金声玉亮2.0模型评测报告

问题

Complete the following python code:
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""

答案

模型	模型答案	测试结果
金声玉亮2.0	Sure! Here is the complete Python code for the `factorize` function: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors. >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] # Check for number of 2s in the factorization while n % 2 == 0: factors.append(2) n //= 2 # Check for odd factors starting from 3 for i in range(3, int(n**0.5) + 1, 2): while n % i == 0: factors.append(i) n //= i # If n is a prime number greater than 2 if n > 2: factors.append(n) return factors # Example usage if __name__ == "__main__": import doctest doctest.testmod() ``` This function works by first checking for the number of times 2 can divide the number `n`, then it checks for other prime factors starting from 3 and going up to the square root of `n`. If there is any remaining prime factor greater than 2, it is appended to the list of factors. The function returns a list of prime factors in ascending order, with each factor repeated as many times as it appears in the factorization.	passed
deepseekr1	To solve this problem, we need to factorize a given integer into its prime factors and return these factors in a list from smallest to largest. Each factor should appear as many times as it divides the given integer. ### Approach The approach to solve this problem involves the following steps: 1. Handle the smallest prime factor (2) separately: Since 2 is the smallest prime number and the only even prime, we first check how many times 2 divides the given integer. Each time it divides, we add 2 to the list of factors and divide the integer by 2. 2. Check for odd factors starting from 3: After handling all factors of 2, we check for odd factors starting from 3. For each odd number, we repeatedly divide the integer by this number as long as it is divisible, adding the factor to the list each time. 3. Check for remaining prime factors: If after processing all possible factors up to the square root of the remaining integer, the integer is still greater than 1, it means this remaining integer is a prime factor and should be added to the list. ### Solution Code ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appeares in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] # Handle the factor 2 while n % 2 == 0: factors.append(2) n = n // 2 # Now check for odd factors starting from 3 i = 3 while i * i <= n: while n % i == 0: factors.append(i) n = n // i i += 2 # If remaining n is a prime if n > 1: factors.append(n) return factors ``` ### Explanation 1. Handling Factor 2: The loop continues dividing the integer by 2 as long as it is even. This ensures all factors of 2 are captured first. 2. Odd Factors Check: Starting from 3, we check each odd number up to the square root of the remaining integer. For each odd number, we divide the integer as long as it is divisible by that number. This ensures we capture all occurrences of each prime factor. 3. Remaining Prime Check: After processing all possible factors up to the square root of the remaining integer, if the integer is still greater than 1, it must be a prime number itself and is added to the list. This method efficiently captures all prime factors by reducing the problem size through division and checking only up to the square root of the remaining integer, ensuring optimal performance.	passed
gpto1	```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] divisor = 2 while divisor * divisor <= n: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 if n > 1: factors.append(n) return factors ``` This function uses trial division to find all prime factors of the given number `n`. It iteratively divides `n` by all integers starting from 2 up to the square root of `n`. If `n` is divisible by the current divisor, it appends the divisor to the `factors` list and updates `n` by dividing it by the divisor. If after the loop `n` is greater than 1, it means that `n` itself is a prime number and is appended to the `factors` list. Example Usage: ```python print(factorize(8)) # Output: [2, 2, 2] print(factorize(25)) # Output: [5, 5] print(factorize(70)) # Output: [2, 5, 7] ```	passed
GPT-4o	Certainly! Here is the complete Python code for the `factorize` function: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] divisor = 2 while n > 1: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 return factors ``` This code defines a function that computes the prime factors of a given integer `n`. It starts with the smallest prime number (2) and continues dividing `n` by the current divisor as long as it is divisible. When it is no longer divisible by the current divisor, it increments the divisor and continues the process. The factors are collected in a list and returned.	passed
llama3.1-70B-Instruct	def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appeares in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] i = 2 while i * i <= n: if n % i: i += 1 else: n //= i factors.append(i) if n > 1: factors.append(n) return factors	passed
llama3.1-405B-Instruct	Here is the completed Python code: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appeares in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] i = 2 while i * i <= n: if n % i: i += 1 else: n //= i factors.append(i) if n > 1: factors.append(n) return factors ``` This code uses a simple trial division method	passed
