TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import math

N=int(input())

ans=0
arr=[]

tmp=1

#print(int(math.sqrt(N))+5)

for i in range(int(math.sqrt(N))+5):
    tmp=1

    while N % ((i+2)**tmp) == 0:
        N=N / ((i+2)**tmp)
        print((i+2)**tmp)
        ans+=1
        tmp+=1

    while N % (i+2) == 0:
        N = N // (i+2)

if N > 1 :
    ans+1

print(ans)


import math

N=int(input())

ans=0
arr=[]

tmp=1

print(int(math.sqrt(N))+5)

for i in range(int(math.sqrt(N))+5):
    tmp=1

    while N % ((i+2)**tmp) == 0:
        N=N // ((i+2)**tmp)
        print((i+2)**tmp)
        ans+=1
        tmp+=1

    while N % (i+2) == 0:
        N = N // (i+2)

if N > 1 :
    ans+1

print(ans)


import math

N=int(input())

ans=0
arr=[]

tmp=1

#print(int(math.sqrt(N))+5)

for i in range(int(math.sqrt(N))+5):
    tmp=1

    while N % ((i+2)**tmp) == 0:
        N=N / ((i+2)**tmp)
        #print((i+2)**tmp)
        ans+=1
        tmp+=1

    while N % (i+2) == 0:
        N = N // (i+2)

if N > 1 :
    ans+=1

print(ans)


Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
def f(i,n):
  ans =0
  cnt = i
  while n%cnt == 0:
    n//=cnt
    cnt*=i
    ans +=1
  while n%i ==0:
    n//=i
  return n,ans

n = int(input())
nn = n
i = 2
ans = 0
while i <= int(n**0.5)+1:
  if n%i ==0:
    n,a = f(i,n)
    ans += a
  i += 1
  
if n == nn:
  ans += 1
print(ans)

def f(i,n):
  ans =0
  cnt = i
  while n%cnt == 0:
    n//=cnt
    cnt*=i
    ans +=1
  while n%i ==0:
    n//=i
  return n,ans

n = int(input())
i = 2
ans = 0
while i <= int(n**0.5)+1:
  if n%i ==0:
    n,a = f(i,n)
    ans += a
  i += 1
  
if n >1:
  ans += 1
print(ans)



Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
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])
    return arr

z = factorization(int(input()))
print(z)
i = 1
ans = 0
flag = True
while flag:
    flag2 = False
    for j in range(len(z)):
        if z[j][1] >= i:
            z[j][1] -= i
            ans += 1
            flag2 = True

    i += 1
    if flag2 == False: flag = False
print(ans)


def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5))+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])
    return arr

z = factorization(int(input()))
i = 1
ans = 0
flag = True
while flag:
    flag2 = False
    for j in range(len(z)):
        if z[j][1] >= i:
            z[j][1] -= i
            ans += 1
            flag2 = True

    i += 1
    if flag2 == False: flag = False
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import collections
def factorization(n):
    if n<2:
        return []
    
    prime_factor=[]
    
    for i in range(2,int(n**0.5)+1):
        while n%i==0:
            prime_factor.append(i)
            n//=i
            
    if n>1:
        prime_factor.append(n)

        
    return prime_factor

n=int(input())

c=collections.Counter(factorization(n))
d=list(c.values())

for v in c.values():
    for i in range(1,n):
        v-=i
        if v>=0:
            count+=1
        else:
            break
print(count)

import collections
def factorization(n):
    if n<2:
        return []
    
    prime_factor=[]
    
    for i in range(2,int(n**0.5)+1):
        while n%i==0:
            prime_factor.append(i)
            n//=i
            
    if n>1:
        prime_factor.append(n)
    return prime_factor

count=0
n=int(input())
 
c=collections.Counter(factorization(n))
d=list(c.values())
 
for v in c.values():
    for i in range(1,n):
        v-=i
        if v>=0:
            count+=1
        else:
            break
print(count)

