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  • nyoj 991-Registration system (map)

    991-Registration system


    内存限制:64MB 时间限制:1000ms 特判: No
    通过数:2 提交数:3 难度:2

    题目描述:

    A new e-mail service "Berlandesk" is going to be opened in Berland in the near future.

     The site administration wants to launch their project as soon as possible, that's why they

     ask you to help. You're suggested to implement the prototype of site registration system. 

    The system should work on the following principle.

    Each time a new user wants to register, he sends to the system a request with his name.

     If such a name does not exist in the system database, it is inserted into the database, and 

    the user gets the response OK, confirming the successful registration. If the name already 

    exists in the system database, the system makes up a new user name, sends it to the user 

    as a prompt and also inserts the prompt into the database. The new name is formed by the

     following rule. Numbers, starting with 1, are appended one after another to name (name1,

     name2, ...), among these numbers the least i is found so that namei does not yet exist in

     the database.


    输入描述:

    The first line contains number n (1 ≤ n ≤ 105). The following n lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 1000 characters, which are all lowercase Latin letters.

    输出描述:

    Print n lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.

    样例输入:

    4
    abacaba
    acaba
    abacaba
    acab
    

    样例输出:

    OK
    OK
    abacaba1
    OK

    C/C++:

    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    #include <stack>
    #include <set>
    #include <map>
    #include <queue>
    #include <climits>
    #include <bitset>
    #define PI 3.1415926
    
    using namespace std;
    
    int main()
    {
        map <string, int> my_map;
        pair <map <string, int> ::iterator, bool> pr;
        int n;
        scanf("%d", &n);
        while (n --)
        {
            string str;
            cin >>str;
            pr = my_map.insert(pair<string, int>(str, 0));
            if (pr.second)
                printf("OK
    ");
            else
            {
                int temp = my_map[str] + 1;
                my_map[str] = temp;
                int A[12], nn = temp, cnt2 = 0;
    
                while (nn)
                {
                    A[cnt2 ++] = nn % 10;
                    nn /= 10;
                }
    
                string temp2 = "";
                for (int i = cnt2 - 1; i >= 0; -- i)
                {
                    temp2 += char(A[i] + '0');
                }
                string str_temp = str + temp2;
                cout <<str_temp <<endl;
                my_map.insert(pair<string, int>(str_temp, 0));
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9362242.html
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