Educational Codeforces Round 11---B-Seating On Bus_seating order csdn-优快云博客

本文介绍了一种用于模拟乘客在公交车上有序上下车的算法，详细解释了乘客上车和下车的规则，并通过代码实现了解决方案。

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Consider 2n rows of the seats in a bus. n rows of the seats on the left and n rows of the seats on the right. Each row can be filled by two people. So the total capacity of the bus is 4n.

Consider that m (m ≤ 4n) people occupy the seats in the bus. The passengers entering the bus are numbered from 1 to m (in the order of their entering the bus). The pattern of the seat occupation is as below:

1-st row left window seat, 1-st row right window seat, 2-nd row left window seat, 2-nd row right window seat, ... , n-th row left window seat, n-th row right window seat.

After occupying all the window seats (for m > 2n) the non-window seats are occupied:

1-st row left non-window seat, 1-st row right non-window seat, ... , n-th row left non-window seat, n-th row right non-window seat.

All the passengers go to a single final destination. In the final destination, the passengers get off in the given order.

1-st row left non-window seat, 1-st row left window seat, 1-st row right non-window seat, 1-st row right window seat, ... , n-th row left non-window seat, n-th row left window seat, n-th row right non-window seat, n-th row right window seat.

$\text{[math]}$ The seating for n = 9 and m = 36.

You are given the values n and m. Output m numbers from 1 to m, the order in which the passengers will get off the bus.

Input

The only line contains two integers, n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 4n) — the number of pairs of rows and the number of passengers.

Output

Print m distinct integers from 1 to m — the order in which the passengers will get off the bus.

Examples

input
2 7

output
5 1 6 2 7 3 4

input
9 36

output
19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18

/*
题意: 一辆车共有 n 行座位，一行有 4 个，左右分别有 2 个，人是先将这 n 行

座位的左右各一个靠窗位置坐满，再坐靠近过道的位置，下车时先下左边靠近

过道位置的人再下靠近窗户的人，同理接着下右边的人，按上车的人先坐的行

开始下车。（先左后右，编号从 1--m 共 m 个人）

思路：直接模拟。
*/

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
#include <utility>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f3f
int main()
{
#ifdef OFFLINE
	freopen("t.txt", "r", stdin);
#endif
	int n, m, i, j, k, tmp;
	while (~scanf("%d%d", &n, &m))
	{
		if (m <= 2 * n){//人只坐靠窗位置
			for (i = 1; i < m; i++){
				printf("%d ", i);
			}
			printf("%d\n", m);
		}
		else{
			int row = (m - 2 * n) / 2;//除了靠窗位置每行有2个过道位置，算出需几行
			if ((m - 2 * n) % 2)  row++;//不整除时加 1
			tmp = 2 * n + 1;//过道位置起始标号
			for (i = 1; i < row; i++){//前 row-1 行都是每行坐满 4 人
				printf("%d %d %d %d", tmp, 2 * i - 1, tmp + 1, 2 * i);
				tmp += 2;
				printf(" ");
			}
			if ((m - 2 * n) % 2 == 0)//第 row 行满人
				printf("%d %d %d %d", tmp, 2 * row - 1, tmp + 1, 2 * row);
			else	if ((m - 2 * n) % 2 == 1)//第 row 行缺一个
				printf("%d %d %d", tmp, 2 * row - 1, 2 * row);
			for (i = row + 1; i <= n; i++){//row 往上的行都只坐靠窗位置
				printf(" %d %d", 2 * i - 1, 2 * i);
			}
			printf("\n");
		}
	}
	return 0;
}