ACM: POJ 1401 Factorial-数论专题-水题

POJ 1401 Factorial

Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

 

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：

n!后缀0的个数;

这里要注意到只有2的倍数和5的倍数相乘会产生后缀0；

毫无疑问2的倍数比5多，所以只要统计5的倍数的个数就行了；

但是要注意，5^n能产生n个后缀0；

所以就要处理一下，把5^i (i=1~n)的所有5的倍数全部加起来，详情看代码；

AC代码
/*/

#include"map"
#include"cmath"
#include"string"
#include"cstdio"
#include"vector"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
typedef long long LL;
const int MX=202;
#define memset(x,y) memset(x,y,sizeof(x))
#define FK(x) cout<<"【"<<x<<"】"<<endl

LL ans(int n) {
	LL ans = 0;
	for (int i=5; n/i>0; i*=5) {  //5的倍数能产生1个0，25的倍数能产生2个0， 125的倍数能产生3个0，以此类推。。。
		ans+=n/i;
	}
	return ans;
}

int main() {
	int T,x;
	scanf("%d",&T);
	while(T--) {
		scanf("%d",&x);
		printf("%d\n",ans(x));
	}
}