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  • 8.1.1 ACboy needs your help again!

    Problem Description
    ACboy was kidnapped!!
    he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
    As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die with ACboy."
    The problems of the monster is shown on the wall:
    Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
    and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
    and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!
     

    Input
    The input contains multiple test cases.
    The first line has one integer,represent the number oftest cases.
    And the input of each subproblem are described above.
     

    Output

                For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.
     

    Sample Input
    4
    4 FIFO
    IN 1
    IN 2
    OUT
    OUT
    4 FILO
    IN 1
    IN 2
    OUT
    OUT
    5 FIFO
    IN 1
    IN 2
    OUT
    OUT
    OUT
    5 FILO
    IN 1
    IN 2
    OUT
    IN 3
    OUT
     

    Sample Output
    1
    2
    2
    1
    1
    2
    None
    2
    3

    栈、队列

     1 #include <cmath>
     2 #include <cstdio>
     3 #include <algorithm>
     4 #include <cstring>
     5 #include <string>
     6 #include <cstdlib>
     7 #include <queue>
     8 #include <stack>
     9 using namespace std;
    10 
    11 const int maxn=210;
    12 queue<int> q;
    13 stack<int> st;
    14 int n,a;
    15 char s[20];
    16 
    17 void close()
    18 {
    19 exit(0);
    20 }
    21 void dostack()
    22 {
    23     for (int i=1;i<=n;i++)
    24     {
    25         scanf("%s",s);
    26         if (s[0]=='I')
    27         {
    28             scanf("%d",&a);
    29             st.push(a);
    30         }
    31         else
    32         {
    33             if (!st.empty())
    34             {
    35                 printf("%d
    ",st.top());
    36                 st.pop();
    37             }
    38             else
    39                 printf("None
    ");
    40         }
    41     }
    42 }
    43 
    44 void doqueue()
    45 {
    46     for (int i=1;i<=n;i++)
    47     {
    48         scanf("%s",s);
    49         if (s[0]=='I')
    50         {
    51             scanf("%d",&a);
    52             q.push(a);
    53         }
    54         else
    55         {
    56             if (!q.empty())
    57             {
    58                 printf("%d
    ",q.front());
    59                 q.pop();
    60             }
    61             else
    62                 printf("None
    ");
    63         }
    64     }
    65 }
    66 
    67 void init()
    68 {
    69     int T;
    70     while (scanf("%d",&T)!=EOF)
    71     {
    72         while (T--)
    73         {
    74             while (!q.empty()) q.pop();
    75             while (!st.empty()) st.pop();
    76             scanf("%d",&n);
    77             scanf("%s",s);
    78             if (s[2]=='F') //queue
    79                 doqueue();
    80             else
    81                 dostack();
    82         }
    83     }
    84 }
    85 
    86 int main ()
    87 {
    88     init();
    89     close();
    90     return 0;
    91 }
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  • 原文地址:https://www.cnblogs.com/cssystem/p/3212442.html
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