Time Limit: 1000MS | | Memory Limit: 262144KB | | 64bit IO Format: %I64d & %I64u |
Description
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
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
Sample Input
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
Source
Educational Codeforces Round 11
//题意：
一辆公交车有四列座位，每个座位有一个编号（座位号），前两列是奇数号，后两列是偶数号，现在告诉你公交车上的座位的排数，和座位的个数，让你输出从第一排到底n排的座位的编号。
//思路：
直接模拟，用dp[i][j]表示第i列第j排的座位的编号，然后从低到高依次输出每个座位的编号。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define ll long long
#define IN __int64
#define N 100010
#define M 1000000007
using namespace std;
int dp[4][110];
int main()
{
int n,m,i,j,k;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(dp,0,sizeof(dp));
k=0;int mm=1;
while(k<n)
{
dp[2][k++]=mm;
mm+=2;
}
k=0;
while(k<n)
{
dp[1][k++]=mm;
mm+=2;
}
k=0;mm=2;
while(k<n)
{
dp[4][k++]=mm;
mm+=2;
}
k=0;
while(k<n)
{
dp[3][k++]=mm;
mm+=2;
}
for(i=0;i<n;i++)
{
for(j=1;j<=4;j++)
{
if(dp[j][i]<=m)
printf("%d ",dp[j][i]);
}
}
printf("\n");
}
return 0;
}