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
n=int(input())
d={2:0}
while n%2<1:  n//=2;  d[2]+=1
for i in range(3,int(n**0.5)+1,2):
  while n%i<1:  n//=i;  d[i]=d.get(i,0)+1
  if n<2: break
if n>1: d[x]=1
a=0
for i in d.values():
  t=c=0
  while t+c<i: c+=1; t+=c
  a+=c
print(a)

n=int(input())

def sieve(x):
  p=[2]
  b=[1]*x
  for i in range(2,x,2):  b[i]=0
  for i in range(3,x,2):
    if b[i]:
      p+=[i]
      for j in range(2*i,x,i): b[j]=0
  return p
def prime_factor(x):
  d={}
  for i in sieve(int(x**0.5)+1):
    while x%i<1:  x//=i;  d[i]=d.get(i,0)+1
    if x<2: break
  if x>1: d[x]=1
  return d

d=prime_factor(n)
a=0
for i in d.values():
  t=c=0
  while t+c<i: c+=1; t+=c
  a+=c
print(a)

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import math

n = int(input())
list = {}
flag = True
while flag:
    flag = False
    for i in range(2,int(math.sqrt(n)+1)):
        if n%i==0:
            if i in list:
                list[i] = list[i]+1
            else:
                list[i] = 1
            n /= i
            flag = True
            break
if n in list:
    list[n] = list[n]+1
else:
    list[n] = 1
ans = 0
for i in list.keys():
    m = list[i]
    for j in range(1,m):
        if not j>m:
            ans += 1
            m -= j
        else:
            break
print(ans)


import math

n = int(input())
list = {}
while n>1:
    for i in range(math.sqrt(n)):
        if n%i==0:
            if i in list:
                list[i] = list[i]+1
            else:
                list[i] = 1
            break
if n%i==0:
    if i in list:
        list[i] = list[i]+1
    else:
        list[i] = 1
ans = 0
for i in list.keys():
    m = list[i]
    for j in range(m):
        if not j>m:
            ans += j
            m -= j
        else:
            break
print(ans)


import math

n = int(input())
list = {}
flag = True
while flag:
    flag = False
    for i in range(2,int(math.sqrt(n)+2)):
        if n%i==0:
            if i in list:
                list[i] = list[i]+1
            else:
                list[i] = 1
            n /= i
            flag = True
            break
if n>1:
    if n in list:
        list[n] = list[n]+1
    else:
        list[n] = 1
ans = 0
for i in list.keys():
    m = list[i]
    for j in range(1,m+1):
        if not j>m:
            ans += 1
            m -= j
        else:
            break
print(ans)


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
from decimal import *
import math
A,B = input().split()
A = Decimal(float(A))
B = Decimal(float(B))
print(math.floor(A*B))

from collections import defaultdict

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 Fib(n):
    a, b = 0, 1
    if n == 1:
        return a
    elif n == 2:
        return b
    else:
        for i in range(n-2):
            a, b = b, a + b
        return b

N = int(input())
if(N == 1):
    print(0)
else:
    fact = factorization(N)
    l = [i for i in range(1,1000)]
    count = 0
    d = {}
    for n in fact:
        num = n[1]
        for f in l:
            num -= f
            if(num < 0):
                break
            count += 1
    print(count)

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
def main():
    n = int(input())
    f = set(prime_factorize(n))
    s = set()
    for p in f:
        e = 1
        z = p ** e
        while z <= n:
            if n % z == 0:
                if z not in s:
                    n //= z
                    s.add(z)
            e += 1
            z = p ** e
    print(len(s))
        
if __name__ == '__main__':
    main()

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

def main():
    n = int(input())
    f = set(prime_factorize(n))
    s = set()
    for p in f:
        e = 1
        z = p ** e
        while z <= n:
            if n % z == 0:
                if z not in s:
                    n //= z
                    s.add(z)
            e += 1
            z = p ** e
    print(len(s))
        
if __name__ == '__main__':
    main()

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

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


def prime_len(n):
    a = set()
    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
    if n != 1:
        a.add(n)
    return len(a)


n = int(input())
c = 0
for z in make_divisors(n):
    if prime_len(z) == 1:
        tmp = n / z
        if tmp.is_integer():
            n = tmp
            c += 1
            print(z, n)
        else:
            continue

print(c)


