Factorial

Factorial

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 8

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

 

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

题目大意：输入一个数字N，求出从1阶层到N的后面一共有多少个0，输出0的个数，即为答案。

该题的，需要的是能够找得出能够使得相乘的值等于10，因为偶数肯定存在，所以，只需要找到关于5的数字即可。

比如，输入N。求出在1~N中有多少个5，即为N/5；然后又在这个范围内继续查找又有多少个关联5的。所以关联5的个数相加，即为5的个总和。每一个5可以和偶数相乘得到10，即可知道有多少个0了的。

#include <stdio.h>
#define Max(a,b) a>b? a:b
#define Min(a,b) a>b? b:a
#include <string.h>
int main()
{
    int T,NUM,sum;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&NUM);
        sum=0;
        while(NUM)/*求出有多少个与5相互关联的个数*/
        {
            sum+=NUM/5;
            NUM/=5;
        }
        printf("%d\n",sum);
    }
    return 0;
}