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 numpy as np
n = int(input())

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
  
fact = factorization(n)
vals = []
for f in fact:
  for e in range(1, f[1]):
    vals.append(f[0]**e)
vals.sort()
if n==1:
  print(0)
else:
  ans = 0
  for d in vals:
    if n%d == 0:
      ans += 1
      n /= d
  print(ans)

import numpy as np
n = int(input())

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
  
fact = factorization(n)
vals = []
for f in fact:
  for e in range(1, f[1]+1):
    vals.append(f[0]**e)
vals.sort()
if n==1:
  print(0)
else:
  ans = 0
  for d in vals:
    if n%d == 0:
      ans += 1
      n /= d
  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 = []
    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
N = int(input())
ans = 0
if N != 1:
    fact_list = factorization(N)
    for _, i in fact_list:
        j = 1
        while i >= j:
            ans += 1
            i = i - j
            j += 1
            
print(ans)


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

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

    return arr
N = int(input())
ans = 0
if N != 1:
    fact_list = factorization(N)
    for _, i in fact_list:
        j = 1
        while i >= j:
            ans += 1
            i = i - j
            j += 1
            
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())
k=n
dic={}
import math
for i in range(2,math.floor(math.sqrt(n))+1):
  for j in range(10**10):
    if n%i==0:
      if i in dic:
        dic[i]+=1
      else:
        dic[i]=1
      n=n//i
    else:
      break

count=0
if k!=1:
  if not dic:
    count+=1
  if n!=1:
    count+=1
print(dic)
while dic:
  keep=1
  x=dic.popitem()
  num=x[1]
  for i in range(1000):
    if num>=keep:
      num-=keep
      count+=1
      keep+=1
    else:
      break
print(count)


n=int(input())
k=n
dic={}
import math
for i in range(2,math.floor(math.sqrt(n))):
  for j in range(10**10):
    if n%i==0:
      if i in dic:
        dic[i]+=1
      else:
        dic[i]=1
      n=n//i
    else:
      break

count=0
if k!=1:
  if not dic or n!=1:
    count+=1

while dic:
  keep=1
  x=dic.popitem()
  num=x[1]
  for i in range(1000):
    if num>=keep:
      num-=keep
      count+=1
      keep+=1
    else:
      break
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 get_prime_fact(n):
    primes = []
    append = primes.append
    while n % 2 == 0:
        append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        append(n)
    return primes


def get_knu(a):
    n = 2
    while True:
        knum = (n * (n + 1)) // 2
        if a >= knum:
            n += 1
        else:
            return n - 1


D = int(input())
p = get_prime_fact(D)

if len(p) != len(set(p)):
    res = 0
    for n in set(p):
        count = p.count(n)
        if count == 1:
            res += 1
        else:
            add = get_knu(count)
            res += add

            print(n, add)

    print(res)
else:
    print(len(p))

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


def get_knu(a):
    before = 1
    n = 2
    while True:
        knum = (n * (n + 1)) // 2
        if a == knum:
            return n
        elif knum > a > before:
            return n - 1
        else:
            n += 1
            before = knum


D = int(input())
p = get_prime_fact(D)

if len(p) != len(set(p)):
    res = 0
    for n in set(p):
        count = p.count(n)
        if count == 1:
            res += 1
        else:
            add = get_knu(count)
            res += add

    print(res)
else:
    print(len(p))

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):

# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A&lang=ja

import math
from collections import Counter
n = int(input())
origin = n
ans = []

i = 2
while i <= math.sqrt(n):
    if n % i == 0:
        n //= i
        ans.append(str(i))
    else:
        i += 1

if n != 1:
    ans.append(str(n))
# print(Counter(ans))
f_ans = 0
for x, y in Counter(ans).items():
    cnt = 0
    for i in range(1, y):
        cnt += i
        if cnt == y:
            f_ans += i
            break
        if cnt > y:
            f_ans += (i - 1)
            break


print(f_ans)




# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A&lang=ja

import math
from collections import Counter
n = int(input())
origin = n
ans = []

i = 2
while i <= math.sqrt(n):
    if n % i == 0:
        n //= i
        ans.append(str(i))
    else:
        i += 1

if n != 1:
    ans.append(str(n))
# print(Counter(ans))
f_ans = 0
for x, y in Counter(ans).items():
    cnt = 0
    for i in range(1, y):
        cnt += i
        if cnt == y:
            f_ans += i
            break
        if cnt > y:
            f_ans += (i - 1)
        

print(f_ans)




# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A&lang=ja

import math
from collections import Counter
n = int(input())
origin = n
ans = []

i = 2
while i <= math.sqrt(n):
    if n % i == 0:
        n //= i
        ans.append(str(i))
    else:
        i += 1

if n != 1:
    ans.append(str(n))
# print(Counter(ans))
f_ans = 0
for x, y in Counter(ans).items():
    cnt = 0
    for i in range(1, y+1):
        cnt += i
        if cnt == y:
            f_ans += i
            break
        if cnt > y:
            f_ans += (i - 1)
            break


print(f_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):



import math
N = int(input())
s = {}

for i in range(2, math.ceil(math.sqrt(N))):
    mod = N % i
    while mod == 0:
        s[i] = s.get(i, 0) + 1
        N = N // i

        if N == 0:
            break

        mod = N % i
if not N == 1:
    s[N] = s.get(N, 0) + 1

count = 0
for prime in s:
    pow = s[prime]

    n = 1
    while pow >= n:
        if pow == n*(n+1)/2:
            count += n
            break
        if pow < n*(n+1)/2:
            count += n-1
            break
        n += 1

print(s)
print(count)





import math
N = int(input())
s = {}

for i in range(2, math.ceil(math.sqrt(N))):
    mod = N % i
    while mod == 0:
        s[i] = s.get(i, 0) + 1
        N = N // i

        if N == 0:
            break

        mod = N % i
if not N == 1:
    s[N] = s.get(N, 0) + 1

count = 0

for prime in s:
    pow = s[prime]

    if pow == 1:
        count += 1
        continue

    for i in range(1, pow+1):
        pow -= i
        if pow < 0:
            count += i - 1
            break

        if pow == 0:
            count += i
            break

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):
import sys
sys.setrecursionlimit(10 ** 9)
a = int(input())
count = 0
#used = []

#  if a%first == 0:
#    #used.append(first)
#    a /= first
#    count += 1
#  if a==1:
#    return count
#  else:
#    div_battle(a,first+1)
count = 0
first = a
while a!=1:
  if a % first == 0:
    a /= first
    count += 1
    first -= 1
print(count)

from collections import Counter

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

def get_count(c):
  ans = 0
  _max = c
  for i in range(1,c+1):
    if i <= _max:
      ans += 1
      _max -= i
    else:
      break
  return ans
a = int(input())
counter = Counter(prime_factor(a))
ans = 0
for i in counter:
  ans += get_count(counter[i])
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])
    if arr==[]:
        arr.append([n, 1])
    return arr

n=int(input())
x=factorization(n)
if n=1:
  print(0)
  exit()
  
ans=0
for i in range(len(x)):
  p=x[i][1]
  j=1
  while True:
    if p>=j:
      p-=j
      ans+=1
      j+=1
    if p<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

n=int(input())
x=factorization(n)
ans=0

if n==1:
  print(0)
  exit()
for i in range(len(x)):
  p=x[i][1]
  j=1
  while True:
    if p>=j:
      p-=j
      ans+=1
      j+=1
    if p<j:
      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):
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

p = [1,1,2,2,2,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15]

n = int(input())
if n == 0:
    print(0)
else:
    ans = 0
    print(factorization(n))
    for j in factorization(n):
        ans += p[j[1]-1]