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


def prime_len(n):
    a = set()
    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
    if n != 1:
        a.add(n)
    return len(a)


n = int(input())
c = 0
for z in make_divisors(n):
    if prime_len(z) == 1:
        tmp = n / z
        if tmp.is_integer():
            n = tmp
            c += 1
        else:
            continue

print(c)


Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
n=int(input())

from math import sqrt

div = int(sqrt(n))
ans=0
for i in range(2,div+1):
    if n%i==0:
        ans+=1
        n=int(n//i)

if ans==0:ans=1

if n==1:ans=0
print(ans)

n=int(input())

import math

def make_prime_list_2(num):
    if num < 2:
        return []

    
    prime_list = [i for i in range(num + 1)]
    prime_list[1] = 0 
    num_sqrt = math.sqrt(num)

    for prime in prime_list:
        if prime == 0:
            continue
        if prime > num_sqrt:
            break

        for non_prime in range(2 * prime, num, prime):
            prime_list[non_prime] = 0

    return [prime for prime in prime_list if prime != 0]

def prime_factorization_2(num):
    if num <= 1:
        return False
    else:
        num_sqrt = math.floor(math.sqrt(num))
        prime_list = make_prime_list_2(num_sqrt)

        dict_counter = {}
        for prime in prime_list:
            while num % prime == 0:
                if prime in dict_counter:
                    dict_counter[prime] += 1
                else:
                    dict_counter[prime] = 1
                num //= prime
        if num != 1:
            if num in dict_counter:
                dict_counter[num] += 1
            else:
                dict_counter[num] = 1

        return dict_counter

l=prime_factorization_2(n)
ans=0

if l==False:pass
else:

    for div,nums in l.items():
        num=nums
        num-=1
        i=2
        while num>=0:
            ans+=1
            num-=i
            i+=1


print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
n = int(input())

count = 0
i = 1
num = n
while num % (2 ** i) == 0:
  num //= 2 ** i
  count += 1
  #print('{}: {}'.format(2**i, count))
  i += 1
  #print(n)
f = 3
i = 1
while f * f <= num:
  if num % (f ** i) == 0:
    num //= f ** i
    count += 1
    #print('{}: {}'.format(f ** i, count))
    i += 1
    #print(n)
  else:
    f += 2
    i = 1
if num == n:
  if num != 1:
    count += 1
print(count)

n = int(input())

lst = []
while n % 2 == 0:
  lst.append(2)
  n //= 2
f = 3
while f * f <= n:
  if n % f == 0:
    lst.append(f)
    n //= f
  else:
    f += 2
if n != 1:
  lst.append(n)
#print(lst)
count = 0
while lst:
  s = 0
  i = 1
  num = lst[0]
  for ele in lst:
    if ele == num:
      s += 1
  for j in range(s):
    lst.remove(num)
  while s > 0:
    s -= i
    i += 1
    if s >= 0:
      count += 1
  #print(lst)
print(count)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
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

from collections import Counter

n = int(input())
if n == 1:
    print(0)
else:
    a = prime_factorize(n)
    c = Counter(a)
    ans = 0
    for v in c.values():
        for i in range(1,100):
            if v == 2:
                ans += 1
                break
            elif (i-1) * i < 2 * v <= i*(i+1):
                print(i,ans)
                ans += i
                break
            else:
                continue
    print(ans)

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

from collections import Counter

n = int(input())
if n == 1:
    print(0)
else:
    a = prime_factorize(n)
    c = Counter(a)
    ans = 0
    for v in c.values():
        temp = 1
        while True:
            v -= temp
            if v >= 0:
                ans +=1
                temp += 1
            else:
                break
    print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
a=int(input())
import math
def div(N):
   ans=0
   for i in range(2,int(N**0.5)+1):
      B=0
      sw=0
      if N%i==0:
         if N//i != i:
            sw=1
            c=N//i
         while N%i==0:
            N//=i
            B+=1
         for j in range(1,10000):
            B-=j
            if B >= 0:
               ans+=1
            else:
               break
         if sw==1:
            B=0
            while N%c==0:
               N//=i
               B+=1
            for j in range(1,10000):
               B-=j
               if B >= 0:
                  ans+=1
               else:
                  break
   return ans
ans=div(a)
if a != 1 and ans == 0:
   print(1)
else:
   print(ans)

N=int(input())
def divisor_enu(N):
   l=[]
   for i in range(2,int(N**0.5)+1):
      cnt=0
      if N%i==0:
         while N%i==0:
            cnt+=1
            N//=i
         l.append(cnt)
   if N != 1:
      l.append(1)
   return l
l=divisor_enu(N)
ans=0
for i in l:
   for j in range(1,11):
      if i >= (1+j)*j//2:
         ans+=1
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
n = int(input())

def factorization(n):
    d = {}
    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
            d[i] = cnt

    if temp!=1:
        d[temp] = 1

    if d.keys()==[]:
        d[n] = 1

    return d

def get_a(n):
    N = n
    if n == 1:
        return [1]
    elif n==2:
        return [2]
    else:
        a = [1]

    while 1:
        n_temp = n - a[-1]
        #print(n_temp, a)
        if n_temp<=0:
            a.pop()
            a[-1] = N - sum(a[:-1])
            break
        else:
            n = n_temp
        a.append(a[-1]+1)
    #print(n_temp, a)
    return a

ans = 0
fact_dict = factorization(n)
#print(fact_dict)

for key, val in fact_dict.items():
    
    ans += len(get_a(val))

if ans>0:
    print(ans)
else:
    print(0)


n = int(input())

def factorization(n):
    d = {}
    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
            d[i] = cnt

    if temp!=1:
        d[temp] = 1

    if d.keys()==[]:
        d[n] = 1

    return d

def get_a(n):
    N = n
    if n == 1:
        return [1]
    elif n==2:
        return [2]
    else:
        a = [1]

    while 1:
        n_temp = n - a[-1]
        #print(n_temp, a)
        if n_temp<0:
            a.pop()
            a[-1] = N - sum(a[:-1])
            break
        elif n_temp == 0:
            break
        else:
            n = n_temp
        a.append(a[-1]+1)
    #print(n_temp, a)
    return a

ans = 0
fact_dict = factorization(n)
#print(fact_dict)

for key, val in fact_dict.items():
    
    ans += len(get_a(val))

if ans>0:
    print(ans)
else:
    print(0)


Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
from collections import Counter

def 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


N = int(input())
arr = dict(Counter(factorize(N)))
print(arr)
ans = 0
for v, c in arr.items():
    ans += 1
    z = 1
    for _ in range(2, c):
        z += z
        if c - z < 0:
            break
        ans += 1
print(ans)

from collections import Counter

def 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


N = int(input())
arr = dict(Counter(factorize(N)))
ans = 0

for v, c in arr.items():
    z = 0
    for i in range(1, c+1):
        z += i
        if c - z < 0:
            break
        ans += 1
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import math

N = int(input())

ans = 0
check = [True for _ in range(10 ** 6)]

for p in range(2, 10 ** 6):
    if check[p] == True:
        for j in range(2, (10 ** 6) // p + 1):
            check[p * j] = False

        e = 1
        while N % (p ** e) == 0:
            N = N // (p ** e)
            e += 1
            ans += 1

if ans == 0:
    if N == 1:
        print(0)
    else:
        print(1)
else:
    print(ans)

import math

N = int(input())

M = int(math.sqrt(N))

ans = 0
check = [True for _ in range(M + 1)]

for p in range(2, M + 1):
    if check[p] == True:
        for j in range(2, M // p + 1):
            check[p * j] = False

        e = 1
        while N % (p ** e) == 0:
            print(N, end=' ')
            print(p ** e)
            N = N // (p ** e)
            e += 1
            ans += 1

if ans == 0:
    if N == 1:
        print(0)
    else:
        print(1)
else:
    print(ans)

import math
 
N = int(input())
 
M = int(math.sqrt(N))

ans = 0 
check = [True for _ in range(M + 1)]
e = [0 for _ in range(M + 1)]

for p in range(2, M + 1):
    if check[p] == True:
        for j in range(2, M // p + 1):
            check[p * j] = False

        while N % p == 0:
            N = N // p
            e[p] += 1
        
        ans += int((math.sqrt(1 + 8 * e[p]) - 1)/2)

if N > 1:
    ans += 1

print(ans)

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
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 = int(input())
b = factorization(a)
countation = 0
for i in range(len(b)):
    t = b[i][1]
    k = 1
    n = 1
    while t >= n:
        countation += 1
        k = k + 1
        n = k*(k+1)/2
if a != 1
    print(countation)
else:
    print(0)

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 = int(input())
b = factorization(a)
countation = 0
for i in range(len(b)):
    t = b[i][1]
    k = 1
    n = 1
    while t >= n:
        countation += 1
        k = k + 1
        n = k*(k+1)/2
if a != 1:
    print(countation)
else:
    print(0)

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
N = int(input())
ans = 0
n = int(N**0.5)
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
for i in range(2,n):
    A = factorization(i)
    if len(A) == 1:
        ans = 1
        N = N // i
    if N == 1:
        break
if N != 1:
    ans += 1
print(ans)

N = int(input())
if N == 1:
    print(0)
    quit()
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 sousa(n):
    x = [1,3,6,10,15,21,28,36,45,55]
    for i in range(10):
        a = x[i]
        if n < a:
            return i
            break
bunkai = factorization(N)
ans = 0
for i in range(len(bunkai)):
    c = bunkai[i][1]
    ans += sousa(c)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import numpy as np
import math
from decimal import *
#from numba import njit

def getVar():
    return map(int, input().split())
def getArray():
    return list(map(int, input().split()))
def getNumpy():
    return np.array(list(map(int, input().split())), dtype='int64')

def factorization(n):
    d = {}
    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
            d.append({i, cnt})

    if temp!=1:
        d.append({temp, 1})

    if d==[]:
        d.append({n, 1})

    return d

def main():
    N = int(input())
    d = factorization(N)
    count = 0

    for v in d.values():
        i = 1
        while(v >= i):
            v -= i
            i += 1
            count += 1
        
    print(count)


main()

import numpy as np
import math
from decimal import *
#from numba import njit

def getVar():
    return map(int, input().split())
def getArray():
    return list(map(int, input().split()))
def getNumpy():
    return np.array(list(map(int, input().split())), dtype='int64')

def factorization(n):
    d = {}
    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
            d.update({i: cnt})

    if temp!=1:
        d.update({temp: 1})

    if d==[]:
        d.update({n:1})

    return d

def main():
    N = int(input())
    d = factorization(N)
    count = 0

    for v in d.values():
        i = 1
        while(v >= i):
            v -= i
            i += 1
            count += 1
        
    print(count)


main()

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import math
N = int(input())
if N == 1:
    print(0)
    exit()


M = int(math.sqrt(N))+1
sosu = [2]
a = 0
for i in range(3,M,2):
  if sosu[a] <= math.sqrt(i):
    a += 1
  hantei = (i%x for x in sosu[:a+1])
  if 0 not in hantei:
    sosu.append(i)

soin = []
for p in sosu:
    for j in range(1,N+1):
        pe = p**j
        if N % pe != 0:
            if pe != p:
                soin.append(j-1)
            break

Ans = 0
for k in soin:
    for n in range(1,k+1):
        if (n*(n+1))//2 > k:
            Ans += n-1
            break
print(max(Ans,1))

import math
N = int(input())
if N == 1:
    print(0)
    exit()
 

M = int(math.sqrt(N))+1
sosu = [2]
a = 0
for i in range(3,M,2):
  if sosu[a] <= math.sqrt(i):
    a += 1
  hantei = (i%x for x in sosu[:a+1])
  if 0 not in hantei:
    sosu.append(i)
 
soin = []
for p in sosu:
    for j in range(1,N+1):
        pe = p**j
        if N % pe != 0:
            if pe != p:
                soin.append(j-1)
                N //= p**(j-1)
            break

Ans = 0
if N != 1:
    Ans += 1

for k in soin:
    if k == 1:
        Ans += 1
        continue
    for n in range(1,k+1):
        if (n*(n+1))//2 > k:
            Ans += n-1
            break
print(max(Ans,1))

