ACM: n!末尾0的个数 数论题 poj 14…

Factorial
 

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

 

题意: 求n!的末尾0的个数.

 

解题思路:

         
1. n! = 1 * 2 * 3 * 4 * ... * n;

               
= 1 * 2 * 3 * (2*2) * 5 * (2*2*2) * 7 * ... * n;

          2.
末尾要出现0, 就必须要乘10, 10 = 2 * 5; 显然2的个数必须大于等5的个数.

         
3. 只需要求1 ~ n中含有5的倍数的个数.

          4.
可得公式: count = ∑(i = 1) (n / 5^i);

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

int n;

int main()

{

// freopen("input.txt","r",stdin);

 int caseNum;

 scanf("%d",&caseNum);

 while(caseNum--)

 {

  scanf("%d",&n);

  int count = 0;

  do

  {

   n /= 5;

   count +=
n;

  }while(n);

  printf("%d\n",count);

 }

 return 0;

}