Codeforces Round #442 (Div. 2) C - Slava and tanks

最新推荐文章于 2022-11-22 23:15:14 发布

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本文介绍了一个轰炸策略游戏的解决方案，玩家需要使用最少的炸弹来摧毁未知数量和位置的坦克。文章详细描述了如何通过特定的轰炸顺序（先攻击偶数格子再攻击奇数格子最后再次攻击偶数格子）来确保所有坦克被摧毁的方法。

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 celln - 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.

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — the size of the map.

Output

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.

Examples
input
2
output
3
2 1 2 
input
3
output
4
2 1 3 2 

题意有m个水平放置且相邻的格子，每个格子中至少有一个坦克，你需要用导弹消灭这些坦克，如果一个坦克承受伤害的第一时间，瞬间移动到一个相邻细胞（坦克n只能移动到n - 1处，1坦克只能移动到2处）。如果坦克第二次受到伤害，它被计算为毁灭，不再移动。坦克只有在它们第一次被损坏时才会移动，它们不会自行移动。
帮助Slava用几颗炸弹可能摧毁所有的坦克。

在第一行打印M炸弹”需要消灭所有坦克的最小数量。
在第二行打印M的K1，K2 ， …，KM。数奇意味着第i个炸弹应该被丢弃在Ki处。

先攻击偶数的格子然后攻击奇数的格子再攻击偶数的格子就可以把所有坦克消灭掉了

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;

int main()
{
	int n;
	scanf("%d", &n);
	int sum = n & 1 ? n + n / 2 : n + n / 2;
	printf("%d\n", sum);
	for (int i = 1; i <= n; i ++) {
		if (!(i & 1)) printf("%d ", i);
	}
	for (int i = 1; i <= n; i ++) {
		if (i & 1) printf("%d ", i);
	}
	for (int i = 2; i <= n; i ++) {
		if (!(i & 1)) printf("%d ", i);
	}
	printf("\n");
}