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):
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
dum = prime_factorize(int(input()))
print(dum)
print(len(dum))
dum1 = 0
ans = 0
for i in range(len(dum)):
    if dum[i] == dum[i+1]:
        cnt += 1
    else:
        dum1 = 0
        cnt = 0
    if cnt > dum1:
        ans += 1
        cnt = 0
        dum1 += 1
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
dum = prime_factorize(int(input()))
dum_len = len(dum)

num = 0
ans = 0
cnt = 0
check = 0
for i in range(dum_len):
    if num == dum[i]:
        cnt += 1
        if cnt >= check:
            check += 1
            ans += 1
            cnt = 0
    else:
        num = dum[i]
        ans += 1
        cnt = 0
        check = 2
print(ans)

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())
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(cnt)

if temp!=1:
    arr.append(1)
if arr==[]:
    arr.append(1)
for i in range(len(arr)):
    if arr[i] == 1:
        arr[i] = 1
    else:
        s = (-1+math.sqrt(1+8*arr[i]))/2
        arr[i] = s
if n = 1:
    print(0)
else:
    print(int(sum(arr)))

import math
N = int(input())
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(cnt)

    if temp!=1:
        arr.append(1)

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

    return arr
A = factorization(N)
if N == 1:
    print(0)
else:
    for i in range(len(A)):
        if A[i] == 1:
            A[i] = 1
        else:
            s = (-1+int(math.sqrt(1+8*A[i])))/2
            A[i] = int(s)
    print(int(sum(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 sys
n=int(input())

if n == 1:
    print(0)
    sys.exit()

orgn=n
counter=0
for i in range(2,int((n**0.5) + 1) ):
    if n % i == 0:
        counter += 1
        n //= i
        wari=i**2
        while n % wari == 0:
            counter +=1
            n //= wari
            wari *= i
        
        while n % i ==0:
            n //= i

if n == orgn:
    print(1)

if n >= int(n ** 0.5):
        counter+=1
print(counter)

import sys
n=int(input())

if n == 1:
    print(0)
    sys.exit()

orgn=n
counter=0
for i in range(2,int((n**0.5) + 1) ):
    if n % i == 0:
        counter += 1
        n //= i
        wari=i**2
        while n % wari == 0:
            counter +=1
            n //= wari
            wari *= i
        
        while n % i ==0:
            n //= i

if n == orgn:
    print(1)
    sys.exit()

if n > int(n ** 0.5):
        counter+=1
print(counter)

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
from collections import Counter
A= int(input())
resArray = []
tmpA = A
i = 1
while i <= math.sqrt(A):
    #print (i)
    i += 1
    if tmpA == 1:
        break

    if i >2 and i %2 == 0:
        continue


    while tmpA % i == 0:
        tmpA = tmpA / i
        resArray.append(i)

res = 0
if tmpA == A:
    print(1)
    exit()
else tmpA > 1:
    res += 1


cArray = Counter(resArray)

for i in cArray.values():
    if i > 0:
        
        tmp = 1
        while tmp*(tmp+1)//2 <= i:
            tmp += 1
        res += tmp -1
print(res)

import math
from collections import Counter
A= int(input())
resArray = []
tmpA = A
i = 1
while i <= math.sqrt(A):
    #print (i)
    i += 1
    if tmpA == 1:
        break

    if i >2 and i %2 == 0:
        continue


    while tmpA % i == 0:
        tmpA = tmpA / i
        resArray.append(i)

res = 0
if tmpA == A:
    print(1)
    exit()
else:
    res += 1


cArray = Counter(resArray)

for i in cArray.values():
    if i > 0:
        
        tmp = 1
        while tmp*(tmp+1)//2 <= i:
            tmp += 1
        res += tmp -1
print(res)

