• 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
    ",seat[i][j]);
    42                 else
    43                     printf("%d ",seat[i][j]);
    44                 k++;
    45             }
    46 
    47     return 0;
    48 }
    View Code
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  • 原文地址:https://www.cnblogs.com/weeping/p/5371884.html
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