DeepSeek-Coder-V2	Here's the completed Python code for the `factorize` function: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] # Start with the smallest prime number divisor = 2 while n > 1: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 return factors # Test cases print(factorize(8)) # [2, 2, 2] print(factorize(25)) # [5, 5] print(factorize(70)) # [2, 5, 7] ``` This function works by starting with the smallest prime number (2) and dividing the input number `n` by this prime number as long as it is divisible. It then increments the divisor and repeats the process until `n` is reduced to 1. The prime factors are collected in the `factors` list, which is returned at the end.	passed
Qwen2-72B-Instruct	Here is the completed Python code that defines a function `factorize` which factorizes an integer into its prime factors: ```python from typing import List def factorize(n: int) -> List[int]: """Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors """ i = 2 factors = [] while i * i <= n: if n % i: i += 1 else: n //= i factors.append(i) if n > 1: factors.append(n) return factors print(factorize(8)) # [2, 2, 2] print(factorize(25)) # [5, 5] print(factorize(70)) # [2, 5, 7] ``` This function works by testing divisibility of the input number `n` starting from the smallest prime number 2. If `n` is divisible by `i`, then `i` is a factor and is appended to the list of factors. The function continues dividing `n` by `i` as long as it is divisible, ensuring that `i` is listed the correct number of times in the factorization. When `n` is no longer divisible by `i`, the function increments `i` and repeats the process until `i * i > n`. If `n` is greater than 1 after this process, it means that `n` is a prime number and is appended to the list of factors.	passed
baichuan4	To complete the Python code for the `factorize` function, we'll use a straightforward approach to find all prime factors of the given integer `n`. We'll iterate over all numbers from 2 to `n` and divide `n` by the current number as long as it is divisible. Each time we find a divisor, we'll add it to the list of factors. Here's the completed function: ```python from typing import List def factorize(n: int) -> List[int]: """Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors. >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] divisor = 2 while divisor * divisor <= n: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 if n > 1: factors.append(n) return factors # Example usage: print(factorize(8)) # Output: [2, 2, 2] print(factorize(25)) # Output: [5, 5] print(factorize(70)) # Output: [2, 5, 7] ``` In this code, we initialize an empty list called `factors` to store the prime factors. We then start with a divisor of 2 (the smallest prime number) and continue dividing `n` by this divisor as long as the remainder is 0, indicating that the divisor is a factor of `n`. For each successful division, we append the divisor to the `factors` list. After each division, we update `n` by integer division (`//=`) to remove the just-found factor. We increment the divisor and repeat the process until the divisor squared is greater than `n`. This condition ensures that we do not miss any factors and efficiently terminates the loop. If, after the loop, `n` is greater than 1, it means `n` itself is a prime number and is not divisible by any number other than itself and 1. In this case, we append `n` to the `factors` list. Finally, we return the `factors` list, which contains the prime factorization of the original input `n`. The provided examples show how the function can be used and demonstrate its correctness with a few test cases.	passed
doubao	Here's the completed Python code: ```python from typing import List def factorize(n: int) -> List[int]: factors = [] i = 2 while i * i <= n: if n % i: i += 1 else: n //= i factors.append(i) if n > 1: factors.append(n) return factors ```	passed
minimax	Here's the completed Python code for the `factorize` function: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors. >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] # Start with the smallest prime factor divisor = 2 while n >= divisor: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 # Optimization: if divisor squared is greater than n, n is prime if divisor * divisor > n: if n > 1: factors.append(n) break return factors # Example usage: print(factorize(8)) # Output: [2, 2, 2] print(factorize(25)) # Output: [5, 5] print(factorize(70)) # Output: [2, 5, 7] ``` This function works by dividing the number `n` by the smallest possible prime factor starting from 2, and continues to divide `n` by the same factor until it is no longer divisible by that factor. It then moves on to the next potential factor and repeats the process. When the divisor squared is greater than `n`, it means that `n` is a prime number itself and should be added to the factors list. The function returns the list of factors, each included the number of times it divides `n`.	passed
kimi	Here's the completed Python function that factorizes an integer into its prime factors: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ i = 2 factors = [] while i * i <= n: if n % i: i += 1 else: n //= i factors.append(i) if n > 1: factors.append(n) return factors # Example usage: print(factorize(8)) # Output: [2, 2, 2] print(factorize(25)) # Output: [5, 5] print(factorize(70)) # Output: [2, 5, 7] ``` This function works by checking divisibility starting from the smallest prime number (2) and continues to divide the number by the current factor until it is no longer divisible. Then it moves on to the next integer. If the number is still greater than 1 after the loop, it means that `n` is a prime number itself and should be added to the list of factors.	passed
