SPOJ FCTRL

题目链接：http://www.spoj.com/problems/FCTRL/

题目大意：询问N的阶乘的结果尾数有几个0.

解题思路：考虑问题：N的阶乘的结果能被2^m整除，这个m最大为多少。

我们对前N个数除以2，忽略奇数，会得到N/2个数字。那么相当于我们得到了2^N/2

对之后的N/2个数字继续除以2，同样忽略奇数。我们会再得到2^N/4

...

所以m=N/2 + N/4 + N/8 +...+0。

那么对于这个问题，我们计算出N的记成被2^m1整除的最大m1以及被5^m2整除的最大m2，由于2*5=10，而10在进行阶乘计算的时候会使得结果尾数多一个0，因此我们只需要计算出N被(2*5)^m整除的最大m即可。由于m1一定大于m2，所以结果就是m2.而计算m2则可以通过N/5 + N/25 +...+ 0计算得到。

代码：

const int maxn = 1e5 + 5;
int n;

void solve(){
    int ans = 0;
    int x = n / 5;
    while(x > 0){
        ans += x;
        x /= 5;
    }
    printf("%d
", ans);
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        solve();
    }
}

题目：

FCTRL - Factorial

#math

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 (equal to about 100000). It stands for the number of numbers to follow. Then there are 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).

Example

Sample Input:

6
3
60
100
1024
23456
8735373

Sample Output:

0
14
24
253
5861
2183837