Codeforces 633B A Trivial Problem 【数论】

B. A Trivial Problem

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

题意：问你那些数的阶乘末尾有连续m个0。

lightoj上基本原题：链接

思路：二分一下m和m+1就可以了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
int Count(int n)
{
int cnt = 0;
while(n)
{
cnt += n / 5;
n /= 5;
}
return cnt;
}
int Two(int l, int r, int m)
{
int ans = 0;
while(r >= l)
{
int mid = (l + r) >> 1;
int res = Count(mid);
if(res < m)
l = mid + 1;
else if(res >= m) {
r = mid - 1;
ans = mid;
}
}
return ans;
}
const int MAXN = 1e9;
int main()
{
int m; cin >> m;
int l = 1, r = MAXN;
int L = Two(l, r, m), R = Two(l, r, m+1);
//cout << L << " " << R << endl;
if(L == 0 || Count(R-1) != m) cout << 0 << endl;
else {
cout << R - L << endl;
for(int i = L; i <= R-1; i++) {
if(i > L) cout << " ";
cout << i;
}
cout << endl;
}
return 0;
}