TJU Problem 1065 Factorial

注意数据范围，十位数以上就可以考虑long long 了，断点调试也十分重要。

原题：

1065.   Factorial

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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 6067   Accepted Runs: 2679

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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: Central European
2000

源代码：

 #include <iostream>
 #include <cmath>
 using namespace std;

 long long int a[];

 int main()    {
     for (int i = ; i < ; i++)    {
         a[i] = pow(, i+);
     }
     int N;    cin >> N;
     while(N--)    {
         long long int num, temp, count = -;    cin >> num;
         temp = num;
         while (temp != )    {
             temp /= ;
             //cout << "temp " << temp << endl;
             count++;
         }
         //cout << "count " << count << endl;
         if (count == )    {
             cout <<  << endl;
             continue;
         }
         //cout << "num " << num << endl;
         long long int temp2 = num / a[count - ];
         //cout << "temp2 " << temp2 << endl;
         long long int res, sum = temp2 * count;
         for (int i = count - ; i >= ; i--)    {
             int temp1 = num / a[i];
             sum += (temp1 - temp2) * (i+);
             temp2 = temp1;
         }
         cout << sum << endl;
     }
     return ;
 }