import math
A= int(input())
resArray = []
tmpA = A
for i in range(2, int(math.sqrt(A))):
    if tmpA == 1:
        break
    tmp = tmpA%i
    resArray.append(0)
    while tmp == 0:
        print(i)
        tmpA = tmpA/i
        resArray[-1] += 1
        tmp = tmpA%i
if tmpA != 1:
    print(1)
    exit()
res = 0
for i in resArray:
    if i != 0:
        for j in range(i + 1):
            if j*(j+1)/2 > i:
                res += j - 1
                break
print(res)

import math
A= int(input())
resArray = []
tmpA = A
for i in range(2, int(math.sqrt(A))):
    if tmpA == 1:
        break
    tmp = tmpA%i
    resArray.append(0)
    while tmp == 0:
        tmpA = tmpA/i
        resArray[-1] += 1
        tmp = tmpA%i
if tmpA != 1:
    print(1)
    exit()
res = 0
for i in resArray:
    if i != 0:
        for j in range(i + 1):
            if j*(j+1)/2 > i:
                res += j - 1
                break
print(res)

import math
from collections import Counter
A= int(input())
if A == 1:
    print(0)
    exit()
resArray = []
tmpA = A
i = 1
while i <= math.sqrt(A):
    #print (i)
    i += 1
    if tmpA == 1:
        break

    if i >2 and i %2 == 0:
        continue


    while tmpA % i == 0:
        tmpA = tmpA / i
        resArray.append(i)

res = 0
if tmpA == A:
    print(1)
    exit()
elif tmpA > 1:
    res += 1


cArray = Counter(resArray)

for i in cArray.values():
    if i > 0:
        
        tmp = 1
        while tmp*(tmp+1)//2 <= i:
            tmp += 1
        res += tmp -1
print(res)


Verdicts: Runtime Error

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):
import collections

N = int(input())

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

c = collections.Counter(prime_factorize(N))
print(c)

ans=0
for i in c.values():
    tmp = i
    cnt=0
    val=1
    while True:
        tmp = tmp-val
        val+=1

        if tmp > 0:
            cnt+=1
        
        if tmp == 0:
            cnt+=1
            ans+=cnt
            break
        
        if tmp < 0:
            ans+=cnt
            break

print(ans)


import collections

N = int(input())

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

c = collections.Counter(prime_factorize(N))

ans=0
for i in c.values():
    tmp = i
    cnt=0
    val=1
    while True:
        tmp = tmp-val
        val+=1

        if tmp > 0:
            cnt+=1
        
        if tmp == 0:
            cnt+=1
            ans+=cnt
            break
        
        if tmp < 0:
            ans+=cnt
            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):
def solve(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

n = int(input())
print(solve(n))

n = int(input())

def factorization(n):
    arr = []
    if n == 1:
        return 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 solve(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

n = int(input())
arr = solve(n)

cnt = 0
for n in arr:
    a=1
    if n[0]==1:
        break
    while n[1]>=a:
        n[1] -= a
        a += 1
        cnt += 1

print(cnt)

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):
import copy
n=int(input())

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

L=prime_factorize(n)
M=L.copy()
M=list(set(M))

cnt=0
for x in M:
    m=L.count(x)
    for i in range(L.count(x)):
        if m-i-1 > 0:
            m -= i+1
            cnt += 1

print(cnt)


import copy
n=int(input())

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

L=prime_factorize(n)
M=L.copy()
M=list(set(M))

cnt=0
for x in M:
    m=L.count(x)
    for i in range(L.count(x)):
        if m-i-1 >= 0:
            m -= i+1
            cnt += 1

print(cnt)


