Problem Description
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.
6 3 60 100 1024 23456 8735373
0 14 24 253 5861 2183837
题意就是让你求出N!中0的个数;我们知道对于任何数都可以用质因数来表示,如N!=1*2*3*(2*2)*5*(2*3)....,要求得该数中0的个数就一定要求出有多少个2*5,由于2的个数肯定比5的个数多,所以只要求出5的个数即可。
运用到数论的一个知识:
求 N! (1*2*3*4*5*...*N)里有多少个5其实可以转化成:
N!中:是5的倍数的数+是5^2的倍数的数+5^3.....
如50!:
含有10个5的倍数的数:5,15,20,25,30,35,40,45,50 【50/5=10】
含有2个5^2的倍数的数:25,50【50/(5^2)=2】
可见N!中一共有12个5相乘,那么尾0也必有12个
代码:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int t,n,sum;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
sum=0;
while(n!=0)
{
n=n/5;
sum+=n;
}
printf("%d\n",sum);
}
return 0;
}