Codeforces 660B Seating On Bus 【模拟】

题目链接：Codeforces 660B Seating On Bus

B. Seating On Bus

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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：m个人，给定一个座位号，输出下车顺序。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
int l1[110], l2[110], r2[110], r1[110];
int main()
{
int n, m;
while(scanf("%d%d", &n, &m) != EOF) {
int s = 1;
for(int i = 1; i <= n; i++) {
l1[i] = s;
s += 2;
}
for(int i = 1; i <= n; i++) {
l2[i] = s;
s += 2;
}
s = 2;
for(int i = 1; i <= n; i++) {
r1[i] = s;
s += 2;
}
for(int i = 1; i <= n; i++) {
r2[i] = s;
s += 2;
}
int use = 0;
for(int i = 1; i <= n; i++) {
if(l2[i] <= m) {
if(use) printf(" ");
use++;
printf("%d", l2[i]);
}
if(l1[i] <= m) {
if(use) printf(" ");
use++;
printf("%d", l1[i]);
}
if(r2[i] <= m) {
if(use) printf(" ");
use++;
printf("%d", r2[i]);
}
if(r1[i] <= m) {
if(use) printf(" ");
use++;
printf("%d", r1[i]);
}
}
printf("\n");
}
return 0;
}