    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

p = [1,1,2,2,2,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15]

n = int(input())
if n == 1:
    print(0)
else:
    ans = 0
    for j in factorization(n):
        ans += p[j[1]-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())
pri_cnt = {}
n = N
i = 2

while i ** 2 <= N
  while n % i == 0:
    if i not in div_cnt:
      pri_cnt[i] = 1
    else:
      pri_cnt[i] += 1
    
    n = n / i

if n > 1:
  pri_cnt[N] = 1

ans = 0
for pri, cnt in pri_knt:
  count = 1
  remain = cnt
  while remain > count:
    ans += 1
    remain -= count
    count += 1
print(ans)

N = int(input())
pri_cnt = {}
n = N
i = 2

while i ** 2 <= N:
  while n % i == 0:
    if i not in pri_cnt:
      pri_cnt[i] = 1
    else:
      pri_cnt[i] += 1
    n = n / i
  if (i == 2):
    i += 1
  else:
    i += 2

if n > 1:
  pri_cnt[int(n)] = 1

ans = 0
for pri, cnt in pri_cnt.items():
  count = 1
  remain = cnt
  while remain >= count:
    ans += 1
    remain -= count
    count += 1
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())
d = n
m = {}

for i in range(2, int(math.sqrt(n)) + 1):
    m[i] = 0
    while (d % i == 0):
        d //= i
        m[i]+=1
res = 0
print(m)
for i in m.values():
    res+=int(math.sqrt(i*2))
if d > 1:
    res += 1
print(res)


import math

n = int(input())
d = n
m = {}

for i in range(2, int(math.sqrt(n)) + 1):
    m[i] = 0
    while (d % i == 0):
        d //= i
        m[i]+=1
res = 0
for i in m.values():
    res+=int(math.sqrt(i*2+0.25)-0.5)

if d > 1:
    res += 1
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 numpy as np
import sys
sys.setrecursionlimit(10**6)

n=int(input())

m=int(np.sqrt(n))+1

if n==1 or n==0:
    print(0)
    exit()

ans=0
for i in range(2,m):
    cnt=0
    for j in range(44):
        if n%i==0:
            n=n/i
            print("i=",i,"n=",n)
            cnt+=1
        else:
            break
    if 1<=cnt<3:
        ans+=1
    if 3<=cnt<6:
        ans+=2
    if 6<=cnt<10:
        ans+=3
    if 10<=cnt<15:
        ans+=4
    if 15<=cnt<21:
        ans+=5
    if 21<=cnt<28:
        ans+=6
    if 28<=cnt<36:
        ans+=7
    if 36<=cnt<45:
        ans+=8

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

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

print(ans)        


import numpy as np
import sys
sys.setrecursionlimit(10**6)

n=int(input())

m=int(np.sqrt(n))+1

if n==1 or n==0:
    print(0)
    exit()

ans=0
for i in range(2,m):
    cnt=0
    for j in range(44):
        if n%i==0:
            n=n/i
            print("i=",i,"n=",n)
            cnt+=1
        else:
            break
    if 1<=cnt<3:
        ans+=1
    if 3<=cnt<6:
        ans+=2
    if 6<=cnt<10:
        ans+=3
    if 10<=cnt<15:
        ans+=4
    if 15<=cnt<21:
        ans+=5
    if 21<=cnt<28:
        ans+=6
    if 28<=cnt<36:
        ans+=7
    if 36<=cnt<45:
        ans+=8

if n!=1:
    ans+=1

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

print(ans)        


import numpy as np
import sys
sys.setrecursionlimit(10**6)

n=int(input())

m=int(np.sqrt(n))+1

if n==1 or n==0:
    print(0)
    exit()

ans=0
for i in range(2,m):
    cnt=0
    for j in range(44):
        if n%i==0:
            n=n/i
#            print("i=",i,"n=",n)
            cnt+=1
        else:
            break
    if 1<=cnt<3:
        ans+=1
    if 3<=cnt<6:
        ans+=2
    if 6<=cnt<10:
        ans+=3
    if 10<=cnt<15:
        ans+=4
    if 15<=cnt<21:
        ans+=5
    if 21<=cnt<28:
        ans+=6
    if 28<=cnt<36:
        ans+=7
    if 36<=cnt<45:
        ans+=8

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

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

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):
N = int(input())


def factorize_dict(n):
    b = 2
    fct = dict()
    while b * b <= n:
        while n % b == 0:
            n //= b
            #fct.append(b)
            if b not in fct:
                fct[b] = 1
            else:
                fct[b] += 1
        b = b + 1
    if n > 1:
        #fct.append(n)
        if b not in fct:
            fct[n] = 1
        else:
            fct[n] += 1
            
    return fct

factorized_N = factorize_dict(N)
print(factorized_N)

z_list = []

for f_dict in factorized_N.items():
    prime_number = f_dict[0]
    how_many = f_dict[1]
    z_list += [prime_number**i for i in range(1,how_many+1)]
    
z_list.sort(reverse=False)
print(z_list)

ans = 0
for z in z_list:
    if N%z == 0:
        ans += 1
        N /= z
        if N == 1:
            break
        
print(ans)    

N = int(input())


def factorize_dict(n):
    b = 2
    fct = dict()
    while b * b <= n:
        while n % b == 0:
            n //= b
            #fct.append(b)
            if b not in fct:
                fct[b] = 1
            else:
                fct[b] += 1
        b = b + 1
    if n > 1:
        #fct.append(n)
        if n not in fct:
            fct[n] = 1
        else:
            fct[n] += 1
            
    return fct

factorized_N = factorize_dict(N)
print(factorized_N)

z_list = []

for f_dict in factorized_N.items():
    prime_number = f_dict[0]
    how_many = f_dict[1]
    z_list += [prime_number**i for i in range(1,how_many+1)]
    
z_list.sort(reverse=False)
print(z_list)

ans = 0
for z in z_list:
    if N%z == 0:
        ans += 1
        N /= z
        if N == 1:
            break
        
print(ans)    

N = int(input())


def factorize_dict(n):
    b = 2
    fct = dict()
    while b * b <= n:
        while n % b == 0:
            n //= b
            #fct.append(b)
            if b not in fct:
                fct[b] = 1
            else:
                fct[b] += 1
        b = b + 1
    if n > 1:
        #fct.append(n)
        if n not in fct:
            fct[n] = 1
        else:
            fct[n] += 1
            
    return fct

factorized_N = factorize_dict(N)
#print(factorized_N)

z_list = []

for f_dict in factorized_N.items():
    prime_number = f_dict[0]
    how_many = f_dict[1]
    z_list += [prime_number**i for i in range(1,how_many+1)]
    
z_list.sort(reverse=False)
#print(z_list)

ans = 0
for z in z_list:
    if N%z == 0:
        ans += 1
        N /= z
        if N == 1:
            break
        
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):
N = int(input())
if N == 1:
    print("0")
    exit()

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

import collections

c = collections.Counter(prime_factorize(N))

ans = 0
for i in c.keys():
    num = c[i]
    m = 1
    while num > 1:
        num = num - m
        m += 1
        ans += 1

print(ans)

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

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

import collections

c = collections.Counter(prime_factorize(N))

ans = 0
for i in c.keys():
    num = c[i]
    m = 2
    while num > 0:
        num = num - m
        m += 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):
n=int(input())
if n == 1:
    print(0)
    exit()

def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)
    divisors.sort()
    return divisors
divisors = make_divisors(n)
if divisors == [1]:
    print(1)
    exit()
ans = 0
for i in divisors:
    if i == 1:
        continue
    j = i
    c = 0
    while n%j == 0:
        n = n //j
        j += i
        c += 1
        d = 0
    for count in range(1,1000):
        if c >= d+count:
            ans += 1
            d+=count
        else:
            break
print(ans)

