CodeForces 660 B. Seating On Bus【模拟】

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

题意：

按题目要求上车选位置以及下车，问最后下车的顺序

题解：

无脑栈模拟...

/* http://blog.csdn.net/liuke19950717 */
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
stack<int> seat[205];
int cnt=0;
for(int i=0;i<m;++i)
{
seat[i%(2*n)].push(++cnt);
}
printf("%d",seat[0].top());
seat[0].pop();
for(int i=0;i<2*n;++i)
{
while(!seat[i].empty())
{
printf(" %d",seat[i].top());
seat[i].pop();
}
}
printf("\n");
}
return 0;
}