wenxinyiyan	To complete the given Python code for the `factorize` function, you can use the following implementation: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appeares in factorization. Input number should be equal to the product of all factors """ factors = [] d = 2 while d * d <= n: while (n % d) == 0: factors.append(d) n //= d d += 1 if n > 1: factors.append(n) return factors ``` This implementation uses a simple algorithm to find the prime factors of a given integer `n`. It starts with a divisor `d` initialized to 2 and iterates until `d * d` is less than or equal to `n`. Within each iteration, it checks if `n` is divisible by `d` using the modulo operator `%`. If it is divisible, `d` is appended to the `factors` list, and `n` is divided by `d` using integer division `//`. This process continues until `n` is no longer divisible by `d`, and then `d` is incremented by 1. Finally, if `n` is greater than 1 after the loop, it means `n` is a prime factor itself, so it is appended to the `factors` list. The resulting list of prime factors is returned.	passed
zhipu	Certainly! Below is the completed Python code for the `factorize` function that returns the list of prime factors of a given integer: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors. >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] # Start with the smallest prime factor divisor = 2 while n > 1: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 # Optimization: if n is not divisible by 2, it won't be divisible by any other even number if divisor > 2 and divisor % 2 == 0: divisor += 1 return factors # Example usage: if __name__ == "__main__": import doctest doctest.testmod() ``` ### Explanation: 1. Initialization: - `factors`: An empty list to store the prime factors. - `divisor`: Starts at 2, the smallest prime number. 2. Main Loop: - The outer `while` loop continues until `n` is reduced to 1. - The inner `while` loop checks if `n` is divisible by the current `divisor`. If it is, the `divisor` is appended to the `factors` list, and `n` is divided by the `divisor`. 3. Divisor Increment: - After checking the current `divisor`, increment it by 1. - An optimization step skips even numbers greater than 2, as they cannot be prime factors (except for 2 itself). 4. Return: - The list of prime factors is returned. ### Testing: - The `if __name__ == "__main__":` block allows the code to run doctest, which tests the function using the examples provided in the docstring. This code should work correctly for the given examples and similar cases. You can run the script to verify the correctness of the function using the doctest module.	passed
gpto1	```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appears in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] divisor = 2 while divisor * divisor <= n: while n % divisor == 0: factors.append(divisor) n //= divisor divisor += 1 if n > 1: factors.append(n) return factors ``` This function uses trial division to find all prime factors of the given number `n`. It iteratively divides `n` by all integers starting from 2 up to the square root of `n`. If `n` is divisible by the current divisor, it appends the divisor to the `factors` list and updates `n` by dividing it by the divisor. If after the loop `n` is greater than 1, it means that `n` itself is a prime number and is appended to the `factors` list. Example Usage: ```python print(factorize(8)) # Output: [2, 2, 2] print(factorize(25)) # Output: [5, 5] print(factorize(70)) # Output: [2, 5, 7] ```	passed
deepseekv3	Here's the completed `factorize` function: ```python from typing import List def factorize(n: int) -> List[int]: """ Return list of prime factors of given integer in the order from smallest to largest. Each of the factors should be listed number of times corresponding to how many times it appeares in factorization. Input number should be equal to the product of all factors >>> factorize(8) [2, 2, 2] >>> factorize(25) [5, 5] >>> factorize(70) [2, 5, 7] """ factors = [] # Start with the smallest prime number divisor = 2 while divisor * divisor <= n: while n % divisor == 0: factors.append(divisor) n = n // divisor divisor += 1 if n > 1: factors.append(n) return factors # Test cases print(factorize(8)) # Output: [2, 2, 2] print(factorize(25)) # Output: [5, 5] print(factorize(70)) # Output: [2, 5, 7] ``` ### Explanation: 1. Initialization: Start with the smallest prime number, `2`. 2. Factorization Loop: - While `divisor * divisor <= n`, check if `n` is divisible by `divisor`. - If it is, add `divisor` to the `factors` list and divide `n` by `divisor`. - Repeat this until `n` is no longer divisible by `divisor`. - Increment `divisor` and repeat the process. 3. Final Check: If after the loop `n` is still greater than `1`, it means `n` itself is a prime factor and should be added to the list. 4. Return: The list of factors is returned, which contains all prime factors in ascending order, repeated according to their multiplicity. ### Example Walkthrough: - For `n = 8`: - 8 is divisible by 2, so add 2 to the list and divide 8 by 2, resulting in 4. - 4 is divisible by 2, so add another 2 and divide 4 by 2, resulting in 2. - 2 is divisible by 2, so add another 2 and divide 2 by 2, resulting in 1. - The final list is `[2, 2, 2]`. - For `n = 70`: - 70 is divisible by 2, so add 2 and divide 70 by 2, resulting in 35. - 35 is divisible by 5, so add 5 and divide 35 by 5, resulting in 7. - 7 is a prime number, so add it to the list. - The final list is `[2, 5, 7]`. This function should work correctly for any positive integer input.	passed

