• 杭电1702--ACboy needs your help again!


    ACboy needs your help again!

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4412    Accepted Submission(s): 2249


    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
     

     

    Source
     

     

    Recommend
    lcy   |   We have carefully selected several similar problems for you:  1701 1700 1703 1704 1706 
     1 #include <stdio.h>
     2 #include <string.h>
     3 int main()
     4 {
     5     int n;
     6     scanf("%d", &n);
     7     while(n--)
     8     {
     9         int i, m;  
    10         char ch[5], str[5];
    11         int s[110];
    12         scanf("%d %s", &m, ch) ;
    13         int  front = 0, rear = 0;
    14         for(i=0; i<m; i++)
    15         {
    16             scanf("%s", str);
    17             if(strcmp(str, "IN") == 0)
    18             {
    19                 scanf("%d", &s[rear]);
    20                 ++rear;
    21             }
    22             if(strcmp(str, "OUT") == 0)
    23             {
    24                 if(strcmp(ch, "FIFO") == 0)
    25                 {
    26                     if(front != rear)
    27                     {
    28                         printf("%d
    ", s[front]);
    29                         ++front;
    30                     }
    31                     else
    32                     printf("None
    ");
    33                 }
    34                 else if(strcmp(ch, "FILO") == 0)
    35                 {
    36                     if(front != rear )
    37                     {
    38                         printf("%d
    ", s[rear-1]) ;
    39                         --rear;
    40                     }
    41                     else
    42                     printf("None
    ");
    43                 }
    44             }
    45         }
    46     }
    47     return 0;  
    48 }
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  • 原文地址:https://www.cnblogs.com/soTired/p/4681053.html
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