B. Seating On Bus

最新推荐文章于 2018-07-31 21:16:11 发布

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最新推荐文章于 2018-07-31 21:16:11 发布

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分类专栏： ——题解——

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本文探讨了一种解决乘车座位分配问题的算法，通过详细解释输入、输出和实例，展示了如何根据人数和座位布局合理安排乘客座位。面对复杂的分配逻辑，作者通过代码实现了算法，并分享了解题过程中的经验和教训。

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B. Seating On Bus

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

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.

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
我的：这道题刚开始的时候感觉就是一个水题，但是第一次提交之后就WA了，之后感觉补了好多次，从比赛开始三十分钟到距离结束前几分钟都在不漏洞，最后终于AC了，汗呀！
接下来是我写的代码：

#include<iostream>
using namespace std;
const int maxn=100+5;
int main()
{
    int a[maxn][2],b[maxn][2];
    int n,m;
    while(cin>>n>>m)
    {
        int x=0,y=0;
        int t=1;
        int i;
        for(i=0;i<n;i++)
        {
            a[i][0]=0;a[i][1]=0;
            b[i][0]=0;b[i][1]=0;
        }
        if(m>=2*n)
        {
           while(t<=2*n)
           {
            a[x++][0]=t++;
            if(t>2*n)break;
            b[y++][1]=t++;
           }
        //cout<<b[y-1][1]<<endl;
           x=0;y=0;
           while(t<=m)
           {
            a[x++][1]=t++;
            if(t>m)break;
            b[y++][0]=t++;
           }
        }
        else 
        {
            t=1;
            x=0;y=0;
            while(t<=m)
            {
                a[x++][0]=t++;
                if(t>m)break;
                b[y++][1]=t++;
            }
        }

        for(i=0;i<n;i++)
            if(i!=n-1)
            {
                if(a[i][1])cout<<a[i][1]<<" ";
                if(a[i][0])cout<<a[i][0]<<" ";
                if(b[i][0])cout<<b[i][0]<<" ";
                if(b[i][1])cout<<b[i][1]<<" ";
            }
            else 
            {
                if(a[i][1])cout<<a[i][1]<<" ";
                if(a[i][0])cout<<a[i][0]<<" ";
                if(b[i][0])cout<<b[i][0]<<" ";
                if(b[i][1])cout<<b[i][1]<<endl;
            }
    }
    return 0;
}