模型

模型答案

测试结果

金声玉亮2.0

Sure! Here is the complete Python code for the `factorize` function:

```python
from typing import List

def factorize(n: int) -> List[int]:
"""
Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors.

>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
# Check for number of 2s in the factorization
while n % 2 == 0:
factors.append(2)
n //= 2

# Check for odd factors starting from 3
for i in range(3, int(n**0.5) + 1, 2):
while n % i == 0:
factors.append(i)
n //= i

# If n is a prime number greater than 2
if n > 2:
factors.append(n)

return factors

# Example usage
if __name__ == "__main__":
import doctest
doctest.testmod()
```

This function works by first checking for the number of times 2 can divide the number `n`, then it checks for other prime factors starting from 3 and going up to the square root of `n`. If there is any remaining prime factor greater than 2, it is appended to the list of factors. The function returns a list of prime factors in ascending order, with each factor repeated as many times as it appears in the factorization.

passed

deepseekr1

To solve this problem, we need to factorize a given integer into its prime factors and return these factors in a list from smallest to largest. Each factor should appear as many times as it divides the given integer.

### Approach
The approach to solve this problem involves the following steps:
1. **Handle the smallest prime factor (2) separately**: Since 2 is the smallest prime number and the only even prime, we first check how many times 2 divides the given integer. Each time it divides, we add 2 to the list of factors and divide the integer by 2.
2. **Check for odd factors starting from 3**: After handling all factors of 2, we check for odd factors starting from 3. For each odd number, we repeatedly divide the integer by this number as long as it is divisible, adding the factor to the list each time.
3. **Check for remaining prime factors**: If after processing all possible factors up to the square root of the remaining integer, the integer is still greater than 1, it means this remaining integer is a prime factor and should be added to the list.

