HDU：1124 Factorial

这题真悲剧啊，这么简单的一个题，题意竟然没看懂!

题意：求N！的结果中0的个数，就是求2和5的组合一共有几个，因为2的个数永远大于5的个数，所以只用求出5的因子个数即可，首先算5的因子，然后是5*5的因子……

相加即得结果。

Factorial

Time Limit: 2000/1000 MS
(Java/Others)    Memory
Limit: 65536/32768 K (Java/Others)

Total Submission(s):
1152    Accepted
Submission(s): 732


[align=left]Problem Description[/align]
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.

 

 

[align=left]Input[/align]
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.

 

 

[align=left]Output[/align]
For every number N, output a single line
containing the single non-negative integer Z(N).

 

 

[align=left]Sample Input[/align]

6 3 60 100
1024 23456 8735373

 

 

[align=left]Sample Output[/align]

0 14 24 253
5861 2183837

 

 

[align=left]Source[/align]

Central Europe 2000
代码如下：

#include<stdio.h>

int main()

{

 int m,n,num;

 scanf("%d",&n);

 while(n--)

 {

  scanf("%d",&m);

  num=0;

  while(m!=0)

  {

   m=m/5;

   num+=m;

  }

  printf("%d\n",num);

 }

 return 0;

}