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

def factorization(n):
    factor_list = []
    tmp = int(math.sqrt(n)) + 1
    for num in range(2, tmp+1):
        count = 0
        while n % num == 0:
            count += 1
            n //= num
            
        if count != 0:
            factor = [num, count]
            factor_list.append(factor)
        
    if n != 1:
        factor = [n, 1]
        factor_list.append(factor)
    #print(n)
    return factor_list
            

def main():
    n = int(input())
    count = 0
    
    if n == 1:
        print(0)
        return
    
    factor_list = factorization(n)
    print(factor_list)
    if len(factor_list) == 0:
        print(1)
        return
    
    for factor in factor_list:
        exponent = 1
        n /= factor[0] ** exponent
        count += 1
        factor[1] -= exponent
        exponent += 1
        
        while factor[1] > 1:
            #print(factor_list)
            n /= factor[0] ** exponent
            count += 1
            factor[1] -= exponent
            exponent += 1
    
    print(count)
    return
    
if __name__ == '__main__':
    main()


import math

def factorization(n):
    factor_list = []
    tmp = int(math.sqrt(n)) + 1
    for num in range(2, tmp+1):
        if n % num == 0:
            count = 0
            while n % num == 0:
                count += 1
                n //= num
                
            factor_list.append([num, count])
        
    if n != 1:
        factor = [n, 1]
        factor_list.append(factor)
    #print(n)
    return factor_list
            

def main():
    n = int(input())
    count = 0
    
    factor_list = factorization(n)
    
    for _, exponent in factor_list:
        
        
        #count += 1
        
        
        
        i = 1
        while exponent >= i:
            #print(factor_list)
            
            count += 1
            exponent = exponent - i
            i += 1
    
    print(count)
    return
    
if __name__ == '__main__':
    main()

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())

import math

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

    if temp!=1:
        arr.append(1)

#    if arr==[]:


    return arr

z=factorization(n)

ans=0
for i in range(len(z)):
    for j in range(1,z[i]):
        if (z[i]-j)>=0:
            z[i]-=j
            ans+=1
        else:
            break
print(ans)

n=int(input())

import math

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

    if temp!=1:
        arr.append(1)

#    if arr==[]:


    return arr

z=factorization(n)

ans=0
for i in range(len(z)):
    for j in range(1,z[i]+1):
        if (z[i]-j)>=0:
            z[i]-=j
            ans+=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):
N = int(input())

p = [0]*10**6
for i in range(2, 10**6):
    if p[i] == 0:
        p[i] = 1
        j = 2
        while i*j < 10**6:
            p[i*j] = -1
            j += 1

L = []
for i in range(10**6):
    if p[i] < 1:
        continue
    j = 1
    while pow(i, j) <= N:
        L.append(pow(i, j))
        j += 1

L.sort()
ans = 0
temp = []
for i in L:
    if N%i == 0:
        N //= i
        ans += 1
        temp.append(i)

i = 2
f = True
while i <= N**0.5:
    if N%i == 0 and i in temp:
        f = False
        break
    i += 1
if N in temp:
    f = False
if f:
    ans += 1
print(ans)

N = int(input())

p = [0]*10**6
for i in range(2, 10**6):
    if p[i] == 0:
        p[i] = 1
        j = 2
        while i*j < 10**6:
            p[i*j] = -1
            j += 1

ans = 0
for i in range(10**6):
    if p[i] < 1:
        continue
    j = 1
    while N%pow(i, j) == 0:
        N //= pow(i, j)
        j += 1
        ans += 1

for i in range(10**6):
    if p[i] < 1:
        continue
    while N%i == 0:
        N //= i
    
if N > 10**6:
    i = 2
    f = True
    while i <= N**0.5:
        if N%i == 0:
            f = False
            break
        i += 1
    if f:
        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 divisor(n): 
    i = 1
    table = []
    while i * i <= n:
        if n%i == 0:
            table.append(i)
            table.append(n//i)
        i += 1
    table = list(set(table))
    return table
def 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

n = int(input())
a = divisors(n)
cnt = 0
for i in range(1, len(a)):
    b = prime_decomposition(a[i])
    if len(list(set(b))) == 1:
        if n % a[i] == 0:
            n //= a[i]
            cnt += 1
print(cnt)

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

n = int(input())
a = sorted(divisor(n))
cnt = 0
for i in range(1, len(a)):
    b = prime_decomposition(a[i])
    if len(list(set(b))) == 1:
        if n % a[i] == 0:
            n //= a[i]
            cnt += 1
print(cnt)

