【codechef】 Zeroes（大数灵活题）

Virat loves mathematical questions. The other day he came across an interesting question which required him to find out the number of trailing zeroes for the
function. F(n) = 1¹*2²......N^N,where N is an integer. Virat solved the problem after a few attempts. He asked the same
question from his friend Rohit to see if he can solve it. Rohit couldn’t arrive at a solution and has asked for your help. Can you help him out?

Input

• The first line contains a single integer T, the number of test cases. T test cases follow.
• Each line contains a single integer

  N.

Output

For each test case, output a single line containing a single integer which denotes the number of trailing zeroes for the value of the function computed for

N.

Constraints

• 1 <= T <= 1000
• 1 <= N <= 10000

Example

Input:
3
5
10
2


Output:
5
15
0
http://www.codechef.com/problems/ZEROES
F(n) = 11*22......NN，求F(n)的末位有几个0。这种题目就是要找规律。因为值实在是太大了，只能算出前几个，于是我就用了字符串大数模板，也只能勉强算出前200个，规律找得不够彻底。。。
0000 55555 1515151515（为方便我把5个数缩成1个）30 50 100 130 165 205 250 350 405 465 530 600 750 830 915 1005 1100 1300 1405 1515 1630 1750 2125....
我们知道 0 1 3 6 10 15....这个数列的前n项和为 n*(n+1)/2 ，上面白色背景的数字就是它的5倍，间隔为5（前面省去一样的了），红色部分减去白色的值是它的25倍，间隔为25，绿色部分减去红色的值是它的125倍，间隔为125，后面以此类推。
#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<vector>  
#include<cmath>  
#include<string.h>  
#include<stdlib.h>  
#include<cstdio>  
#define ll long long  
using namespace std;
int main(void){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		cin>>n;
		ll s=0;
		for(int i=1;pow(5.0,i)<=n;++i){
			int m=n/pow(5.0,i);
			s+=pow(5.0,i)*(m+1)*m/2;
		}
		printf("%I64d\n",s);
	}
	return 0;
}  

#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<vector>  
#include<cmath>  
#include<string.h>  
#include<stdlib.h>  
#include<cstdio>  
#define ll long long  
using namespace std;
int main (void)
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,i;
		long long int zero=0;
		scanf("%d",&n);
		for(i=5;i<=n;i+=5)
		{
			if(i%78125 == 0)
				zero += i*7;
			else if(i%12625 == 0)
				zero += i*6;
			else if(i%3125 == 0)
				zero += i*5;
			else if(i%625 == 0)
				zero += i*4;
			else if(i%125 == 0)
				zero += i*3;
			else if(i%25 == 0)
				zero += i*2;
			else
				zero += i;
		}
		printf("%lld\n",zero);
	}
}