
"""
def gcd(a, b):
    
    if (a % b == 0):
        return b
    return gcd(b, a % b)

def factorize(num):
    i = 2
    factorize_list = []
    num_tmp = num
    while i * i <= num_tmp:
        count = 0
        if num % i == 0:
            while num % i == 0:
                count += 1
                num /= i
            factorize_list.append([i, count])
        i += 1
    return factorize_list 
                


if __name__ == "__main__":
    A, B =  map(int, input().split())
    big_num = max(A, B)
    short_num = min(A, B)
    gcd_num = gcd(big_num, short_num)
    print(len(factorize(gcd_num)) + 1)
            


# -*- coding: utf-8 -*-
"""
Created on Wed Jun  3 14:11:35 2020

@author: NEC-PCuser
"""
def gcd(a, b):
    
    if (a % b == 0):
        return b
    return gcd(b, a % b)

def factorize(num):
    i = 2
    factorize_list = []
    num_tmp = num
    while i * i <= num_tmp:
        count = 0
        if num % i == 0:
            while num % i == 0:
                count += 1
                num //= i
            factorize_list.append([i, count])
        i += 1
    if num != 1:
        factorize_list.append([num, 1])
    return factorize_list 
                


if __name__ == "__main__":
    A, B =  map(int, input().split())
    big_num = max(A, B)
    short_num = min(A, B)
    gcd_num = gcd(big_num, short_num)
    print(len(factorize(gcd_num)) + 1)
            


Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: 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.

Submission attempts (in order):
A, B = map(int, input().split())
 
if A>=B:
  C = A%B
else:
  C = B%A
 
count = 0
 
now = 0
 
a = 2
 
while C >= a*a:
  if C%a == 0:
    if a != now:
      count += 1
      now = a
    C /= a
  else:
    a +=1
 
if C > 2 & C != now:
  count +=1
 
count += 1
 
print(count)

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long A,B;

    cin >> A >> B;

    long long C;

    if(A>=B){
        C = A%B;
    }else{
        C = B%A;
    }

    long long count = 0LL;

    long long now = 0; 
    long long a = 2;

    vector<int> yakusu;
    while(C >= a*a){
        if(C%a == 0){
            if (a != now){
                count++;
                yakusu.push_back(a);
                now = a;
            }
            C /= a;
        }else{
            a++;
        }
    }

    if(std::count(yakusu.begin(), yakusu.end(), C) == 0) count++;

    count++;

    cout << count << endl;
    return 0;


}

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


A_yaku = []

now = 0

a = 2

while A >= a*a:
  if A%a == 0:
    if a != now:
      A_yaku.append(a)
      now = a
    A /= a
  else:
    a +=1

if A != now and A != 1:
  A_yaku.append(A)

B_yaku = []

now = 0

a = 2

while B >= a*a:
  if B%a == 0:
    if a != now:
      B_yaku.append(a)
      now = a
    B /= a
  else:
    a +=1

if B != now and B != 1:
  B_yaku.append(B)

count = 0

if(len(A_yaku) >= len(B_yaku)):
  for i in range(0,len(B_yaku)):
    if(B_yaku[i] in A_yaku):
      count += 1
else:
  for i in range(0,len(A_yaku)):
    if(A_yaku[i] in B_yaku):
      count += 1

count += 1

print(count)




Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
import fractions
a,b=map(int,input().split())
c=list()
cnt=1
if fractions.gcd(a,b)==1:
  print(1)
elif fractions.gcd(a,b)==2:
  print(2)
else:
  for i in range(2,fractions.gcd(a,b)+1):
    if a%i==0 and b%i==0:
      if i==2:
        cnt+=1
      else:
        c.append(i)
  for j in c:
      ok=True
      for k in range(2,j):
        if j%k==0:
            ok=False
      if ok:
        cnt+=1
  print(cnt)

a,b=map(int,input().split())
ans=1
for i in range(2,int(min(a,b)**0.5)+1):
  if a%i==0 and b%i==0:
    ans+=1
  while a%i==0:
    a//=i
  while b%i==0:
    b//=i
if a!=1 and b!=1 and (a%b==0 or b%a==0):
  ans+=1
print(ans)

Verdicts: Time Limit Exceeded

Accepted
Problem: 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.

