HDU 1124 Factorial&&nyoj 84 阶乘的 0【数学】

Factorial

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3356 Accepted Submission(s): 2168

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

63601001024234568735373

Sample Output

0142425358612183837

/*2015.05.27 19:41 这个题和之前一个题的解法一样，那个题是素因数分解，只要找到特殊的就行这个解法的思想真的很巧妙，呵呵，没完全想出来，还是看着思路慢慢想出来的这样算的的原因是，只要所有的数可以拆成多少个五，那么肯定会有多少个零因为某个数（>1）的阶乘肯定是偶数，那么偶数乘以5，个位肯定是0，哈哈所以也就是相当于求分解素因数那个题，只是没那个题很好理解了点毕竟转换了方式，以后做题还是要多想思路，很多东西也没有自己想的那么难只要好好想了，至少不会损失什么，好了，就这样！ */#include<stdio.h>int main(){int t,i,n,s;scanf("%d",&t);while(t--){s=0;scanf("%d",&n);for(i=n;i>0;i/=5){s=s+i/5;}printf("%d\n",s);}return 0;}

以前做过的题，结论记下了.....懒得再推一遍.......

#include<stdio.h>int slove(int n){int ans=0,tp=n;for(int i=5;i<=n;i*=5){ans+=tp/i;}return ans;}int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);printf("%d\n",slove(n));}return 0;}