Educational Codeforces Round 11B. Seating On Bus 模拟

地址:http://codeforces.com/contest/660/problem/B

题目:

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

 思路:直接模拟就好了,座位位置找规律就好了,别告诉我你找不到规律、、、、

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <queue>
 7 #include <stack>
 8 #include <map>
 9 #include <vector>
10 
11 #define PI acos((double)-1)
12 #define E exp(double(1))
13 using namespace std;
14 int seat[110][5];
15 int main (void)
16 {
17     int n,m;
18     cin>>n>>m;
19     memset(seat,0,sizeof(seat));
20     for(int i=1;i<=m;i++)
21         if(i<=2*n)
22         {
23             if(i%2 == 1)
24                 seat[(i+1)/2][2] = i;
25             else
26                 seat[i/2][4] = i;
27         }
28         else
29         {
30             int t = i-2*n;
31             if(t%2 == 1)
32                 seat[(t+1)/2][1]=i;
33             else
34                 seat[t/2][3] = i;
35         }
36     for(int i=1,k=1;k<=m && i<=n;i++)
37         for(int j = 1;j<=4;j++)
38             if(seat[i][j])
39             {
40                 if(k == m)
41                     printf("%d\n",seat[i][j]);
42                 else
43                     printf("%d ",seat[i][j]);
44                 k++;
45             }
46 
47     return 0;
48 }
View Code

 

 

posted @ 2016-04-09 16:08  weeping  阅读(229)  评论(0编辑  收藏  举报