n = int(input())
if n == 1:
    print(0)
    exit()

def primes_factraizar(n):
    rn = int(n**0.5)
    is_prime = [True]*(rn+1)
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2,rn+1):
        if not is_prime[i]:
            continue
        for j in range(i*2,rn+1,i):
            is_prime[j] = False
        return [i for i in range(rn+1) if is_prime[i]]
primes = primes_factraizar(n)

divisors = []
for i in primes:
    if n % i == 0:
        divisors.append(i)

divisors.sort()

if divisors == []:
    print(1)
    exit()
ans = 0

for i in divisors:
    if i == 1:
        continue

    c = 0

    while n%i == 0:
        n = n // i
        c += 1
    
    d=0

    for count in range(1,1000):
        if c >= d+count:
            ans += 1
            d+=count
        else:
            break
if not 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):
import math
N = int(input())

def furui(n):
    if n < 2:
        return []
    plist = [1 for i in range(n)]
    plist[0] = plist[1] = 0
    for i in range(2, n):
        if plist[i]:
            for j in range(i*2, n, i):
                plist[j] = 0
    ret = []
    for i in range(n):
        if plist[i] != 0:
            ret.append(i)
    return ret


def isprime(n):
    if n == 1:
        return False
    if n % 2 == 0 and n > 2:
        return False
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        if n % i == 0:
            return False
    return True


primes = furui(1000000)
count = 0

if N == 1:
    print(0)
else:
    for p in primes:
        used = set([])
        e = 1
        z = p ** e
        while N >= z:
            if e not in used and N % z == 0:
                N //= z
                count += 1
                used.add(e)
                e = 1
                # print(z, N)
            else:
                e += 1
            z = p ** e
    if N not in primes:
        count += 1
    print(count)


def furui(n):
    if n < 2:
        return []
    plist = [1 for i in range(n)]
    plist[0] = plist[1] = 0
    for i in range(2, n):
        if plist[i]:
            for j in range(i*2, n, i):
                plist[j] = 0
    ret = []
    for i in range(n):
        if plist[i] != 0:
            ret.append(i)
    return ret


primes = furui(1000000)
count = 0
N = int(input())


if N == 1:
    print(0)
else:
    
    factors = []
    for p in primes:
        if N % p == 0:
            e = 1
            while N % (p ** (e+1)) == 0:
                e += 1
            factors.append((p, e))
            N /= p ** e
    if N != 1:
        
        factors.append((N, 1))
    # print(factors)

    
    count = 0
    for p, e in factors:
        s = 0
        for i in range(1, e+1):
            s += i
            if s <= e:
                # print(p ** i)
                count += 1
            else:
                break
    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):
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
a = prime_factorize(n)
b = list(set(a))
total = 0
for j in range(len(b)):
    c  = len([i for i, x in enumerate(a) if x == list(set(a))[j]])
    y = 1
    while c>0:
        c -= y
        y += 1
        if c>=0:
            total +=1
 

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
a = prime_factorize(n)
b = list(set(a))
total = 0
for j in range(len(b)):
    c  = len([i for i, x in enumerate(a) if x == list(set(a))[j]])
    y = 1
    while c>0:
        c -= y
        y += 1
        if c>=0:
            total +=1
print(total)

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 factorize(n):
    
    res = {}
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp % i == 0:
            res[i] = 0
            while temp % i ==0:
                res[i] += 1
                temp //= i

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

    if len(res) == 0:
        res[n] = 1

    return res

def factorize_exp(n):
    """
    Return only exponents of factorization.
    :return: e.g. factorize(24) -> [3, 1] (24 = 2**3 * 3**1)
    """
    res = []
    temp = n

    if n == 1:
        
        return res

    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
            res.append(cnt)
    if temp != 1:
        res.append(1)

    if len(res) == 0:
        res.append(1)

    return res


n = int(input())
factors = factorize_exp(n)
cnt = 0
print(factors)

for i in factors:
    rem = i
    j = 1
    while rem >= j:
        rem -= j
        cnt += 1
        j += 1

print(cnt)

def factorize(n):
    
    res = {}
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp % i == 0:
            res[i] = 0
            while temp % i ==0:
                res[i] += 1
                temp //= i