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(n**0.5)+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 = factorization(int(input()))
b = [1, 3, 6, 10, 15, 21, 28, 36, 43]

ans = 0

if len(a) == 1 and a[0][0] == 1:
  print(0)
  exit()
  
for i in a:
  for j in range(len(b)):
    if 2 < b[j]:
      ans+=j
      break
    
print(ans)


def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(n**0.5)+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 = factorization(int(input()))
b = [1, 3, 6, 10, 15, 21, 28, 36, 43]

ans = 0

if len(a) == 1 and a[0][0] == 1:
  print(0)
  exit()
  
for i in a:
  for j in range(len(b)):
    if i[1] < b[j]:
      ans+=j
      break
    
print(ans)


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 = factorization(int(input()))
b = [1, 3, 6, 10, 15, 21, 28, 36, 43]

ans = 0

if len(a) == 1 and a[0][0] == 1:
  print(0)
  exit()
  
for i in a:
  for j in range(len(b)):
    if i[1] < b[j]:
      ans+=j
      break
    
print(ans)

def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(n**0.5)+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 = factorization(int(input()))
b = [1, 3, 6, 10, 15, 21, 28, 36, 45]

ans = 0

if len(a) == 1 and a[0][0] == 1:
  print(0)
  exit()
  
for i in a:
  for j in range(len(b)):
    if i[1] < b[j]:
      ans+=j
      break
    
print(ans)


Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
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 calcN(num):
    if num >= 45:
        return 9
    elif num >= 36:
        return 8
    elif num >= 28:
        return 7
    elif num >= 21:
        return 6
    elif num >= 15:
        return 5
    elif num >= 10:
        return 4
    elif num >= 6:
        return 3
    elif num >= 3:
        return 2
    elif num >= 1:
        return 1
    

A = int(input().split())
insuu =  factorization(A)
ret = 0
for i in insuu:
    ret += calcN(i[1])
print(ret)

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 calcN(num):
    if num >= 45:
        return 9
    elif num >= 36:
        return 8
    elif num >= 28:
        return 7
    elif num >= 21:
        return 6
    elif num >= 15:
        return 5
    elif num >= 10:
        return 4
    elif num >= 6:
        return 3
    elif num >= 3:
        return 2
    elif num >= 1:
        return 1
    

A = int(input())
if A == 1:
    print(0)
else:
    insuu =  factorization(A)
    ret = 0
    for i in insuu:
        ret += calcN(i[1])
    print(ret)

Verdicts: Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import math

def primelist(num):
    if num < 2:
        return []
    ls = [2]
    n = int(math.sqrt(num))
    for i in range(3, n+1, 2):
        prime = True
        for j in ls:
            if i % j ==0:
                prime = False
                break
        if prime:
            ls += [i]
    prime = True
    for j in ls:
        if num % j ==0:
            prime = False
            break
    if prime:
        ls += [num]
    return ls

n = int(input())
ls = primelist(n)

ans = 0
for i in ls:
    j = i
    while n % j == 0:
        ans += 1
        j *= i
print(ans)


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

from itertools import groupby
N = int(input())
P = prime_decomposition(N)
ans = 0
for k, g in groupby(P):
    n = len(g)
    for i in range(1, 10000000):
        n -= i
        if n >=0:
            ans += 1
        else:
            break
print(ans)

        

import sys
import time
st = time.perf_counter()
# ------------------------------


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

from itertools import groupby
N = int(input())
P = prime_decomposition(N)
ans = 0
for k, g in groupby(P):
    n = len(list(g))
    for i in range(1, 10000000):
        n -= i
        if n >=0:
            ans += 1
        else:
            break
print(ans)

        
        



# ------------------------------
ed = time.perf_counter()
print('time:', ed-st, file=sys.stderr)


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
a,n=0,int(input())
for i in range(2,10**6):
  x=j=0
  while n%i<1:n/=i;x+=1
  while x>=j:a+=1;j+=1;x-=j
print(a+(n>1))

a,n=0,int(input())
for i in range(2,int(n**0.5+1)):x=0
  while(n%i==0):n//=i;x+=1
  j=1
  while(x>=j):a+=1;x-=j;j+=1
print(1 if n-1 else a)

a,n=0,int(input())
for i in range(2,10**6):
  x=j=0
  while n%i<1:n//=i;x+=1
  while x>j:a+=1;j+=1;x-=j
print(a+(n>1))

