程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> Codeforces Round #Pi (Div. 2) B. Berland National Library

Codeforces Round #Pi (Div. 2) B. Berland National Library

編輯:C++入門知識

Codeforces Round #Pi (Div. 2) B. Berland National Library


B. Berland National Library time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading room.

Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form reader entered room, reader left room. Every reader is assigned aregistration number during the registration procedure at the library — it's a unique integer from 1 to 106. Thus, the system logs events of two forms:

  • + ri — the reader with registration number ri entered the room;
  • - ri — the reader with registration number ri left the room.

    The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.

    Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation will bring. Now, the developers of the system need to urgently come up with reasons for its existence.

    Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you.

    Input

    The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of records in the system log. Next follow n events from the system journal in the order in which the were made. Each event was written on a single line and looks as + ri or - ri, where ri is an integer from 1 to 106, the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).

    It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors.

    Output

    Print a single integer — the minimum possible capacity of the reading room.

    Sample test(s) input
    6
    + 12001
    - 12001
    - 1
    - 1200
    + 1
    + 7
    
    output
    3
    input
    2
    - 1
    - 2
    
    output
    2
    input
    2
    + 1
    - 1
    
    output
    1
    Note

    In the first sample test, the system log will ensure that at some point in the reading room were visitors with registration numbers 1, 1200and 12001. More people were not in the room at the same time based on the log. Therefore, the answer to the test is 3.

    題意,有個圖書館,+表示進人,-表示出人,記錄前有人,所以可以沒有+ 就有-,問可能的最大值是多少人。

    先直接統計先 +再-的總人數,這+統計出來的人就是有記錄後的最大值,然後,如果,有人,沒有+就直接-,那麼這個人,是記錄之前就有的人,會直接影響當前的結果,由於當前的結果並沒有統計上,所以直接加上這個人就可以了。

     

    int n,num,ans,tt;
    bool in[N];
    char str[10];
    int main()
    {
        while(S(n)!=EOF)
        {
            fill(in,false);
            ans = 0;tt = 0;
            FI(n){
                SS(str);S(num);
                if(str[0] == '+'){
                    in[num] = true;
                    tt++;
                    ans = max(tt,ans);
                }
                else {
                   if(in[num]){
                    in[num] = false;
                    tt--;
                   }
                   else {
                        ans ++;
                        in[num] = false;
                   }
                }
            }
            printf(%d
    ,ans);
        }
        return 0;
    }
    


     

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved