Factorial

Time Limit: 1500MS	Memory Limit: 65536K
Total Submissions: 12332	Accepted: 7682

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

題意：求一個階乘數末尾連續的0的個數。

分析：首先要知道，末尾出現0的話，一定是某個偶數和5相乘得到的，所以整個數含有5（包含拆開，即25=5*5，出現兩次）的5的個數就是所求末尾0的個數。

思路分析（轉）：

每次只計算最多含有5，5^2,5^3……的數字個數
每次含有5的[n/5]
........25.[n/25]

注意，當統計最多含有5^2的因子的時候，5的個數應該為[n/25]而不是[n/25]*2因為，在含有[n/5]的時候已經統計過一次了
所以，只需要把[n/5],[n/25]...加起來就可以了

例如1~100中
（1）含有5的：5,10,15....總計100/5=20個，每個之中含有1個5，總5因子數為20
（2）含有25的：25,50,75,100，總計100/25=4個，每個之中含有2個5，但是因為在（1）中含5的數字已經統計過一次因子，所以這裡仍然記錄因子個數1*5個
。。。。所以1~100中含有5的因子一共100/5+100/25=24個

1~N中的所有數字相乘，能拆分成的最簡形式為素數的相乘（1忽略）,其中2的個數多於5的個數（2的倍數多於5的倍數），所以，N！中，因子5有多少個，最後N！的末尾的0就有多少個。

這題在前面知識點的時候已經提到具體演算法了。。。

#include
#include
using namespace std;
int solve(int n)
{
    int count=0;
    while(n>=5)
    {
        n=n/5;
        count+=n;
    }
    return count;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        //sum=sovle(n);
        printf("%d\n",solve(n));
    }
    return 0;
}