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import collections

def primes(n): 
 cnt=collections.defaultdict(int)
 for i in range(2,int(n**0.5)+1):
   if n%i==0:
     while n%i==0:
       cnt[i]+=1
       n//=i
 if n!=1:
   cnt[n]+=1
 return cnt

n=int(input())
cnt=primes(n)
ans=0
for key in cnt.keys():
    val=cnt[key]
    print(key,val)
    for i in range(100): 
        if (i*(i+1))//2>val:
            ans+=i-1
            break
print(ans)

import collections
 
def primes(n): 
 cnt=collections.defaultdict(int)
 for i in range(2,int(n**0.5)+1):
   if n%i==0:
     while n%i==0:
       cnt[i]+=1
       n//=i
 if n!=1:
   cnt[n]+=1
 return cnt
 
n=int(input())
cnt=primes(n)
ans=0
for key in cnt.keys():
    val=cnt[key]
    for i in range(100): 
        if (i*(i+1))//2>val:
            ans+=i-1
            break
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
import math
import collections


def trial_division(n):
    factor = []
    tmp = int(math.sqrt(n)) + 1
    for num in range(2, tmp):
        while n % num == 0:
            n //= num
            if num != 1:
                factor.append(num)
    if not factor:
        return -1
    else:
        factor.append(n)
        return factor


n = int(input())
l = trial_division(n)
if n == 1:
    print(0)
    exit(0)
if l == -1:
    print(1)
    exit(0)
if l[len(l) - 1] == 1:
    l = l[:-1]
c = collections.Counter(l)

print('debug')
ans = 0
for k, v in c.items():
    s = 1
    for i in range(1, v):
        if i * (i + 1) <= v * 2:
            s = max(s, i)
    ans += s
print(ans)


import math
import collections


def trial_division(n):
    factor = []
    tmp = int(math.sqrt(n)) + 1
    for num in range(2, tmp):
        while n % num == 0:
            n //= num
            if num != 1:
                factor.append(num)
    if not factor:
        return -1
    else:
        factor.append(n)
        return factor


n = int(input())
l = trial_division(n)
if n == 1:
    print(0)
    exit(0)
if l == -1:
    print(1)
    exit(0)
if l[len(l) - 1] == 1:
    l = l[:-1]
c = collections.Counter(l)

ans = 0
for k, v in c.items():
    s = 1
    for i in range(1, v):
        if i * (i + 1) <= v * 2:
            s = max(s, i)
    ans += s
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
def primefact(n):
    l=[]
    if n%2==0:
        while n%2==0:
            n=n//2
        l.append(2)
    for i in range(3,int(n**0.5)+1,2):
        if n%i==0:
            while n%i==0:
                n=n//i
            l.append(i)
    if n>2:
        l.append(n)
    print(len(l))
n=int(input())
primefact(n)



def primefact(n):
    l=[]
    if n%2==0:
        while n%2==0:
            n=n//2
        l.append(2)
    for i in range(3,int(n**0.5)+1,2):
        if n%i==0:
            while n%i==0:
                n=n//i
            l.append(i)
    if n>2:
        l.append(n)
    return l

n=int(input())
c=0
a=(primefact(n))
for i in a:
    x=i
    while n%x==0:
        n=n//x
        c+=1
        x=x*i
print(c)
        


Verdicts: Wrong Answer

Accepted
Problem: Given is a positive integer N. Consider repeatedly applying the operation below on N:    * First, choose a positive integer z satisfying all of the conditions below:     * z can be represented as z=p^e, where p is a prime number and e is a positive integer;     * z divides N;     * z is different from all integers chosen in previous operations.   * Then, replace N with N/z.  Find the maximum number of times the operation can be applied.

Submission attempts (in order):
def prime(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 list(set(a))
x=int(input())
p=prime(x)
s=0

for j in p:
    i=1
    while x%(j**i)==0:
        x=x//(j**i)
        s+=1
        i+=1
print(p)
print(s)

def prime(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 list(set(a))
x=int(input())
p=prime(x)
s=0

for j in p:
    i=1
    while x%(j**i)==0:
        x=x//(j**i)
        s+=1
        i+=1
print(s)

Verdicts: Wrong Answer

Accepted