### Solution Code
```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
# Handle the factor 2
while n % 2 == 0:
factors.append(2)
n = n // 2
# Now check for odd factors starting from 3
i = 3
while i * i <= n:
while n % i == 0:
factors.append(i)
n = n // i
i += 2
# If remaining n is a prime
if n > 1:
factors.append(n)
return factors
```

### Explanation
1. **Handling Factor 2**: The loop continues dividing the integer by 2 as long as it is even. This ensures all factors of 2 are captured first.
2. **Odd Factors Check**: Starting from 3, we check each odd number up to the square root of the remaining integer. For each odd number, we divide the integer as long as it is divisible by that number. This ensures we capture all occurrences of each prime factor.
3. **Remaining Prime Check**: After processing all possible factors up to the square root of the remaining integer, if the integer is still greater than 1, it must be a prime number itself and is added to the list.

This method efficiently captures all prime factors by reducing the problem size through division and checking only up to the square root of the remaining integer, ensuring optimal performance.

passed

gpto1

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
divisor = 2
while divisor * divisor <= n:
while n % divisor == 0:
factors.append(divisor)
n //= divisor
divisor += 1
if n > 1:
factors.append(n)
return factors
```

This function uses trial division to find all prime factors of the given number `n`. It iteratively divides `n` by all integers starting from 2 up to the square root of `n`. If `n` is divisible by the current divisor, it appends the divisor to the `factors` list and updates `n` by dividing it by the divisor. If after the loop `n` is greater than 1, it means that `n` itself is a prime number and is appended to the `factors` list.

**Example Usage:**

```python
print(factorize(8)) # Output: [2, 2, 2]
print(factorize(25)) # Output: [5, 5]
print(factorize(70)) # Output: [2, 5, 7]
```

passed

GPT-4o

Certainly! Here is the complete Python code for the `factorize` function:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
divisor = 2

while n > 1:
while n % divisor == 0:
factors.append(divisor)
n //= divisor
divisor += 1

return factors
```

This code defines a function that computes the prime factors of a given integer `n`. It starts with the smallest prime number (2) and continues dividing `n` by the current divisor as long as it is divisible. When it is no longer divisible by the current divisor, it increments the divisor and continues the process. The factors are collected in a list and returned.

passed

llama3.1-70B-Instruct

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors

passed

llama3.1-405B-Instruct

Here is the completed Python code:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
```

This code uses a simple trial division method

passed

DeepSeek-Coder-V2

Here's the completed Python code for the `factorize` function:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
# Start with the smallest prime number
divisor = 2
while n > 1:
while n % divisor == 0:
factors.append(divisor)
n //= divisor
divisor += 1
return factors

# Test cases
print(factorize(8)) # [2, 2, 2]
print(factorize(25)) # [5, 5]
print(factorize(70)) # [2, 5, 7]
```

This function works by starting with the smallest prime number (2) and dividing the input number `n` by this prime number as long as it is divisible. It then increments the divisor and repeats the process until `n` is reduced to 1. The prime factors are collected in the `factors` list, which is returned at the end.

passed

Qwen2-72B-Instruct

Here is the completed Python code that defines a function `factorize` which factorizes an integer into its prime factors:

```python
from typing import List

def factorize(n: int) -> List[int]:
"""Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors
"""
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors

print(factorize(8)) # [2, 2, 2]
print(factorize(25)) # [5, 5]
print(factorize(70)) # [2, 5, 7]
```

This function works by testing divisibility of the input number `n` starting from the smallest prime number 2. If `n` is divisible by `i`, then `i` is a factor and is appended to the list of factors. The function continues dividing `n` by `i` as long as it is divisible, ensuring that `i` is listed the correct number of times in the factorization. When `n` is no longer divisible by `i`, the function increments `i` and repeats the process until `i * i > n`. If `n` is greater than 1 after this process, it means that `n` is a prime number and is appended to the list of factors.

passed

baichuan4

To complete the Python code for the `factorize` function, we'll use a straightforward approach to find all prime factors of the given integer `n`. We'll iterate over all numbers from 2 to `n` and divide `n` by the current number as long as it is divisible. Each time we find a divisor, we'll add it to the list of factors.

