Factorial

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2493    Accepted Submission(s): 1590

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

63601001024234568735373

 

Sample Output

0142425358612183837

 

题意：N阶乘有多少个尾0？(1 <= N <= 1000000000) 

Sample Input 
6    //表示数据个数 
3 
60 
100 
1024 
23456 
8735373 

Sample Output 
0 
14 
24 
253 
5861 
2183837 

N! = 1 * 2 * 3 * (2*2) * 5 * (2*3) * 7... 

产生10的原因是有2,5的因子，显然在N!中2的个数大于5的个数，所以只需求出5的个数即可 

求 N! (1*2*3*4*5*...*N)里有多少个5其实可以转化成： 
N!中：是5的倍数的数+是5^2的倍数的数+5^3..... 

如50!： 
含有10个5的倍数的数：5,15,20,25,30,35,40,45,50 【50/5=10】 
含有2个5^2的倍数的数：25,50【50/(5^2)=2】 
可见N!中一共有12个5相乘，那么尾0也必有12个 

#include <iostream>#include <fstream>#include <algorithm>#include <string>#include <set>#include <map>#include <queue>#include <utility>#include <iomanip>#include <stack>#include <list>#include <vector>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <ctype.h>using namespace std;int main(){    int t, res, n;    scanf ("%d", &t);    while (t--)    {        res = 0;        scanf ("%d", &n);        while (n)        {            res += n / 5;            n /= 5;        }        printf ("%d\n", res);    }    return 0;}

原文来自：http://972169909-qq-com.iteye.com/blog/1126188