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

代码：

#include<cstdio>#include<cstring>int main(){int n,m,s;int shu[4][105];int ge[4]={0};scanf("%d%d",&n,&m);for (int i=1;i<=m;i++){if (i<=2*n){if(i%2==1){shu[0][i/2]=i;ge[0]++;}else{shu[3][i/2-1]=i;ge[3]++;}}else{if (i%2==1){shu[1][(i-2*n)/2]=i;ge[1]++;}else{shu[2][(i-2*n)/2-1]=i;ge[2]++;}}}int pp=0;for (int i=1;i<=n;i++){if (ge[1]>=i){if (pp==0){printf("%d",shu[1][i-1]);pp=1;}elseprintf(" %d",shu[1][i-1]);}if (ge[0]>=i){if (pp==0){printf("%d",shu[0][i-1]);pp=1;}elseprintf(" %d",shu[0][i-1]);}if (ge[2]>=i){if (pp==0){printf("%d",shu[2][i-1]);pp=1;}elseprintf(" %d",shu[2][i-1]);}if (ge[3]>=i){if (pp==0){printf("%d",shu[3][i-1]);pp=1;}elseprintf(" %d",shu[3][i-1]);}}printf("\n");return 0;}