Here's the completed function:

```python
from typing import List

def factorize(n: int) -> List[int]:
"""Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors.
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
divisor = 2
while divisor * divisor <= n:
while n % divisor == 0:
factors.append(divisor)
n //= divisor
divisor += 1
if n > 1:
factors.append(n)
return factors

# Example usage:
print(factorize(8)) # Output: [2, 2, 2]
print(factorize(25)) # Output: [5, 5]
print(factorize(70)) # Output: [2, 5, 7]
```

In this code, we initialize an empty list called `factors` to store the prime factors. We then start with a divisor of 2 (the smallest prime number) and continue dividing `n` by this divisor as long as the remainder is 0, indicating that the divisor is a factor of `n`. For each successful division, we append the divisor to the `factors` list.

After each division, we update `n` by integer division (`//=`) to remove the just-found factor. We increment the divisor and repeat the process until the divisor squared is greater than `n`. This condition ensures that we do not miss any factors and efficiently terminates the loop.

If, after the loop, `n` is greater than 1, it means `n` itself is a prime number and is not divisible by any number other than itself and 1. In this case, we append `n` to the `factors` list.

Finally, we return the `factors` list, which contains the prime factorization of the original input `n`. The provided examples show how the function can be used and demonstrate its correctness with a few test cases.

passed

doubao

Here's the completed Python code:

```python
from typing import List

def factorize(n: int) -> List[int]:
factors = []
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
```

passed

minimax

Here's the completed Python code for the `factorize` function:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors.
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
# Start with the smallest prime factor
divisor = 2
while n >= divisor:
while n % divisor == 0:
factors.append(divisor)
n //= divisor
divisor += 1
# Optimization: if divisor squared is greater than n, n is prime
if divisor * divisor > n:
if n > 1:
factors.append(n)
break
return factors

# Example usage:
print(factorize(8)) # Output: [2, 2, 2]
print(factorize(25)) # Output: [5, 5]
print(factorize(70)) # Output: [2, 5, 7]
```

This function works by dividing the number `n` by the smallest possible prime factor starting from 2, and continues to divide `n` by the same factor until it is no longer divisible by that factor. It then moves on to the next potential factor and repeats the process. When the divisor squared is greater than `n`, it means that `n` is a prime number itself and should be added to the factors list. The function returns the list of factors, each included the number of times it divides `n`.

passed

kimi

Here's the completed Python function that factorizes an integer into its prime factors:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors

# Example usage:
print(factorize(8)) # Output: [2, 2, 2]
print(factorize(25)) # Output: [5, 5]
print(factorize(70)) # Output: [2, 5, 7]
```

This function works by checking divisibility starting from the smallest prime number (2) and continues to divide the number by the current factor until it is no longer divisible. Then it moves on to the next integer. If the number is still greater than 1 after the loop, it means that `n` is a prime number itself and should be added to the list of factors.

