Poj 1401 Factorial(计算N!尾数0的个数——质因数分解）

一、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).

二、题解

       这个题目看了很久还是没看懂，后来发现要求阶乘后的数尾数中0的个数。于是就傻逼的算阶乘，这个根本没法算，数太大了，会发生溢出的。于是就想到要有0就只有1 * 10, 2 * 5而10也可以化成2 * 5，而显然能被5整除的个数远小于能被2整除的个数，于是就计算1~60中能被5整除的个数，后来发现还是不对，结果要比答案少。后来发现，原来25 * 4=100，能增加两个0，而25=5 * 5，能分开来乘。所以，所以在<N的数里面，能分出多少个5来，末尾就多少个0。

       所以应该这样计算：

---

100/5=20->20/5=4
那么20+4=24
1024/5=204  ->  204/5=40  ->  40/5=8  ->  8/5=1
那么204+40+8+1=253

这里的严谨分析应该是进行质因数分解N ! =(2^x) * (3^y) * (5^z)...,由于10=2*5,所以M只跟x和z有关，每一对2和5相乘可以得到一个10，于是M=min(x,z)。不难看出x大于z,因为能被2整除的数出现的频率能被5整除的数高得多，所以把公式简化为M=z.

       根据上面的分析，只要计算出Z的值，就可以得到N!末尾0的个数了。

三、java代码

import java.util.Scanner;

  public class Main {

	  public static void main(String[] args)  {
       Scanner sc = new Scanner(System.in);
       int n,m,out;
       n=sc.nextInt();
       while(n!=0){
    	 m=sc.nextInt();
    	 out=0;
    	 while (m>=5){
    	        m/=5;
    	        out+=m;
    	    }
    	System.out.println(out);
    	n--;
       }
    }
 }     

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