SPOJ – FCTRL – Factorial

The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form the areas called cells (this term gave the name to the cellular phone) and every phone connects to the BTS with the strongest signal (in a little-simplified view). Of course, BTSes need some attention and technicians need to check their function periodically.

ACM technicians faced a very interesting problem recently. Given a set of BTSes to visit, they needed to find the shortest path to visit all of the given points and return back to the central company building. Programmers have spent several months studying this problem but with no results. They were unable to find the solution fast enough. After a long time, one of the programmers found this problem in a conference article. Unfortunately, he found that the problem is so-called “Travelling Salesman Problem” and it is very hard to solve. If we have N BTSes to be visited, we can visit them in any order, giving us N! possibilities to examine. The function expressing that number is called factorial and can be computed as a product….N. The number is very high even for a relatively small N.

The programmers understood they had no chance to solve the problem. But because they have already received the research grant from the government, they needed to continue with their studies and produce at least some results. So they started to study the behavior of the factorial function.

For example, they defined the function Z. For any positive integer N, Z(N) is the number of zeros at the end of the decimal form of number N!. They noticed that this function never decreases. If we have two numbers N1<N2, then Z(N1) <= Z(N2). It is because we can never “lose” any trailing zero by multiplying by any positive number. We can only get new and new zeros. The function Z is very interesting, so we need a computer program that can determine its value efficiently.

There is a single positive integer T on the first line of input (equal to about 100000). It stands for the number of numbers to follow. Then there are T lines, each containing exactly one positive integer number N, 1 <= N <= 1000000000.

For every number N, output a single line containing the single non-negative integer Z(N).

Sample Input:


Sample Output:



Approach –

If we observe the problem statement properly we come to the solution that it expects the number of trailing zeros in factorial of n.

Now we can approach it in 2 ways, the first approach is to calculate factorial of a number and then calculate the number of trailing zeros from the result. But for any value greater than 12 we will see integer exceed. So let’s go with the second approach and Find a relation between number and trailing zeros and n. The number of zeros will be directly related to the number of 10’s available in result and 10 will be the result of 5*2. The number of 2’s in the pair can be taken from any even number. So the problem is now reduced to getting the count of 5’s in factorial or Count of 5s in prime factors of n!

find no of 5’s –

  • Take the number that you’ve been given the factorial of.
  • Divide by 5; if you get a decimal, truncate to a whole number.
  • Divide by 52 = 25; if you get a decimal, truncate to a whole number.
  • Divide by 53 = 125; if you get a decimal, truncate to a whole number.
  • Continue with ever-higher powers of 5, until your division results in a number less than 1. Once the division is less than 1, stop.
  • Sum all the whole numbers you got in your divisions. This is the number of trailing zeroes
Trailing 0s or 5s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + floor(n/625) + ....

Solution CODE

Resources –

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