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 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 using_sqrt(limit):
    n = []
    for k in range(2, limit + 1):
        factor = 0

        
        if k % 2 == 0 and k != 2:
            continue

        
        for divisor in range(2, math.floor(math.sqrt(k)) + 1):
            if k % divisor == 0:
                factor += 1

        if factor == 0:
            n.append(k)

    return n


def isPrime(limit):
    for k in range(limit, limit + 1):
        factor = 0

        
        if k % 2 == 0 and k != 2:
            continue

        
        for divisor in range(2, math.floor(math.sqrt(k)) + 1):
            if k % divisor == 0:
                factor += 1

        if factor == 0:
            return True
    return False


n = int(input())
prime = using_sqrt(50**5)
prev = []

if n > 1 and isPrime(n):
    print(1)
    exit()


def check(n, prev):
    divs = make_divisors(int(n))
    flag = False
    for i in divs[1:]:
        for p in prime:
            e = math.log(i, p)
            if e.is_integer() and i not in prev:
                prev.append(i)
                flag = True
                check(n/i, prev)
                break
        if flag:
            break


check(n, prev)
print(len(prev))

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


n = int(input())
if n == 1:
    print(0)
    exit()
z = factorization(n)
cnt = 0
for item in z:
    p = item[0]
    e = item[1]
    num = 1
    while e >= num:
        e -= num
        cnt += 1
        num += 1

print(cnt)

Verdicts: Time Limit Exceeded

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 itertools import combinations
from functools import reduce
from operator import mul

N = int(input())

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

facts = prime_factorize(N)

used = set()
cnt = 0

while facts:
  #print(used)
  #print(facts)
  for i in range(1, len(facts)):
    for s in combinations(facts, i):
      m = reduce(mul, s)
      if not m in used:
        used |= {m}
        cnt += 1
        for j in s:
          facts.remove(j)
        break
    else:
      continue
    break
  else:
    break

print(cnt)

from collections import Counter

N = int(input())

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

facts = Counter(prime_factorize(N))
cnt = 0

def pattern_cnt(n):
  res = 0
  cur = 1
  while n - cur >= 0:
    n -= cur
    res += 1
    cur += 1
  return res

for k in facts:
  cnt += pattern_cnt(facts[k])
  #print(k, cnt)

print(cnt)

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 defaultdict
from itertools import accumulate


def main():
    n = int(input())
    prime_counter = defaultdict(int)
    acc = tuple(accumulate(range(1, 42)))  # 10^12 < 2^39

    f = 2
    while f * f <= n:
        if n % f == 0:
            n //= f
            prime_counter[f] += 1
        else:
            f += 1
    if n != 1:
        prime_counter[f] += 1

    ans = 0
    for c in prime_counter.values():
        for i, x in enumerate(acc):
            if c < x:
                tmp = i
                break
        ans += tmp
    print(ans)


if __name__ == '__main__':
    main()


def main():
    n = int(input())
    ans = 0

    e = 1
    for fac in (2, *(x * 2 + 3 for x in range(pow(10, 12)))):
        while n % (div := pow(fac, e)) == 0:
            n //= div
            ans += 1
            e += 1

        while n % fac == 0:
            n //= fac

        if fac * fac > n:
            break
        e = 1

    if n != 1:
        ans += 1

    print(ans)


if __name__ == '__main__':
    main()


from itertools import accumulate


def main():
    n = int(input())
    prime_counter = [0] * int(n ** 0.5 + 2)
    acc = tuple(accumulate(range(1, 42)))  # 10^12 < 2^39

    f = 2
    while f * f <= n:
        if n % f == 0:
            n //= f
            prime_counter[f] += 1
        else:
            f += 1
    if n != 1:
        prime_counter[f] += 1

    ans = 0
    for c in prime_counter:
        for i, x in enumerate(acc):
            if c < x:
                tmp = i
                break
        ans += tmp
    print(ans)


if __name__ == '__main__':
    main()


from itertools import accumulate
from collections import defaultdict


def main():
    n = int(input())
    prime_counter = defaultdict(int)  # key ^ value
    # 10 because 2^39 < 10^12 < 2^40 and accum[9] = 45
    accum = tuple(accumulate(range(10)))

    def facs(n):
        yield 2
        max_prime = int(pow(n, 0.5)) + 1
        for x in range(3, max_prime, 2):
            yield x

    for fac in facs(n):
        while n % fac == 0:
            n //= fac
            prime_counter[fac] += 1
    if n != 1:
        prime_counter[n] += 1

    ans = 0
    for c in prime_counter.values():
        for accum_count, accum_value in enumerate(accum):
            if c < accum_value:
                ans += accum_count - 1
                break
    print(ans)


if __name__ == '__main__':
    main()


Verdicts: Wrong Answer

Time Limit Exceeded

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())
n = N
ans = 0
num = [2]
for i in range(3, int(N ** 0.5)):
  if all(i % x != 0 for x in num):
    num.append(i)
  if n % i == 0:
    index = 1
    while n % i**index == 0:
      ans += 1
      n = n // (i**index)
      index += 1
  if n == 1:
    break
print(ans)

N = int(input())
ans = 0
n = N
counter = []
for i in range(2, int(N**0.5) + 2):
  if n % i == 0:
    counter.append(1)
    n = n // i
    while n % i == 0:
      counter[-1] += 1
      n = n // i
if n != 1:
  counter.append(1)
for c in counter:
  a = 1
  while c >= a:
    c -= a
    a += 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):
history = set([])


import sympy

def func(n):
    if n == 1:
        return 
    l = sympy.divisors(n)
    for i in sorted(l[1:]):
        if i in history:
            continue
        if len(sympy.factorint(i)) > 1:
            continue
        print(i, n)
        history.add(i)
        _ = n // i
        func(_)
        break
    return

N = int(input())
if N == 0:
    print(0)
elif N == 1:
    print(1)
else:
    func(N)
    print(len(history))

input_count = 0

import math

A, B = input().split()

A = int(A)
if '.' in B:
    B_a = int(B.split('.')[0])
    B_b_size = len(B.split('.')[1])
    B_b =  int(B.split('.')[1])
else:
    B_a = int(B)
    B_b_size = 0
    B_b = 0
ans_a = A * B_a
ans_b = A * B_b 
ans = ans_a + int(ans_b * pow(10, B_b_size))
print(ans)

history = set([])

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 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 func(n):
    if n == 1:
        return 
    l = make_divisors(n)
    for i in sorted(l[1:]):
        if i in history:
            continue
        if len(factorization(i)) > 1:
            continue
        history.add(i)
        _ = n // i
        func(_)
        break
    return

N = int(input())
if N == 0:
    print(0)
elif N == 1:
    print(0)
else:
    func(N)
    print(len(history))

Verdicts: Runtime Error

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())
def prime_dict(n):
    a = {}
    if n%2 == 0:
        a[2] = 1
        n//=2
    while n%2 == 0:
        a[2] += 1
        n //= 2
    f = 3
    while f*f <= n:
        if n%f == 0:
            if f not in a:
                a[f] = 1
            else:
                a[f] += 1
        else:
            f += 2
    if n!= 1:
        a[n] = 1
    return a
primes = prime_dict(N)
ans = 0

print(ans)
 

from collections import Counter
N = int(input())
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
ans = 0
primes = prime_factorize(N)
pc = Counter(primes)
for v in pc.values():
    now = 1
    while now*(now+1)//2 <= v:
        now += 1
    ans += now-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())
n=N
p=2
dcmp={}
while n>1:
    if n%p==0:
        dcmp[p]=1+(dcmp[p] if p in dcmp else 0)
        n//=p
    else:
        p+=1
cnt=0
for k,v in dcmp.items():
    i=1
    while True:
        dcmp[k]-=i
        if dcmp[k]<0:
            break
        i+=1
        cnt+=1



N=int(input())
n=N
p=2
dcmp={}
while n>1:
    if n%p==0:
        dcmp[p]=1+(dcmp[p] if p in dcmp else 0)
    else:
        p+=1
cnt=0
for k,v in dcmp.items():
    i=1
    while True:
        dcmp[k]-=i
        if dcmp[k]<0:
            break
        i+=1
        cnt+=1

print(cnt)


N=int(input())
n=N
p=2
dcmp={}
while n>1:
    if n%p==0:
        dcmp[p]=1+(dcmp[p] if p in dcmp else 0)
    else:
        p+=1
    n//=p
cnt=0
for k,v in dcmp.items():
    i=1
    while True:
        dcmp[k]-=i
        if dcmp[k]<0:
            break
        i+=1
        cnt+=1

print(cnt)


import math
N=int(input())
n=N
p=2
dp=1
thrP=math.sqrt(N)
dcmp={}
while n>1:
    if n%p==0:
        dcmp[p]=1+(dcmp[p] if p in dcmp else 0)
        n//=p
    else:
        p+=dp
        dp=2
    if p>thrP:
        if n>1:
            dcmp[n]=1
        break
cnt=0
for k,v in dcmp.items():
    i=1
    while True:
        dcmp[k]-=i
        if dcmp[k]<0:
            break
        i+=1
        cnt+=1

print(cnt)


Verdicts: Wrong Answer

Time Limit Exceeded

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())


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

factorization(a)


a=int(input())
ans =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
b = factorization(a)
for i in range(len(b)):
    count=0
    c = b[i][1]

    for j in range(1,40):
        if c - j >= 0:
            c -= j
            count += 1
        else:
            ans += count
            break
if a ==1:
    print(0)
else:
    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 sys
input = sys.stdin.readline
import numpy as np
from decimal import *

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


n = int(input())
if n == 1:
    print(0)
else:
    arr = factorization(n)

    arr = np.array(arr)

    kazu = arr[:,1]
    sum = 0
    for i in range(1, 1000000):
        if len(kazu) ==0:
            break
        sum += len(kazu)
        kazu -= (i)
        tugi =[]
        for k in kazu:
            if k <= 0:
                sum -=1
                pass
            else:
                tugi.append(k)
        kazu = np.array(tugi)
    print(sum)


import sys
input = sys.stdin.readline
import numpy as np
from decimal import *

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


n = int(input())
if n == 1:
    print(0)
else:
    arr = factorization(n)

    arr = np.array(arr)

    kazu = arr[:,1]
    sum = 0
    for i in range(1, 1000000):
        if len(kazu) ==0:
            break
        sum += len(kazu)
        kazu -= (i)
        tugi =[]
        for k in kazu:
            if k < 0:
                sum -=1
                pass
            elif k == 0:
                pass
            else:
                tugi.append(k)
        kazu = np.array(tugi)
    print(sum)


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 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
l=prime_factorize(N)
L=set(l)
print(l,L)
z=0
for j in L:
  i=1
  while i*(i+1)//2<=l.count(j):
    i+=1
  z+=i-1;print(z)
print(z)

N=int(input())
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
l=prime_factorize(N)
L=set(l)
z=0
for j in L:
  i=1
  while i*(i+1)//2<=l.count(j):
    i+=1
  z+=i-1
print(z)

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
 
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



N=int(input())
if N==1:
  print(0)
  exit(0)
c=0
p=factorization(N)
print(p)
for i in range(len(p)):
    r=1
    s=0
    while 1:
        s+=r
        if p[i][1]>=s:
            N/=pow(p[i][0],r)
            print(N)
            c+=1
            r+=1
        else:
            break
            
print(c)