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

    if len(res) == 0:
        res[n] = 1

    return res

def factorize_exp(n):
    """
    Return only exponents of factorization.
    :return: e.g. factorize(24) -> [3, 1] (24 = 2**3 * 3**1)
    """
    res = []
    temp = n

    if n == 1:
        
        return res

    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
            res.append(cnt)
    if temp != 1:
        res.append(1)

    if len(res) == 0:
        res.append(1)

    return res


n = int(input())
factors = factorize_exp(n)
cnt = 0

for i in factors:
    rem = i
    j = 1
    while rem >= j:
        rem -= j
        cnt += 1
        j += 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

N = int(input())

fact = set()
counter = 0

for i in range(2, int(math.sqrt(N))+1):
    if N % i != 0:
        continue
    powt = 0
    while N % i == 0:
        powt += 1
        N /= i
    fact.add((i, powt))

if N != 1:
    fact.add((N, 1))

for base, e in fact:
    for t in range(1, e+1):
        if t <= e:
            e -= t
            counter += 1
print(fact)
print(counter)


import math

N = int(input())

fact = set()
counter = 0

for i in range(2, int(math.sqrt(N))+1):
    if N % i != 0:
        continue
    powt = 0
    while N % i == 0:
        powt += 1
        N /= i
    fact.add((i, powt))

if N != 1:
    fact.add((N, 1))

for base, e in fact:
    for t in range(1, e+1):
        if t <= e:
            e -= t
            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

def f(n):
    arr = list()
    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(tuple(i, cnt))

    if temp != 1:
        tup = tuple(temp, 1)
        arr.append(tup)

    if len(arr) == 0:
        tup = tuple(n, 1)
        arr.append(tup)

    return arr

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

t = f(N)
res = 0
A = [0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6]
for a in t:
    res += int(A[a[1]])
print(res)


import math

def f(n):
    arr = list()
    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(tuple([i, cnt]))

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

    if len(arr) == 0:
        tup = tuple([n, 1])
        arr.append(tup)

    return arr

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

t = f(N)
res = 0

for a in t:
    cnt = 0
    c = 0
    while cnt <= a[1]:
        c += 1
        cnt += c
    res += c - 1
    
print(res)


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 numpy as np
from functools import *
import sys
sys.setrecursionlimit(100000)
input = sys.stdin.readline


import numpy as np

def acinput():
    return list(map(int, input().split(" ")))

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



s=factorize(float(input()))

res=0

for ss,n in s:
    tmp=np.floor((np.sqrt(8*n+1)-1)/2)
    print(tmp)
    res+=tmp

print(int(np.floor(res)))


import numpy as np
from functools import *
import sys
sys.setrecursionlimit(100000)
input = sys.stdin.readline


import numpy as np

def acinput():
    return list(map(int, input().split(" ")))

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



s=factorize(float(input()))

res=0

for ss,n in s:
    tmp=np.floor((np.sqrt(8*n+1)-1)/2)
    #print(tmp)
    res+=tmp

print(int(np.floor(res)))

#print(s)

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 trial_division(n):
    
    factor = []
    
    tmp = int(math.sqrt(n)) + 1
    for num in range(2,tmp):
        while n % num == 0:
            n //= num
            factor.append(num)
    
    if not factor:
        return 'prime number'
    else:
        factor.append(n)
        return factor
      
n = int(input())
n_list = trial_division(n)
ans = 0
z = []
for i in range(len(n_list)):
  if n_list[i] == 1:
    break
  else:
    if n % n_list[i] == 0:
      z.append(n_list[i])
      n = n / n_list[i]
      ans += 1
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
      
