Factorial
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2807 Accepted Submission(s): 1804
Problem 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
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题目的意思就是求出N!的末尾有多少个0
N很大，求出N!然后再统计0个数明显不可能
10=2*5，末尾的0是由2*5产生。
所以，统计2和5的个数即可。
但是因为N!中2的个数比5多，所以只用统计5的个数即可。
数论知识：
若p是质数，p<=N，则N是p的倍数，设p^x是p在N内的最高幂，则
x=[n/p]+[n/p^2]+[n/p^3]+............;
1 #include<cstdio>
2 #include<cstring>
3 #include<stdlib.h>
4 #include<algorithm>
5 using namespace std;
6 int main()
7 {
8 //freopen("in.txt","r",stdin);
9 int kase,n;
10 scanf("%d",&kase);
11 while(kase--)
12 {
13 int ans=0;
14 scanf("%d",&n);
15 while(n)
16 {
17 ans=ans+n/5;
18 n=n/5;
19 }
20 printf("%d\n",ans);
21 }
22 return 0;
23 }