ZZULI_SUMMER_PRACTICE(8) POJ  10…

Factorial

Time Limit: 1500MS		Memory Limit: 65536K
Total Submissions: 10952		Accepted: 6791

Description
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. It stands for the number of numbers to follow. Then
there is 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

6
3
60
100
1024
23456
8735373

Sample Output

0
14
24
253
5861
2183837

Source
Central Europe
2000

额……也不知道这题思路怎么来的，正在纠结的时候突然灵光乍现，就想试一下，没想到几个Sample全符合，就做了下去，嘿嘿，1A
代码：

C语言: 临时自用代码

#include<stdio.h>
int main()
{

    int t,n,sum;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        sum=0;

        while(n!=0){

            sum+=n/5;

           
n=n/5;

        }

        printf("%d\n",sum);

    }

    return 0;
}