n = int(input())
n_list = factorization(n)
oh = []
for i in range(len(n_list)):
  num = n_list[i][0] 
  kata = n_list[i][1] 
  for i in range(1, kata+1):
    oh.append(num ** i)
oh.sort()
ans = 0
z = []
for i in range(len(oh)):
  if oh[i] == 1:
    break
  else:
    if n % oh[i] == 0 and oh[i] not in z:
      z.append(oh[i])
      n = n / oh[i]
      ans += 1
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):
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(N) 

count = 0
for i in A:
  judge = 1
  temp = i[1]
  while temp >= judge:
    count += 1
    temp -=judge
    judge += 1

if N == 1:
  count = 0

print(count)

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())
A  = factorization(N) 

count = 0
for i in A:
  judge = 1
  temp = i[1]
  while temp >= judge:
    count += 1
    temp -=judge
    judge += 1

if N == 1:
  count = 0

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):
a =int(input())
list= []
n = 0
for i in range(2,1000000):
    if (a % i == 0):
        warikaisi = True
        while(warikaisi):
            if a % i == 0:
                list.append(i)
                a = a / i
            else:
                n = n + 1
                warikaisi = False
print(n)

a =int(input())
list = []
b = 1
for i in range(2,44):
    list.append(b)
    b = b + i
n = 0
e = 0
k = 0
for i in range(2,1000000):
    if (a % i == 0):
        warikaisi = True
        while(warikaisi):
            if a % i == 0:
                list.append(i)
                a = a / i
                e = e + 1
                if (e == list[k]):
                    n = n + 1
                    k = k + 1
            else:
                e = 0
                k = 0
                warikaisi = False
if (a >= 1000000):
    n = n + 1
print(n)

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 = 2
    fct = []
    while a <= n:
        print('-------')
        while n % a == 0:
            n = n // a
            fct.append(a)
        a += 1
    if n > 1:
        fct.append(n)
    return fct
 
 
n = int(input())
c = Counter(factorize(n))
# print(factorize(n))
# print(c.values())


ans = 0
for v in c.values():
    t = 1
    while v >= t:
        v -= t
        t += 1
        ans += 1
print(ans)

from collections import Counter
 
def factorize(n):
    a = 2
    fct = []
    while a*a <= n:
        # print('-------')
        while n % a == 0:
            n = n // a
            fct.append(a)
        a += 1
    if n > 1:
        fct.append(n)
    return fct
 
 
n = int(input())
c = Counter(factorize(n))
# print(factorize(n))
# print(c.values())


ans = 0
for v in c.values():
    t = 1
    while v >= t:
        v -= t
        t += 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):
n = int(input())

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

aoinsu = factorize(n)

kaijo = [1]
b = 1
for i in range(1,10):
    b = b+i
    kaijo.append(b)



#print(aoinsu)

ans = 0
for i in range(len(aoinsu)):
    a = aoinsu[i][1]
    for j in range(10):
        if a <= kaijo[j]:
            ans += j
            break


print(ans)

n = int(input())

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

aoinsu = factorize(n)

kaijo = [1]
b = 1
for i in range(2,10):
    b = b+i
    kaijo.append(b)



#print(aoinsu)

ans = 0
for i in range(len(aoinsu)):
    a = aoinsu[i][1]
    for j in range(10):
        if a < kaijo[j]:
            ans += j
            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())
if N == 1:
    print(0)
else:
    result = 0
    for i in range(len(factorization(N))):
        tempnumber = factorization(N)[i][1]
        plusnumber = 1
        while tempnumber > 0:
            tempnumber -= plusnumber
            result += 1
            plusnumber += 1
    print(result)

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:
    result = 0
    for i in range(len(factorization(N))):
        tempnumber = factorization(N)[i][1]
        plusnumber = 1
        while tempnumber > 0:
            result += 1
            plusnumber += 1
            tempnumber -= plusnumber
    print(result)

Verdicts: Runtime Error

Accepted