Educational Codeforces Round 11（B）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：看图就可以了。。不解释了。。。。

题解：我是找了下规律，按照x=（2*n+1）,y=1,x++,y++填充数组，输出是只要是<=m的都输出，就是答案啦

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
using namespace std;

#define N int(1e5)
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
typedef long long LL;

int ans
;
int main()
{
#ifdef CDZSC
freopen("i.txt", "r", stdin);
//freopen("o.txt","w",stdout);
int _time_jc = clock();
#endif

int n, m;
while (~scanf("%d%d", &n, &m))
{
int fst = 2 * n + 1;
int sec = 1;
memset(ans, 0, sizeof(ans));
for (int i = 1; i <= n * 4; i+=2)
{
ans[i] = fst++;
ans[i + 1] = sec++;
}
for (int i = 1; i <= 4 * n; i++)
{
if (ans[i] <= m)
{
printf("%d ", ans[i]);
}
}
puts("");
}
return 0;
}