Factorial

Time Limit: 1500MS	Memory Limit: 65536K
Total Submissions: 15475	Accepted: 9533

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

這道題挺經典的，這個演算法理解起來挺簡單的。

先說一下原理：後面的0都是由2和5相乘得到的，而2每兩個就有一個，所以個數足夠，不考慮。

所以只需要考慮因數5的個數即可。

我覺得個人參照程式碼比較容易學會吧，如果給出n，小於等於n的5的倍數有 n/5 個，這些數都包括了因數5。

然後把這些數縮小為原來的 1/5 （記得高中語文老師說，小於1的倍數的說法不能說縮小几倍，瞎扯幾句，繼續說），如果剩下的數仍然大於5，則還是存在5的倍數，再次迴圈上面的過程即可。不懂的話可以看看程式碼。

程式碼如下：

#include <cstdio>
int main()
{
	int u;
	int ans,n;
	scanf ("%d",&u);
	while (u--)
	{
		ans = 0;
		scanf ("%d",&n);
		while (n)
		{
			ans += n/5;
			n /= 5;
		}
		printf ("%d\n",ans);
	}
	return 0;
}