Submission attempts (in order):
if a == 1 or b == 1:
  print(1)
def prime(n):
  pf=[1]
  for i in range(2,int(n**0.5)+1):
    while n%i==0:
      pf.append(i)
      n//=i
  list(set(pf))
  return pf
a, b = map(int, input().split())
if a == 1 or b == 1:
  print(1)
else:
  print(len(set(prime(a)+prime(b))))

def prime(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 = map(int, input().split())
if a == 1 or b == 1:
  print(1)
else:
	print(len(set(prime(a))&set(prime(b))))

a, b = map(int, input().split())
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 = set(prime_factorize(a))
B = set(prime_factorize(b))
ans = A&B
print(len(ans)+1)


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
import math

def sosuucheck(val):
    flg = True
    max_search = math.ceil(math.sqrt(val))
    if val <= 3:
        pass
    else:
        for i in range(2, max_search+1):
            if val % i == 0:
                flg = False
                break
    return flg

def yakusuu(a):
    t = []
    for i in range(1,a+1):
        if a%i==0:
            t.append(i)
    return t

def kouyakusuu(a,b):
    t = []
    u = 0
    d = 0
    if a>b:
        u = a
        d = b
    else:
        u = b
        d = a
    for i in range(1,d+1):
        if u%i==0 and d%i==0:
            t.append(i)
    return t

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


t = kouyakusuu(a,b)
print(t)
sn = 0
for i in t:
    if sosuucheck(i):
        sn += 1


ans = sn*(sn-1)//2
print(ans)


from sympy import sieve

def yakusuu(a):
    t = []
    for i in range(1,a+1):
        if a%i==0:
            t.append(i)
    return t

def kouyakusuu(a,b):
    t = []
    u = 0
    d = 0
    if a>b:
        u = a
        d = b
    else:
        u = b
        d = a
    for i in range(1,d+1):
        if u%i==0 and d%i==0:
            t.append(i)
    return t

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


t = kouyakusuu(a,b)
sn = 0
for i in t:
    if i in sieve:
        sn += 1

sn += 1

ans = sn*(sn-1)/2
print(ans)

import math

A, B = [int(item) for item in input().split()]

def gdc(A, B):
    if B == 0:
        return A
    else:
        return gdc(B, A % B)

def chk_pn(num):
    flag = True
    if num  <= 3:
        pass
    else:
        for i in range(2, int(math.sqrt(num)) + 1):
            if num % i == 0:
                flag = False
                break
    return flag

def mk_factor(num):
    max_divisor = int(math.sqrt(num))+1
    divisor = 2
    factor_list = [1]

    while divisor <= max_divisor:
        if num % divisor == 0:
            factor_list.append(divisor)
            num /= divisor
        else:
            divisor += 1
    if num != 1.0:
        factor_list.append(num)
    return factor_list

GDC = gdc(A, B)
#pn_factor = [i for i in mk_factor(GDC) if chk_pn(i) is True]

print(len(set(mk_factor(GDC))))

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
a, b = (int(i) for i in input().split())

common_divisor_set = []
result_set = []

GCD = lambda a, b: (GCD(b, a % b) if a % b else b)

for i in range(min(a, b), 0, -1):
    if a%i == 0 and b%i == 0:
        common_divisor_set.insert(0, i)

left = [0 for x in range(n)]
right = [0 for x in range(n)]
ans = -1

n = len(common_divisor_set)
for i in range(n):
    if (i == 0):
        gcd = a[0]
    else:
        gcd = GCD(a[i], left[i - 1])
    left[i] = gcd

for i in range(n-2, -1, -1):   
    if (i == n - 1):
        gcd = a[n - 1]
    else:
        gcd = GCD(a[i], right[i + 1])
    right[i] = gcd

for i in range(n):
    if (i == 0):
        gcd = right[1]
    elif (i == n - 1):
        gcd = left[n - 2]
    else:
        gcd = GCD(left[i - 1], right[i + 1])
    if (gcd>ans):
        ans = gcd

print(ans)


from math import sqrt
a, b = (int(i) for i in input().split())

GCD = lambda x, y: (GCD(y, x % y) if x % y else y)
ab_gcd = GCD(a, b)

def factoring(gcb,n=2):
    for i in range(n,int(gcb**0.5)+1):
        if gcb%i==0:
            ret = factoring(gcb//i,i)
            ret[i] = ret.get(i,0)+1
            return ret
    return {gcb:1}

result = set(factoring(ab_gcd).keys())
result.add(1)
print(len(result))


Verdicts: Runtime Error

Accepted
Problem: 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.

Submission attempts (in order):
import math
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a


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

g = gcd(a,b)


ans = 1
for i in range(2,int(math.sqrt(g))):
    while g%i == 0:
        g=g//i
        isdevided = True
    if isdevided:
        ans += 1
        isdevided = False
    if g == 1:
        break

print(ans)

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


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

g = gcd(a,b)


arr = []
temp = g
for i in range(2, int(-(-g**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])

print(len(arr)+1)

Verdicts: Runtime Error

Accepted