• Codeforces Beta Round #4 (Div. 2 Only) C. Registration system hash


    C. Registration system

    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://codeforces.com/problemset/problem/4/C

    Description

    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.

    Input

    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 32 characters, which are all lowercase Latin letters.

    Output

    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.

    Sample Input

    4
    abacaba
    acaba
    abacaba
    acab

    Sample Output

    OK
    OK
    abacaba1
    OK

    HINT

    题意

     如果这个词第一次出现,输出ok

    否则输出这个词,并且输出在此之前这个词出现了多少次

    题解:

    双hash+map就好了

    代码

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 200001
    #define mod 10007
    #define eps 1e-9
    int Num;
    char CH[20];
    //const int inf=0x7fffffff;   //нчоч╢С
    const int inf=0x3f3f3f3f;
    /*
    
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    */
    inline ll read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    //**************************************************************************************
    
    int get_hash(char *key)
    {
        ll N=1998585857;
        long long h=0;
        while(*key)
            h=(h*127+(*key++)+N)%N;
        return h%N;
    }
    int get_hash2(char *key)
    {
        ll N=127398127;
        long long h=0;
        while(*key)
            h=(h*127+(*key++)+N)%N;
        return h%N;
    }
    char s[40];
    map< pair<int,int> ,int>H;
    int main()
    {
        int n=read();
        for(int i=0;i<n;i++)
        {
            scanf("%s",s);
            pair<int,int> a;
            a.first=get_hash(s);
            a.second=get_hash2(s);
            H[a]++;
            if(H[a]==1)
                printf("OK
    ");
            else printf("%s%d
    ",s,H[a]-1);
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4593594.html
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