poj1401

转载自http://blog.csdn.net/bingsanchun1/article/details/17098419

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

63601001024234568735373

Sample Output

0142425358612183837

 

 

 

题意是说求N！后面有多少个0.从网上看到是数论知识，说的是

//逢一个因子5会产生一个0，
//如10!=10*9*8*7*6*5*4*3*2*1中有两个因数5，能构成结尾2个0。
//数论：这道题的正解就是对sum=n/5+n/25+n/125......n/(5^k)，直到n<(5^k).复杂度是log5(N)

 简单的说就是求也个数的阶乘的末尾有多少个零的问题，如果用阶乘直接求得话相信大家都会知道那是不可能的了！！

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <algorithm>using namespace std;long long findFacTailZero(int n){long long ans = 0;for (int i = 5; n / i > 0; i *= 5){ans += n / i;}return ans;}void FactorialTrailZero(){int T = 0, N = 0;cin>>T;while (T--){cin>>N;cout<<findFacTailZero(N)<<endl;}}int main(){FactorialTrailZero();return 0;}