a good article click

Hey , I am not sure what went wrong with this logic. it is giving time limit error for almost all the cases.https://calcularrfc.com.mx/ Any help would be appreciated

in the statement it is said that n<=1e16, but in the test cases there are values greater than that. maybe what was meant that n has at most 16 digits

Using Binary Search:

def solve(n): def count(x): res = 0 while x: t = x//5 res+=t x = t return res

l,r = 5,5*(10**16)
res = 5
while l<=r:
    mid = (l+r)//2
    if count(mid)>=n:
        res = mid
        r = mid - 1
    else:
        l = mid + 1
return res 

There are floor(n/5) + floor(n/25) + floor(n/125)+... zeroes at the end of n! Since 1/5 + 1/25 + 1/125 + ... = 1/4, we would expect (4n!) to be a very good lower bound for our answer. Also, the number of zeroes in successive values of n! only increase when n is a multiple of five. Using those two facts, we can create a starting value that requires very few checks even when n is very large.

def fact_zeros(n):
    count = 0
    while n:
        count += n//5
        n //= 5
    return count

def solve(n):
    test = ((4*n)//5) * 5
    while fact_zeros(test) < n:
        test += 5
    return test