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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 1.2.3.4....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 behaviour 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.InputThere 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.OutputFor every number N, output a single line containing the single non-negative integer Z(N).ExampleSample Input:63601001024234568735373Sample Output:0142425358612183837

Question

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 1.2.3.4....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 behaviour 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.InputThere 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.OutputFor every number N, output a single line containing the single non-negative integer Z(N).ExampleSample Input:63601001024234568735373Sample Output:0142425358612183837

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Solution

Para resolver el problema de encontrar la cantidad de ceros al final del factorial de un número N N , debemos entender cómo se forman los ceros en un número factorial. Un cero al final de un número se forma por un factor de 10, y un factor de 10 se compone de un factor de 2 y un factor de 5. En un factorial, siempre hay más factores de 2 que de 5, por lo que la cantidad de ceros al final del factorial de N N está determinada por la cantidad de factores de 5 en los números de 1 a N N .

Para calcular Z(N) Z(N) , contamos cuántos múltiplos de 5, 25, 125, etc., hay en N N . Esto se debe a que cada múltiplo de 5 contribuye al menos con un factor de 5, cada múltiplo de 25 contribuye con un factor adicional de 5, y así sucesivamente.

Aquí está el procedimiento paso a paso:

  1. Leer el número de casos de prueba T T .
  2. Para cada número N N :
    • Inicializar una variable Z Z a 0.
    • Para cada potencia de 5 (5, 25, 125, ...), mientras la potencia sea menor o igual a N N :
      • Sumar Npotencia \left\lfloor \frac{N}{\text{potencia}} \right\rfloor a Z Z .
      • Multiplicar la potencia por 5.
    • Imprimir el valor de Z Z .

Ejemplo de implementación:

def count_trailing_zeros(n):
    count = 0
    power_of_5 = 5
    while n >= power_of_5:
        count += n // power_of_5
        power_of_5 *= 5
    return count

# Leer el número de casos de prueba
T = int(input())
results = []

# Procesar cada caso de prueba
for _ in range(T):
    N = int(input())
    results.append(count_trailing_zeros(N))

# Imprimir los resultados
for result in results:
    print(result)

Este código lee el número de casos de prueba T T , luego para cada número N N calcula la cantidad de ceros al final del factorial de N N utilizando la función count_trailing_zeros, y finalmente imprime los resultados.

This problem has been solved

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