HDU 1124 Factorial
2012-10-12 14:19
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Factorial
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1827 Accepted Submission(s): 1163
Problem 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
Source
Central Europe 2000
Recommend
Eddy
分析:计算N!后面有几个0,简单的数论的题目,只有偶数和5相乘后面才有可能出0,因为在一个数的阶乘中,5的个数一定少于偶数的个数
所以只要找出5的个数即可,但是注意,25乘以偶数有2个0,所以还要加回25的个数,同理125.……
代码:
#include<stdio.h> int main() { int N,M; scanf("%d",&M); while(M--) { int count=0; scanf("%d",&N); while(N) { count+=N/5; N/=5; } printf("%d\n",count); } return 0; }
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