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.
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.
Print m distinct integers from 1 to m — the order in which the passengers will get off the bus.
2 7
5 1 6 2 7 3 4
9 36
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：有2n排座位，n排在左边，n排在右边。有m个人（编号为1-m）依次上车，优先坐左边靠窗的座位，再是右边靠窗的座位，然后是左边不靠窗，最后是右边不靠窗。要输出下车时人的循序编号。
下车时左边第一排不靠窗先下，然后是靠窗，再是右边第一排不靠窗下，然后靠窗，每一排依次下车直到所有人下车。
题解:模拟即可。
1 /*B*/ 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 6 int main() 7 { 8 int n,m; 9 int seat[5][110]; 10 while(scanf("%d%d",&n,&m)!=EOF) 11 { 12 memset(seat,0,sizeof(seat)); 13 seat[1][1]=1;seat[4][1]=2; 14 for(int i=2;i<=n;i++) 15 { 16 seat[1][i]=seat[1][i-1]+2; 17 seat[4][i]=seat[4][i-1]+2; 18 } 19 seat[2][1]=seat[4][n]+1;seat[3][1]=seat[2][1]+1; 20 for(int i=2;i<=n;i++) 21 { 22 seat[2][i]=seat[2][i-1]+2; 23 seat[3][i]=seat[3][i-1]+2; 24 } 25 for(int i=0;i<=4;i++) 26 for(int j=0;j<=n;j++) 27 if(seat[i][j]>m) 28 seat[i][j]=0; 29 int s=0; 30 for(int i=1;i<=n;i++) 31 { 32 if(seat[2][i]) 33 { 34 printf("%d ",seat[2][i]); 35 s++; 36 } 37 if(s==m) break; 38 if(seat[1][i]) 39 { 40 printf("%d ",seat[1][i]); 41 s++; 42 } 43 if(s==m) break; 44 if(seat[3][i]) 45 { 46 printf("%d ",seat[3][i]); 47 s++; 48 } 49 if(s==m) break; 50 if(seat[4][i]) 51 { 52 printf("%d ",seat[4][i]); 53 s++; 54 } 55 if(s==m) break; 56 } 57 printf(" "); 58 } 59 return 0; 60 }