Slava plays his favorite game "Peace Lightning". Now he is flying a bomber on a very specific map.
Formally, map is a checkered field of size 1 × n, the cells of which are numbered from 1 to n, in each cell there can be one or several tanks. Slava doesn't know the number of tanks and their positions, because he flies very high, but he can drop a bomb in any cell. All tanks in this cell will be damaged.
If a tank takes damage for the first time, it instantly moves to one of the neighboring cells (a tank in the cell n can only move to the cell n - 1, a tank in the cell 1 can only move to the cell 2). If a tank takes damage for the second time, it's counted as destroyed and never moves again. The tanks move only when they are damaged for the first time, they do not move by themselves.
Help Slava to destroy all tanks using as few bombs as possible.
The first line contains a single integer n (2 ≤ n ≤ 100 000) — the size of the map.
In the first line print m — the minimum number of bombs Slava needs to destroy all tanks.
In the second line print m integers k1, k2, ..., km. The number ki means that the i-th bomb should be dropped at the cell ki.
If there are multiple answers, you can print any of them.
2
3
2 1 2
3
4
2 1 3 2
翻译就算了吧 谷歌比我翻译的好不少
抽象一下：每个tank两滴血，tank排列在1*n的网格中，每个网格中有多个tank，你可以在空中进行轰炸，每个tank被炸到后，都会往左右两个方向挑一个移动，炸两次就死了，求用多少炸弹能炸死所有的tank
这个题我没想出来，还是看的别人的题解，但是好多题解都没有证明为什么这么炸这是最佳结论，包括官方题解都说it's easy to prove the strategy is optimal 可能这就是差距吧，我想了好长时间都
想不明白
当然现在也想不明白，之后看能不能在补上吧
先输出偶数位，在输出奇数位，再输出偶数位
丑陋的代码：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int i,j,n;
scanf("%d",&n);
printf("%d ",n+n/2);
for(i = 2; i <= n; i += 2)
printf("%d ",i);
for(i = 1; i <= n; i += 2)
printf("%d ",i);
for(i = 2; i <= n - 2; i += 2)
printf("%d ",i);
printf("%d ",i);
}