passed

wenxinyiyan

To complete the given Python code for the `factorize` function, you can use the following implementation:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
"""
factors = []
d = 2
while d * d <= n:
while (n % d) == 0:
factors.append(d)
n //= d
d += 1
if n > 1:
factors.append(n)
return factors
```
This implementation uses a simple algorithm to find the prime factors of a given integer `n`. It starts with a divisor `d` initialized to 2 and iterates until `d * d` is less than or equal to `n`. Within each iteration, it checks if `n` is divisible by `d` using the modulo operator `%`. If it is divisible, `d` is appended to the `factors` list, and `n` is divided by `d` using integer division `//`. This process continues until `n` is no longer divisible by `d`, and then `d` is incremented by 1. Finally, if `n` is greater than 1 after the loop, it means `n` is a prime factor itself, so it is appended to the `factors` list. The resulting list of prime factors is returned.

passed

zhipu

Certainly! Below is the completed Python code for the `factorize` function that returns the list of prime factors of a given integer:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appears in factorization.
Input number should be equal to the product of all factors.
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
# Start with the smallest prime factor
divisor = 2

while n > 1:
while n % divisor == 0:
factors.append(divisor)
n //= divisor
divisor += 1
# Optimization: if n is not divisible by 2, it won't be divisible by any other even number
if divisor > 2 and divisor % 2 == 0:
divisor += 1

return factors

# Example usage:
if __name__ == "__main__":
import doctest
doctest.testmod()
```

### Explanation:
1. **Initialization**:
- `factors`: An empty list to store the prime factors.
- `divisor`: Starts at 2, the smallest prime number.

2. **Main Loop**:
- The outer `while` loop continues until `n` is reduced to 1.
- The inner `while` loop checks if `n` is divisible by the current `divisor`. If it is, the `divisor` is appended to the `factors` list, and `n` is divided by the `divisor`.

3. **Divisor Increment**:
- After checking the current `divisor`, increment it by 1.
- An optimization step skips even numbers greater than 2, as they cannot be prime factors (except for 2 itself).

4. **Return**:
- The list of prime factors is returned.

### Testing:
- The `if __name__ == "__main__":` block allows the code to run doctest, which tests the function using the examples provided in the docstring.

This code should work correctly for the given examples and similar cases. You can run the script to verify the correctness of the function using the doctest module.

passed

gpto1

passed

deepseekv3

Here's the completed `factorize` function:

```python
from typing import List

def factorize(n: int) -> List[int]:
""" Return list of prime factors of given integer in the order from smallest to largest.
Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
Input number should be equal to the product of all factors
>>> factorize(8)
[2, 2, 2]
>>> factorize(25)
[5, 5]
>>> factorize(70)
[2, 5, 7]
"""
factors = []
# Start with the smallest prime number
divisor = 2
while divisor * divisor <= n:
while n % divisor == 0:
factors.append(divisor)
n = n // divisor
divisor += 1
if n > 1:
factors.append(n)
return factors

# Test cases
print(factorize(8)) # Output: [2, 2, 2]
print(factorize(25)) # Output: [5, 5]
print(factorize(70)) # Output: [2, 5, 7]
```

### Explanation:
1. **Initialization**: Start with the smallest prime number, `2`.
2. **Factorization Loop**:
- While `divisor * divisor <= n`, check if `n` is divisible by `divisor`.
- If it is, add `divisor` to the `factors` list and divide `n` by `divisor`.
- Repeat this until `n` is no longer divisible by `divisor`.
- Increment `divisor` and repeat the process.
3. **Final Check**: If after the loop `n` is still greater than `1`, it means `n` itself is a prime factor and should be added to the list.
4. **Return**: The list of factors is returned, which contains all prime factors in ascending order, repeated according to their multiplicity.

### Example Walkthrough:
- For `n = 8`:
- 8 is divisible by 2, so add 2 to the list and divide 8 by 2, resulting in 4.
- 4 is divisible by 2, so add another 2 and divide 4 by 2, resulting in 2.
- 2 is divisible by 2, so add another 2 and divide 2 by 2, resulting in 1.
- The final list is `[2, 2, 2]`.

- For `n = 70`:
- 70 is divisible by 2, so add 2 and divide 70 by 2, resulting in 35.
- 35 is divisible by 5, so add 5 and divide 35 by 5, resulting in 7.
- 7 is a prime number, so add it to the list.
- The final list is `[2, 5, 7]`.

This function should work correctly for any positive integer input.

passed