solutions of problem 21 and problem 20

solutions/0020.FactorialDigitSum/euler_20.py

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+"""
+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()