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?
The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.
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.
1
5
5 6 7 8 9
5
0
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.
题意:给一个m,问哪些数的阶乘的末尾0的个数为m;
思路:末尾0的个数为这些数中有多少个5,2的个数一定大于5,所以只需找5的个数就行;
AC代码:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int m;
scanf("%d",&m);
int num,x,sum=0;
for(int i=1;i<=100000;i++)
{
num=1;
x=i;
while(1)
{
if(x%5==0)
{
num++;
x=x/5;
}
else break;
}
sum+=num;
if(sum==m)
{
cout<<"5"<<"
";
for(int j=5*i;j<5*i+5;j++)
{
printf("%d ",j);
}
return 0;
}
else if(sum>m)
{
break;
}
}
cout<<"0";
return 0;
}