34.py

f=[1]
for i in range(1,10):
    f.append(f[-1]*i)
    
ans=0
for i in range(10,10**7):
    s=0
    for j in str(i):
        s+=f[ord(j)-ord('0')]
    if s==i:
        print("sum",i)
        ans+=i
print(ans)
        
        

import math


def compute():
	# As stated in the problem, 1 = 1! and 2 = 2! are excluded.
	# If a number has at least n >= 8 digits, then even if every digit is 9,
	# n * 9! is still less than the number (which is at least 10^n).
	ans = sum(i for i in range(3, 10000000) if i == factorial_digit_sum(i))
	return str(ans)


def factorial_digit_sum(n):
	result = 0
	while n >= 10000:
		result += FACTORIAL_DIGITS_SUM_WITH_LEADING_ZEROS[n % 10000]
		n //= 10000
	return result + FACTORIAL_DIGITS_SUM_WITHOUT_LEADING_ZEROS[n]

FACTORIAL_DIGITS_SUM_WITHOUT_LEADING_ZEROS = [sum(math.factorial(int(c)) for c in str(i)) for i in range(10000)]
FACTORIAL_DIGITS_SUM_WITH_LEADING_ZEROS = [sum(math.factorial(int(c)) for c in str(i).zfill(4)) for i in range(10000)]


if __name__ == "__main__":
	print(compute())