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.
Input.
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).
Output.
For every number n,
output a single line containing the single non-negative integer Z(n).
|
Sample input |
Sample output |
|
6 3 60 100 1024 23456 8735373 |
0 14 24 253 5861 2183837 |
циклы
Для
заданного числа n следует найти
количество нулей, которыми оканчивается значение n!.
Реализация
алгоритма
#include <stdio.h>
int tests,n,res;
int main(void)
{
scanf("%d",&tests);
while(tests--)
{
scanf("%d",&n);
res = 0;
while (n > 0)
{
res += n / 5;
n /= 5;
}
printf("%d\n",res);
}
return 0;
}