ACM-简单题之Factorial——poj1401

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Factorial

Time Limit: 1500MS		Memory Limit: 65536K
Total Submissions: 13993		Accepted: 8678

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

题目：http://poj.org/problem?id=1401

题目是求N！中末尾0的个数。

这道题，是我曾经在经典例题100道中看到的，做出来的，可是，发现理解错了。（感谢一位网友的提醒啊!）

所以，又一次做一下，发现poj上有这道题，正好能够拿来測试一下做的是否正确。

这道题做法，显然不能求出来阶乘再数0的个数。

所以，要换一个思路。

为什么会产生0呢？源于2x5,于是能够通过数有多少个（2,5）因子对来推断有多少个0.

我们又能够发现，5的个数永远会大于2的个数。

又能够简化为数5的个数来做。

当时，还非常年轻，做法让如今的我看，有点纠结啊。。

当时做的是，从1到N（所输入的数）循环，看看能不能被5整除，若能整除，继续除5，直到不能整除为止。

这个做法显然答案正确，可是在这道题里，TLE必须的。。。

#include <stdio.h>
int main()
{
    int i,N,k,t;
    double num;
    bool prime;
    scanf("%d",&t);
    while( t-- )
    {
        scanf("%d",&N);
        k=0;
        for(i=1;i<=N;i++)           //查看有多少个5
        {
            num=i/5.0;
            if(num-int(num)==0)
                prime=true;
            else
                prime=false;
            while(num>=1&&prime==true)
            {
                num/=5.0;
                if(num-int(num)==0)
                    prime=true;
                else
                    prime=false;
                k+=1;
            }
        }
        printf("%d\n",k);
    }
    return 0;
}

然后，如今想想，这道题，不须要这么麻烦啊。

数5的倍数，就能够了。

以698为例：

698/5=139.6→取整→139     （表示有139个数为5的倍数） sum=sum+139 （sum初始化为0）

139/5=27.8 →取整→27        （表示有27个数为25的倍数）sum=sum+27

（为什么25的倍数，仅仅加了一遍，不应该加 27*2的吗，25表示两个5？

由于，25的之前有一个5，在第一遍139个里面算过一次了，所以不须要加两遍。）

27/5=5.4 → 取整→5             （表示有5个数为125的倍数）sum=sum+5

5/5=1 → 取整 → 1
     （表示有1个数为625的倍数） sum=sum+1

1/5=0  结束。

所以答案是   139+27+5+1=172

恩，所以程序，就非常easy了：

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : Factorial                   *
*Source: poj 1401                     *
* Hint  : N!求0的个数                *
***************************************
**************************************/
// scanf 125MS， cin 422MS
#include <stdio.h>
int main()
{
    // sum为答案，存储每次除5后的数（累加）
    int t,n,sum;
    scanf("%d",&t);
    while( t-- )
    {
        sum=0;
        scanf("%d",&n);
        while( n )
        {
            n/=5;
            sum+=n;
        }
        printf("%d\n",sum);
    }
    return 0;
}