CodeForces - 633B A Trivial Problem（找规律）

A Trivial Problem

Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number
of positive integers n, such that the factorial of n ends
with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) —
the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such
that the factorial of n ends with m zeroes.
Then print these k integers in increasing order.

Examples

input

1

output

5
5 6 7 8 9

input

5

output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive,
that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

ps：找规律的大水题，
首先我们想一下，什么时候一个数的后面会出现0，
可以发现如果阶乘中有2*5的话，就会出现一个10，所以后面会出现一个0，然而5又比2大，所以如果有5出现肯定就已经有2出现了，那么显然我们只需要找阶乘因子中5的个数就好了

代码：

#include<stdio.h>

#define maxn 1000000+10
int a[maxn]={0},b[10];//a[i]储存i的阶乘后面有几个0

int main()
{
int k=0;
for(int i=5; i<maxn; i++)
{
if(i%5==0)
{
int t=i;
while(t%5==0)
{
t/=5;
k++;
}
}
a[i]=k;
}
int m,ps=0;
scanf("%d",&m);
for(int i=1;i<maxn;i++)
{
if(a[i]==m)
b[ps++]=i;
}
printf("%d\n",ps);
for(int i=0;i<ps;i++)
printf("%d ",b[i]);
return 0;
}