Write a program in C that reads in an positive integer n where 1 <= n <=1000 and prints the count of trailing zeros in n!, where n! = n(n - 1)(n -2) ... (2)(1):

Hints: the number of trailing zeros equals to the number of 5s in the prime factorization of n!. For example, 10! = 3628800 = 2834527 which has two trailing zeros.

$ ./trailingzeros Please enter the number: 10

Factorial of 10 has 2 trailing zeros