Educational Codeforces Round 11 B. Seating On Bus 水题

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最新推荐文章于 2025-08-23 16:01:06 发布

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文章标签：数据结构与算法

本文介绍了一种模拟公交车乘客上下车顺序的算法。通过输入公交车的座位布局与乘客数量，可以预测乘客离开车辆的顺序。文章包含问题描述、解决思路及C++实现代码。

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B. Seating On Bus

题目连接：

http://www.codeforces.com/contest/660/problem/B

Description

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.

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.

Sample Input

2 7

Sample Output

5 1 6 2 7 3 4

Hint

题意

有一个公交车

如果人数小于2n的话，那么这些人就只会坐在边上，是先坐左边，然后坐右边这样的

如果大于等于2n的话，就会去坐中间，也是先坐左边，再坐右边

走的时候，就先走第二列，然后走第一列，然后走第三列，然后走第四列这样的，依次走一个这样。

问你这些人走的样子是什么样子

题解：

先走的奇数，然后再走偶数位置，先走大于2n的，再走小于的。

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=2*n;i++)
    {
        if(2*n+i<=m)cout<<2*n+i<<" ";
        if(i<=m)cout<<i<<" ";
    }
    cout<<endl;
}