#import math
 
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



N=int(input())
if N==1:
  print(0)
  exit(0)
c=0
p=factorization(N)
#print(p)
for i in range(len(p)):
    r=1
    s=0
    while 1:
        s+=r
        if p[i][1]>=s:
            N/=pow(p[i][0],r)
            #print(N)
            c+=1
            r+=1
        else:
            break
            
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):
from math import sqrt
from collections import defaultdict

N = int(input())
dic = defaultdict(int)

def func(N):
  if N == 1:
    return 1
  
  for i in range(2, max(int(sqrt(N))+1, 3)):
    if N % i == 0:
      N = N // i
      dic[i] += 1
      return func(N)

func(N)

def func2(k):
  n = 1
  count = 1
  while count * (count+1) // 2 <= k:
    count += 1
  return count-1

res = 0

for value in dic.values():
  if value > 0:
    res += func2(value)

print(res)

from math import sqrt
from collections import defaultdict

N = int(input())

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

def func2(k):
  n = 1
  count = 1
  while count * (count+1) // 2 <= k:
    count += 1
  return count-1

res = 0

for value in ps.values():
  if value > 0:
    res += func2(value)

print(res)

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 sys
n=int(input())
if n==1:
    print(0)
    sys.exit()
def f(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
tmp=f(n)
ans=0
for i in tmp:
    dd=i[1]
    tt=1
    while dd>0:
        if dd-tt>=0:
            dd-=tt
            tt+=1
            ans+=1
        else:
            break
print(ans)=0:
            dd-=tt
            tt+=1
            ans+=1
print(ans)

import sys
n=int(input())
if n==1:
    print(0)
    sys.exit()
def f(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
tmp=f(n)
ans=0
for i in tmp:
    dd=i[1]
    tt=1
    while dd>0:
        if dd-tt>=0:
            dd-=tt
            tt+=1
            ans+=1
        else:
            break
print(ans)

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):
m=int(input())
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

l=factorization(m)
ans=0
for i in range(len(l)):
    a=1
    while(l[i][1]>=a):
        ans+=1
        l[i][1]-=a
        a+=1
if m=1:
    print("0")
else:
    print(ans)

m=int(input())
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

l=factorization(m)
ans=0
for i in range(len(l)):
    a=1
    while(l[i][1]>=a):
        ans+=1
        l[i][1]-=a
        a+=1
if m==1:
    print("0")
else:
    print(ans)

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())

p = 2
counter = 0
while N != 1:
  ind = 0
  n = -1
  while N % p == 0:
    N //= p
    ind += 1
  while (n+1)(n+2)//2 <= ind:
    n += 1
  counter += n
  p += 1

print(counter)

import math

N = int(input())

p = 2
counter = 0
while N != 1:
 if p > math.sqrt(N):
  counter += 1
  break
 ind = 0
 n = -1
 while N % p == 0:
  N //= p
  ind += 1
 if ind == 0:
  p += 1
  continue
 while (n+1)*(n+2)//2 <= ind:
  n += 1
 counter += n
 p += 1

print(counter)

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())
ans = 0
k = int(math.sqrt(n)) + 1
prime = set()
n1 = n


for i in range(2,k):
    while n1%i==0:
        prime.add(i)
        n1 = n1//i
    if n1==1: break


for i in range(2,k):
    p = 1
    while n%(i**p)==0 and i in prime:
        ans+=1
        n = n//(i**p)
        p+=1
    if n<i: break

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



import math
n = int(input())
ans = 0
k = int(math.sqrt(n)) + 1
for i in range(2,k):
    p = 1
    while n%(i**p)==0:
        ans+=1
        p+=1
        n = n//(i**p)
    if n==1: break
print(ans)



import math
n = int(input())
ans = 0
k = int(math.sqrt(n)) + 1

for i in range(2,k):
    p = 1
    while n%(i**p)==0:
        ans+=1
        n = n//(i**p)
        p+=1
        
    while n % i == 0:
        n //= i

    if n<i: break

if n!=1: ans+=1
print(ans)



Verdicts: Wrong Answer

Wrong Answer

Accepted