"""
n! means `n! = n × (n-1) × ... 2 × 1`.
For example, 10! = 10 × 9 × ... × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27 .

Find the sum of the digits in the number 100!.
"""

import math
import time
from functools import reduce


def timer(func):
    def wrapper(*args, **kwargs):
        start_time = time.time()
        result = func(*args, **kwargs)
        end_time = time.time()
        print(f"Execution time: {end_time - start_time:.8f} seconds")
        return result

    return wrapper


def large_multiplication(a: str, b: str) -> str:
    result = [0] * (len(a) + len(b))
    for i in range(len(a) - 1, -1, -1):
        carry = 0
        for j in range(len(b) - 1, -1, -1):
            product = int(a[i]) * int(b[j]) + result[i + j + 1] + carry
            result[i + j + 1] = product % 10
            carry = product // 10
        result[i] += carry
    return "".join(map(str, result)).lstrip("0") or "0"


def factorial(n: int) -> int:
    res = reduce(lambda x, y: x * y, range(1, n + 1))
    return res


@timer
def factorial_digit_sum(n: int) -> int:
    if n >= 1000:
        fact = math.factorial(n)
    else:
        fact = factorial(n)
    return sum(int(digit) for digit in str(fact))


@timer
def large_factorial(n: int) -> int:
    result = "1"
    for i in range(2, n + 1):
        result = large_multiplication(result, str(i))
    print(f"{n}!={result}")
    sum_all = sum(int(digit) for digit in result)
    return sum_all


def main():
    print(factorial_digit_sum(100))
    print(large_factorial(100))


if __name__ == "__main__":
    main()